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Multiplication rule and tree diagrams

Learning goals

Represents successive events with a probability tree.
Writes conditional probabilities on branches when appropriate.
Uses the multiplication rule to compute path probabilities.
Adds relevant path probabilities for compound events.
Checks that outgoing branch probabilities sum to 1.

Difficulty: 1 – 5

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Multiplication rule and tree diagrams 1

Arnar has drawn this probability tree for two possible results of a spinner. Explain what is wrong with the branch probabilities, and give the corrected probability for the second branch.

Figure for problem 537500

Hints

- Check the total probability across both possible results. - If the first probability stays fixed, how much is left for the other result?

Solution

1. The probabilities leaving the start add to \(0.6+0.5=1.1\), which is greater than \(1\). 2. The second branch should have probability \(1-0.6=0.4\).

Answer

The two branches incorrectly total \(1.1\). The second probability should be \(0.4\).

Multiplication rule and tree diagrams 2

The tree shows the choice of lunch for one student. What is the probability the student chooses Soup?

Figure for problem 560992

Hints

- Which branches leave the starting point of the tree? - Is reaching Soup a single choice or a sequence of choices?

Solution

1. The Soup branch starts directly at the root. 2. Its label is \(\frac{1}{6}\), so that is the probability of Soup.

Answer

\(\frac{1}{6}\)

Multiplication rule and tree diagrams 3

The tree shows how a student travels and whether the student arrives late. Given that the student cycled, what is the probability of arriving late?

Figure for problem 560993

Hints

- Which first-stage choice has already been made? - Find the matching outcome after that choice in the tree. - How would you describe the group of trips the question is asking about?

Solution

1. Follow the Cycle branch and look at its Late outgoing branch. 2. That conditional branch is labeled \(\frac{1}{10}\).

Answer

\(P(\text{Late}\mid\text{Cycle})=\frac{1}{10}\)

Multiplication rule and tree diagrams 4

A weather tree first shows Rain or Dry, then whether a concert is delayed. What does the branch from Rain to Delayed mean? Does its probability describe all concerts or only rainy concerts?

Figure for problem 560994

Hints

- Which weather outcome must occur before you can reach that branch? - Imagine you follow the tree from the beginning. At which point is the branch available?

Solution

1. The branch is reached only if Rain has already occurred. 2. Its label means \(P(\text{Delayed}\mid\text{Rain})=0.2\). 3. It does not state the fraction of all concerts that are rainy and delayed.

Answer

Among rainy concerts, \(20\%\) are delayed. No, it is not a probability out of all concerts.

Multiplication rule and tree diagrams 5

The tree lists all possible weather outcomes for one day. The Rain and Snow probabilities have been checked and are correct. The Sun probability is wrong. What should the Sun probability be? Explain how you know.

Figure for problem 560995

Hints

- What must the probabilities on all branches from Day add to? - Which probabilities are already known to be correct?

Solution

1. Outgoing branch probabilities from one node must add to \(1\). 2. The Rain and Snow labels total \(0.4+0.3=0.7\). 3. Therefore the Sun branch should be \(1-0.7=0.3\), not \(0.4\).

Answer

Change the Sun branch to \(0.3\).

Multiplication rule and tree diagrams 6

The tree shows whether a bus comes on time or late on a particular route. Find the missing branch probability marked ?.

Figure for problem 560996

Hints

- Focus only on the two branches leaving Route X. - What must their probabilities add to?

Solution

1. The two possible results under Route X are On time and Late. 2. Their probabilities must sum to one, so the missing value is \(1-0.85=0.15\).

Answer

\(0.15\)

Multiplication rule and tree diagrams 7

On a test, each question has four answer choices, exactly one of which is correct. Mia cannot solve three questions, so she guesses randomly on each one. What is the probability that all three answers are correct? a) \(\frac{1}{4}\) b) \(\frac{3}{4}\) c) \(\frac{1}{12}\) d) \(\frac{1}{64}\)

Hints

- Find the probability of one correct random guess. - For three independent correct guesses, multiply the probabilities.

Solution

1. The probability of a correct answer on one question is \(\frac{1}{4}\). 2. The guesses are independent, so multiply the three probabilities. 3. \(P(\text{all three correct}) = \left(\frac{1}{4}\right)^3 = \frac{1}{64}\).

Answer

d) \(\frac{1}{64}\)

Multiplication rule and tree diagrams 8

Tom rolls a standard six-sided die three times. What is the probability that a \(6\) occurs only on the first and second rolls? a) \(\frac{2}{3}\) b) \(\frac{1}{36}\) c) \(\frac{5}{216}\) d) \(\frac{25}{216}\)

Hints

- State the required result of each roll. - The word “only” determines the result of the third roll. - Multiply the probabilities of the independent results in the specified order.

Solution

1. The required sequence is \(6\), \(6\), then not \(6\). 2. The probability of rolling a \(6\) is \(\frac{1}{6}\), and the probability of not rolling a \(6\) is \(\frac{5}{6}\). 3. Multiply along the specified sequence: \(\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}=\frac{5}{216}\).

Answer

c) \(\frac{5}{216}\)

Multiplication rule and tree diagrams 9

A school cafeteria offers a surprise meal with one snack, one drink, and one fruit. The choices are: - Snacks: pretzel, sandwich, muffin, or granola bar - Drinks: water, apple juice, or milk - Fruit: apple, pear, banana, kiwi, or orange One item from each category is selected independently and at random, with all choices within a category equally likely. a) Find the probability of getting exactly a pretzel, water, and an apple. b) Find the probability of getting no pretzel, no water, and no apple.

Hints

- Find the probability of the required choice within each category. - Multiply probabilities because the category selections are independent. - For part b, count the choices that are not the named item in each category.

Solution

1. a) The probability of the specific combination is \(\frac{1}{4} \cdot \frac{1}{3} \cdot \frac{1}{5} = \frac{1}{60}\). 2. b) The probabilities of selecting a different item in each category are \(\frac{3}{4}\), \(\frac{2}{3}\), and \(\frac{4}{5}\). Therefore, \(P(\text{no listed item}) = \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{4}{5} = \frac{24}{60} = \frac{2}{5} = 0.4\).

Answer

a) \(\frac{1}{60}\) b) \(\frac{2}{5} = 40\%\)

Multiplication rule and tree diagrams 10

A bag contains \(12\) marbles: \(4\) red, \(5\) blue, and \(3\) yellow. Three marbles are drawn without replacement. Find the probability of drawing, in this exact order, a red marble, another red marble, and then a yellow marble.

Hints

- Update the total number of marbles after each draw. - Update the number of red marbles after each red draw. - Write each stage probability as a fraction before multiplying. - Multiply the conditional probabilities along the required path.

Solution

1. The probability of red on the first draw is \(\frac{4}{12}\). 2. After one red marble is removed, the probability of red on the second draw is \(\frac{3}{11}\). 3. After two red marbles are removed, the probability of yellow on the third draw is \(\frac{3}{10}\). 4. Multiply along the path: \(P(\text{red, red, yellow}) = \frac{4}{12} \cdot \frac{3}{11} \cdot \frac{3}{10} = \frac{3}{110}\).

Answer

\(\frac{3}{110} \approx 2.7\%\)

Multiplication rule and tree diagrams 11

A spinner has \(5\) equal sections: \(2\) red, \(2\) blue, and \(1\) green. The spinner is spun three times. a) Find the probability of green on all three spins. b) Find the probability of the exact color sequence red-blue-red. c) Find the probability of green on at least one of the three spins.

Hints

- Treat the experiment as a sequence of three stages. - Use the section counts to find each one-spin probability. - Multiply along a fixed three-spin sequence. - For “at least one green,” use the complement of no green.

Solution

1. The single-spin probabilities are \(P(\text{red}) = \frac{2}{5}\), \(P(\text{blue}) = \frac{2}{5}\), and \(P(\text{green}) = \frac{1}{5}\). 2. a) \(P(\text{green, green, green}) = \left(\frac{1}{5}\right)^3 = \frac{1}{125} = 0.008\). 3. b) \(P(\text{red, blue, red}) = \frac{2}{5} \cdot \frac{2}{5} \cdot \frac{2}{5} = \frac{8}{125} = 0.064\). 4. c) The probability of no green on one spin is \(\frac{4}{5}\). Using the complement, \(P(\text{at least one green}) = 1 - \left(\frac{4}{5}\right)^3 = 1 - \frac{64}{125} = \frac{61}{125} = 0.488\).

Answer

a) \(\frac{1}{125} = 0.8\%\) b) \(\frac{8}{125} = 6.4\%\) c) \(\frac{61}{125} = 48.8\%\)

Multiplication rule and tree diagrams 12

A spinner has \(10\) equal sections. Two sections say “win,” and the other \(8\) say “lose.” The spinner is spun three times. Find the probability of each event. a) The result is win on all three spins. b) The result is lose on at least one spin. c) The first win occurs on the third spin.

Hints

- Write the one-spin probabilities for win and lose. - Use the complement of all wins for part b. - For part c, identify the exact three-spin sequence.

Solution

1. On one spin, \(P(\text{win}) = \frac{2}{10} = 0.2\) and \(P(\text{lose}) = 0.8\). 2. a) \(P(\text{win, win, win}) = 0.2^3 = 0.008\). 3. b) The complement of at least one loss is all wins. Therefore, \(P(\text{at least one loss}) = 1 - 0.008 = 0.992\). 4. c) The required sequence is lose-lose-win, so \(P = 0.8 \cdot 0.8 \cdot 0.2 = 0.128\).

Answer

a) \(0.008 = 0.8\%\) b) \(0.992 = 99.2\%\) c) \(0.128 = 12.8\%\)

Multiplication rule and tree diagrams 13

A bag contains \(25\) fruit candies: \(15\) red and \(10\) yellow. Tim selects two candies one at a time without replacement. a) Find the probability that both candies are red. b) Find the probability that at least one candy is yellow.

Hints

- Update the red count and total after the first red candy. - Use the complement of both candies being red.

Solution

1. a) \(P(\text{red, red}) = \frac{15}{25} \cdot \frac{14}{24} = \frac{7}{20} = 0.35\). 2. b) “At least one yellow” is the complement of “both red.” Therefore, \(P(\text{at least one yellow}) = 1 - \frac{7}{20} = \frac{13}{20} = 0.65\).

Answer

a) \(\frac{7}{20} = 35\%\) b) \(\frac{13}{20} = 65\%\)

Multiplication rule and tree diagrams 14

A basketball player makes a free throw with probability \(0.80\). The player takes \(5\) independent free throws. Find the probability that all \(5\) shots are made. Give the answer as a percent.

Hints

- Write the probability of one make as a decimal. - All five events must occur, so multiply their probabilities. - Use the same probability for each independent shot. - Convert the final decimal to a percent.

Solution

1. The probability of making one shot is \(0.80\). 2. Because all five independent shots must be made, multiply the probabilities: \(P(\text{five makes})=(0.80)^5=0.32768\). 3. Convert to a percent: \(0.32768=32.768\%\).

Answer

\(32.768\%\)

Multiplication rule and tree diagrams 15

Decide whether each statement is correct. Give a brief justification. a) If an event contains several paths in a tree diagram, its probability is found by multiplying the probabilities of those paths. b) At every node of a complete probability tree, the probabilities on the outgoing branches add to \(1\).

Hints

- Distinguish multiplication along one path from addition across paths. - Ask whether two different outgoing branches can occur at the same time. - Consider what all outgoing branches from one node represent.

Solution

1. Statement a is false. Multiply branch probabilities within one path, but add the probabilities of the different disjoint paths that make up an event. 2. Statement b is true. The outgoing branches represent all mutually exclusive ways the experiment can continue after reaching that node, so their conditional probabilities add to \(1\).

Answer

a) False. Add the probabilities of the paths belonging to the event. b) True. The outgoing branches cover all possible next outcomes, so their probabilities sum to \(1\).

Multiplication rule and tree diagrams 16

The tree shows two draws from a bag, with the first ball replaced before the second draw. Find the probability that both draws give the same color.

Figure for problem 537495

Hints

- Which two complete paths represent matching colors? - Look at the probabilities leaving each second-stage node. How does replacing the ball affect them? - Multiply along each matching path, then combine those path probabilities.

Solution

1. The red-red path has probability \(\frac{3}{5}\cdot\frac{3}{5}=\frac{9}{25}\). 2. The blue-blue path has probability \(\frac{2}{5}\cdot\frac{2}{5}=\frac{4}{25}\). 3. Add these two disjoint paths: \(\frac{9}{25}+\frac{4}{25}=\frac{13}{25}\).

Answer

\(\frac{13}{25}=52\%\)

Multiplication rule and tree diagrams 17

The tree shows two draws from a bag, without replacement. Find the probability that the two draws give different colors.

Figure for problem 537496

Hints

- Find both orders that produce one red and one blue ball. - Compare the second-draw probabilities after a red ball versus a blue ball is removed. - What do you do with the probabilities of two different favorable paths?

Solution

1. Red then blue has probability \(\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{10}\). 2. Blue then red has probability \(\frac{2}{5}\cdot\frac{3}{4}=\frac{3}{10}\). 3. Add the two different-color paths: \(\frac{3}{10}+\frac{3}{10}=\frac{3}{5}\).

Answer

\(\frac{3}{5}=60\%\)

Multiplication rule and tree diagrams 18

Complete the missing probability in the tree diagram, and then find the probability of “green and star.”

Figure for problem 537498

Hints

- Branch probabilities from a single node add to \(1\). - Multiply the probabilities along the requested path.

Solution

1. The first-stage branch probabilities must add to \(1\), so the missing value is \(1 - 0.35 = 0.65\). 2. Along the green-star path, \(P(\text{green and star}) = 0.65 \cdot 0.4 = 0.26\).

Answer

Missing value: \(0.65\); \(P(\text{green and star}) = 0.26\)

Multiplication rule and tree diagrams 19

The tree shows two draws from a bag, with replacement. Let \(R\) mean a red ball and \(R^c\) mean a ball that is not red. a) Fill in all the missing branch probabilities. b) Describe the complement of “red at least once” and find its probability.

Figure for problem 537499

Hints

- What must the outgoing branch probabilities at each node add to? - Which complete path has no red ball at all? - How does replacement affect the probability on the second draw?

Solution

1. At each node, the red and nonred probabilities total \(1\). Thus, each missing nonred branch has probability \(1-0.3=0.7\). 2. The complement of “red at least once” is “no red on either draw.” 3. The path \(R^c,R^c\) has probability \(0.7\cdot0.7=0.49\).

Answer

a) Each missing branch is \(0.7\). b) Neither draw is red; the probability is \(0.49\).

Multiplication rule and tree diagrams 20

The tree represents two independent spins of a spinner. Each spin gives either an odd number or an even number. Find the probability that the product of the two numbers is even.

Figure for problem 537505

Hints

- When is a product odd rather than even? - Which complete path represents the easier complementary event? - How does the probability of a complement relate to the original event?

Solution

1. A product is odd only when both factors are odd. 2. The probability of two odd results is \(\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\). 3. The complement is an even product, with probability \(1-\frac{1}{4}=\frac{3}{4}\).

Answer

\(\frac{3}{4}=75\%\)

Multiplication rule and tree diagrams 21

Use the partial probability tree to find the missing branch probability and \(P(A\cap B^c)\).

Figure for problem 537514

Hints

- Only the two branches after \(A\) are needed to complete the missing probability. - Follow one full route from the start to \(B^c\) after \(A\).

Solution

1. The two branches after \(A\) must total \(1\), so the missing probability is \(1-0.65=0.35\). 2. Multiply along the \(A,B^c\) path: \(P(A\cap B^c)=0.4\cdot0.35=0.14\).

Answer

Missing branch: \(0.35\); \(P(A\cap B^c)=0.14\).

Multiplication rule and tree diagrams 22

The tree shows two rolls of a fair six-sided die. Find the probability that at least one roll is a \(6\).

Figure for problem 537515

Hints

- How many different paths include at least one \(6\)? - Is there a single path representing the complement? - Read the probabilities on that route before subtracting from \(1\).

Solution

1. The complement of at least one \(6\) is no \(6\) on either roll. 2. The no-\(6\) path has probability \(\frac{5}{6}\cdot\frac{5}{6}=\frac{25}{36}\). 3. Therefore, the required probability is \(1-\frac{25}{36}=\frac{11}{36}\).

Answer

\(\frac{11}{36}\approx30.6\%\)

Multiplication rule and tree diagrams 23

The tree shows two independent penalty kicks by a soccer player. Find the probability that the player scores exactly once.

Figure for problem 537516

Hints

- Which two paths show one goal and one miss? - The kicks are independent. How is that reflected in the second-stage probabilities? - After multiplying along each qualifying path, how should their probabilities be combined?

Solution

1. Goal then miss has probability \(0.8\cdot0.2=0.16\). 2. Miss then goal also has probability \(0.2\cdot0.8=0.16\). 3. Add these two disjoint paths: \(0.16+0.16=0.32\).

Answer

\(0.32=32\%\)

Multiplication rule and tree diagrams 24

Two raffle tickets are drawn from a drum without replacement. Use the tree to find the probability that at least one ticket wins.

Figure for problem 537517

Hints

- Is it easier to follow every path with a win, or the path with no wins? - Read the updated second-stage chance after the first losing ticket is removed. - How does the probability of a complementary event help?

Solution

1. The complement is that both tickets lose. 2. The lose-lose path has probability \(\frac{7}{10}\cdot\frac{6}{9}=\frac{7}{15}\). 3. Therefore, the probability of at least one winning ticket is \(1-\frac{7}{15}=\frac{8}{15}\).

Answer

\(\frac{8}{15}\approx53.3\%\)

Multiplication rule and tree diagrams 25

An inspector checks up to three independently selected components, stopping as soon as a defective component is found. The tree gives the chance that each inspected component is defective or good. Find the probability that the inspector finds at least one defective component.

Figure for problem 537527

Hints

- Why does the inspection stop along some paths but not others? - Which one complete path means that no defective component is found? - Use the complement to avoid adding several paths separately.

Solution

1. The complement is that all three checked components are good, so no defect is found. 2. The probability of that path is \(0.98^3=0.941192\). 3. Therefore, the probability of finding at least one defect is \(1-0.941192=0.058808\).

Answer

\(0.058808\approx5.88\%\)

Multiplication rule and tree diagrams 26

Use the tree diagram of counts to find all six branch probabilities.

Figure for problem 537536

Hints

- Divide each child count by its parent count. - The probabilities on branches from the same node must add to \(1\).

Solution

1. At the first stage, \(P(A)=\frac{300}{500}=0.60\) and \(P(A^c)=\frac{200}{500}=0.40\). 2. After \(A\), \(P(B\mid A)=\frac{180}{300}=0.60\) and \(P(B^c\mid A)=\frac{120}{300}=0.40\). 3. After \(A^c\), \(P(B\mid A^c)=\frac{50}{200}=0.25\) and \(P(B^c\mid A^c)=\frac{150}{200}=0.75\).

Answer

First stage: \(0.60\), \(0.40\) After \(A\): \(0.60\), \(0.40\) After \(A^c\): \(0.25\), \(0.75\)

Multiplication rule and tree diagrams 27

A manufacturer makes parts on two production lines. The tree shows each line's share of production and its defect rate. Find the probability that a randomly selected part is defective.

Figure for problem 537537

Hints

- Which complete tree paths end in a defective part? - For each path, account for both the production-line share and its defect rate. - The two paths are distinct ways for a part to be defective. How should their probabilities be combined?

Solution

1. The defective path from line A has probability \(0.50\cdot0.03=0.015\). 2. The defective path from line B has probability \(0.50\cdot0.01=0.005\). 3. Add the two paths: \(P(\text{defective})=0.015+0.005=0.02\).

Answer

\(0.02=2\%\)

Multiplication rule and tree diagrams 28

Zanele says the probability of the path \(A\) then \(B\) is \(1.1\) because she added the branch probabilities in the tree. Explain her mistake and find the correct path probability.

Figure for problem 537558

Hints

- Do the two edges describe alternatives or successive stages of one route? - Is Zanele's reported probability even in the possible range from \(0\) to \(1\)? - Use the two edge labels along the specified route.

Solution

1. The two branch events happen in sequence, so their probabilities are multiplied, not added. 2. The correct path probability is \(P(A\cap B)=0.4\cdot0.7=0.28\). 3. The claimed value \(1.1\) also cannot be a probability because it exceeds \(1\).

Answer

Zanele incorrectly added probabilities along one path. She should multiply them: \(P(A\cap B)=0.28\).

Multiplication rule and tree diagrams 29

The tree shows two successive draws of a colored ball. Does the probability of red on the second draw depend on the first draw? How can you tell?

Figure for problem 560997

Hints

- Follow each first-stage branch to the later Red branch. - Are the two later Red labels equal?

Solution

1. The probability of second red is \(\frac{1}{3}\) after first red. 2. It is \(\frac{2}{3}\) after first blue. 3. Since the second-red probability changes with the first result, the draws are dependent.

Answer

Yes. The second-red probability is \(\frac{1}{3}\) after red but \(\frac{2}{3}\) after blue.

Multiplication rule and tree diagrams 30

A spinner is used twice. The tree describes whether each spin lands on R or B. Do the branch probabilities show independent spins? Explain without calculating path probabilities.

Figure for problem 560998

Hints

- Compare the two pairs of branches leaving the first-stage nodes. - Would knowing the first spin change the chance of R on the second spin?

Solution

1. After R, the second spin has probability \(\frac{1}{4}\) of R and \(\frac{3}{4}\) of B. 2. After B, those probabilities are the same. 3. The first spin does not change the second-spin probabilities, so the spins are independent.

Answer

Yes. Both second-stage branches have the same probabilities after either first-stage result.

Multiplication rule and tree diagrams 31

The tree shows which ticket a visitor gets and whether it wins a prize. Find the probability of getting Ticket A and winning.

Figure for problem 560999

Hints

- Trace the route that reaches Ticket A and then Win. - How do probabilities combine along a single route?

Solution

1. Follow Ticket A with probability \(0.6\). 2. Along that branch, the chance of a win is \(0.5\). 3. Multiply the two branch probabilities: \(0.6\cdot0.5=0.30\).

Answer

\(0.30\)

Multiplication rule and tree diagrams 32

At a snack stand, students choose a sandwich size, whether to add cheese, and a drink. What is the probability of ordering a regular sandwich with cheese and water?

Figure for problem 561000

Hints

- Trace just the path named in the question, one branch at a time. - How does the multiplication rule extend when a path has three branches?

Solution

1. Read the three successive branches: Regular has probability \(0.6\), Cheese after Regular has probability \(0.5\), and Water along that route has probability \(0.2\). 2. Multiply along the full path: \(0.6\cdot0.5\cdot0.2=0.06\).

Answer

\(0.06\)

Multiplication rule and tree diagrams 33

Two balls are drawn without replacement. The tree shows the changing probabilities. Find the probability that the two balls have different colors.

Figure for problem 561001

Hints

- Which paths end with two different colors? - For each such path, what probabilities must be combined? - How would you combine probabilities of different valid paths?

Solution

1. Different colors can occur as Red then Blue or Blue then Red. 2. The two paths have probabilities \(\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{3}\) and \(\frac{1}{3}\cdot1=\frac{1}{3}\). 3. Add the disjoint paths: \(\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\).

Answer

\(\frac{2}{3}\)

Multiplication rule and tree diagrams 34

A player gets at most two tries at a game. The game ends as soon as the player succeeds. The tree shows the chances at each try. Find the probability that the player succeeds on either try.

Figure for problem 561002

Hints

- Look at the leaves of the tree. Which ones represent success? - Trace the steps needed to reach each successful ending. - Do those successful endings describe the same outcome or different outcomes?

Solution

1. The player can succeed immediately, with probability \(0.3\), or fail first and then succeed, with probability \(0.7\cdot0.4=0.28\). 2. Add these distinct stopping paths: \(0.3+0.28=0.58\).

Answer

\(0.58\)

Multiplication rule and tree diagrams 35

Two balls are drawn in order from the urn. Find the probability of Yellow followed by Green a) with replacement and b) without replacement. Which is greater?

Figure for problem 561003

Hints

- Read the total of each color from the urn. - If a yellow ball is put back, what changes for the next draw? - If it is not put back, what is the new total before drawing green?

Solution

1. The urn contains three yellow and two green balls, five altogether. 2. With replacement, \(P(Y\text{ then }G)=\frac{3}{5}\cdot\frac{2}{5}=\frac{6}{25}\). 3. Without replacement, after removing a yellow ball there are four balls left, two green. Thus \(P(Y\text{ then }G)=\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{10}\). 4. Since \(\frac{3}{10}>\frac{6}{25}\), the ordered event is more likely without replacement.

Answer

a) \(\frac{6}{25}\) b) \(\frac{3}{10}\). Without replacement is more likely.

Multiplication rule and tree diagrams 36

A tree shows an event \(B\) followed by whether \(W\) occurs. The probability that both \(B\) and \(W\) occur is \(0.18\). Find both missing branch probabilities after \(B\).

Figure for problem 561004

Hints

- Which path represents the given probability of both events? - How can the multiplication rule be used backward? - Two outgoing probabilities at the same node must add to one.

Solution

1. The path for B then W has probability \(P(B)P(W\mid B)=0.30\cdot P(W\mid B)\). 2. So \(P(W\mid B)=\frac{0.18}{0.30}=0.60\). 3. The other branch from B has probability \(1-0.60=0.40\).

Answer

\(P(W\mid B)=0.60\), \(P(W^c\mid B)=0.40\).

Multiplication rule and tree diagrams 37

A technician tests up to three sensors, stopping as soon as one sensor fails. Another technician tests all three sensors no matter what happens. Which technician's experiment matches the tree? Explain using the branch that ends after the first test.

Figure for problem 561005

Hints

- Look for a first-stage branch that has no children. - Would the experiment always continue after that result?

Solution

1. In the tree, the first failure is a terminal result: no further tests follow it. 2. The first technician stops after a failure, so that experiment matches the tree. The second would need additional branches after a failure.

Answer

The first technician. The tree stops after a failed sensor, rather than continuing to three tests in every case.

Multiplication rule and tree diagrams 38

A fair coin is tossed exactly twice, and both results are recorded. A student drew the tree shown. Is the tree complete? If not, describe the missing branches.

Figure for problem 561006

Hints

- Does every first-toss result lead to the required second toss? - Compare the number of stages after first H and first T.

Solution

1. A first toss of heads does not end the experiment because the coin must be tossed twice. 2. The H branch needs two further branches, one for second-toss H and one for second-toss T, each with probability \(\frac{1}{2}\).

Answer

No. Add H and T second-toss branches after the first H, each with probability \(\frac{1}{2}\).

Multiplication rule and tree diagrams 39

A multiple-choice test has \(6\) questions. Each question has \(3\) answer choices, exactly one of which is correct. Tiago guesses randomly on every question. a) Find the probability that Tiago answers every question incorrectly. b) Find the probability that Tiago answers at least one question correctly. c) Mei can eliminate one incorrect choice on each question and then guesses between the two remaining choices. Find the probability that Mei answers all \(6\) questions correctly, and compare it with the probability for completely random guessing.

Hints

- Find the probability of an incorrect answer on one question. - Use a complement for “at least one.” - Multiply probabilities across independent questions. - Compare the two all-correct probabilities.

Solution

1. The probability of an incorrect answer on one question is \(\frac{2}{3}\). Because the questions are independent, \(P(\text{all incorrect})=\left(\frac{2}{3}\right)^6=\frac{64}{729}\approx0.0878\). 2. Use the complement: \(P(\text{at least one correct})=1-\frac{64}{729}=\frac{665}{729}\approx0.9122\). 3. After eliminating one choice, the probability of a correct answer on each question is \(\frac{1}{2}\). Thus, \(P(\text{all correct})=\left(\frac{1}{2}\right)^6=\frac{1}{64}=0.015625\). 4. With completely random guessing, \(P(\text{all correct})=\left(\frac{1}{3}\right)^6=\frac{1}{729}\approx0.00137\). Eliminating one choice makes a perfect score about \(\frac{729}{64}\approx11.4\) times as likely.

Answer

a) \(\frac{64}{729}\approx0.0878\) b) \(\frac{665}{729}\approx0.9122\) c) \(\frac{1}{64}=0.015625\), compared with \(\frac{1}{729}\approx0.00137\) for completely random guessing

Multiplication rule and tree diagrams 40

A fair six-sided number cube is rolled three times. a) Find the number of possible ordered roll sequences. b) Find the probability that no \(6\) is rolled. c) Find the probability that at least one \(6\) is rolled. d) Jonas says, “Rolling the specific sequence \((6, 6, 6)\) is much less likely than rolling the specific sequence \((1, 2, 3)\).” Evaluate his claim mathematically.

Hints

- Multiply the number of choices for each roll. - For no \(6\), use the probability of one non-\(6\) on all three rolls. - Use a complement for “at least one.” - Compare two fixed ordered sequences, not broader patterns.

Solution

1. a) There are \(6 \cdot 6 \cdot 6 = 6^3 = 216\) ordered sequences. 2. b) The probability of not rolling a \(6\) on one roll is \(\frac{5}{6}\). Thus, \(P(\text{no 6}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \approx 0.5787\). 3. c) Use the complement: \(P(\text{at least one 6}) = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.4213\). 4. d) Every specific ordered sequence has probability \(\left(\frac{1}{6}\right)^3 = \frac{1}{216}\). Therefore, \((6, 6, 6)\) and \((1, 2, 3)\) are equally likely.

Answer

a) \(216\) b) \(\frac{125}{216} \approx 0.5787\) c) \(\frac{91}{216} \approx 0.4213\) d) The claim is false; each specific sequence has probability \(\frac{1}{216}\).

Multiplication rule and tree diagrams 41

At an amusement park, a prize station has three spinners that are spun independently. - Spinner 1 has \(6\) equal sections, with \(1\) gold section. - Spinner 2 has \(8\) equal sections, with \(1\) gold section. - Spinner 3 has \(10\) equal sections, with \(1\) gold section. a) Find the probability that all three spinners land on gold. b) Find the probability that none of the spinners lands on gold. c) Find the probability that at least one spinner lands on gold.

Hints

- Find the gold and non-gold probability for each spinner. - Multiply across independent spins. - Use the complement of “no gold” for “at least one gold.”

Solution

1. a) \(P(\text{all gold}) = \frac{1}{6} \cdot \frac{1}{8} \cdot \frac{1}{10} = \frac{1}{480}\). 2. b) The non-gold probabilities are \(\frac{5}{6}\), \(\frac{7}{8}\), and \(\frac{9}{10}\). Thus, \(P(\text{no gold}) = \frac{5}{6} \cdot \frac{7}{8} \cdot \frac{9}{10} = \frac{315}{480} = \frac{21}{32} = 0.65625\). 3. c) Use the complement: \(P(\text{at least one gold}) = 1 - \frac{21}{32} = \frac{11}{32} = 0.34375\).

Answer

a) \(\frac{1}{480}\) b) \(\frac{21}{32} = 0.65625\) c) \(\frac{11}{32} = 0.34375\)

Multiplication rule and tree diagrams 42

A drawer contains \(4\) black socks and \(4\) white socks mixed together. Lucas selects two socks one at a time without replacement. Find the probability of each event. a) Both socks are black. b) Both socks are the same color. c) The socks are different colors.

Hints

- A tree diagram can organize the possible two-draw paths. - The second-draw probabilities change because the first sock is not replaced. - For the same color, include black-black and white-white. - Use a complement for different colors if helpful.

Solution

1. a) \(P(\text{black, black}) = \frac{4}{8} \cdot \frac{3}{7} = \frac{12}{56} = \frac{3}{14}\). 2. The probability of two white socks is also \(\frac{4}{8} \cdot \frac{3}{7} = \frac{3}{14}\). 3. b) \(P(\text{same color}) = \frac{3}{14} + \frac{3}{14} = \frac{3}{7}\). 4. c) Different colors is the complement of same color, so \(P(\text{different colors}) = 1 - \frac{3}{7} = \frac{4}{7}\).

Answer

a) \(\frac{3}{14}\) b) \(\frac{3}{7}\) c) \(\frac{4}{7}\)

Multiplication rule and tree diagrams 43

A bag contains \(20\) cards numbered \(1\) through \(20\). Three cards are drawn one at a time without replacement. What is the probability that all three cards show even numbers?

Hints

- Count the even numbers from \(1\) through \(20\). - Update both the favorable count and the total after each draw. - Multiply the conditional probabilities.

Solution

1. There are \(10\) even-numbered cards among the \(20\) cards. 2. The probabilities of drawing an even number on the three successive draws are \(\frac{10}{20}\), \(\frac{9}{19}\), and \(\frac{8}{18}\). 3. Therefore, \(P(\text{three even numbers}) = \frac{10}{20} \cdot \frac{9}{19} \cdot \frac{8}{18} = \frac{2}{19}\).

Answer

\(\frac{2}{19} \approx 10.5\%\)

Multiplication rule and tree diagrams 44

A combination lock has \(4\) independently rotating dials. Each dial shows a digit from \(0\) through \(9\). A person has forgotten the code and sets each dial randomly. a) Find the probability of entering the exact code on the first try. b) Find the probability that only the first dial is correct and the other three dials are incorrect. c) Find the probability that at least one of the four dials shows the correct digit.

Hints

- Find the probability of one correct digit and one incorrect digit. - Use the fundamental counting principle to find the number of possible four-digit settings. - Multiply along the required four-dial pattern. - For part c, use the complement of no correct digits.

Solution

1. For one dial, \(P(\text{correct}) = 0.1\) and \(P(\text{incorrect}) = 0.9\). 2. a) \(P(\text{all correct}) = 0.1^4 = 0.0001 = 0.01\%\). 3. b) \(P(\text{correct, incorrect, incorrect, incorrect}) = 0.1 \cdot 0.9^3 = 0.0729 = 7.29\%\). 4. c) Use the complement of all four dials being incorrect: \(P(\text{at least one correct}) = 1 - 0.9^4 = 1 - 0.6561 = 0.3439 = 34.39\%\).

Answer

a) \(0.0001 = 0.01\%\) b) \(0.0729 = 7.29\%\) c) \(0.3439 = 34.39\%\)

Multiplication rule and tree diagrams 45

A factory produces LED bulbs, and \(10\%\) are defective. Four bulbs are selected from a very large production run. Model the selections as independent. a) Find the probability that all four bulbs are not defective. b) Find the probability that exactly one of the four bulbs is defective.

Hints

- Find the probability that one bulb is not defective. - Count the possible positions of the one defective bulb. - Multiply one sequence probability by the number of such sequences.

Solution

1. The probabilities are \(P(\text{defective}) = 0.1\) and \(P(\text{not defective}) = 0.9\). 2. a) \(P(\text{all not defective}) = 0.9^4 = 0.6561\). 3. b) Exactly one defective bulb can occur in \(4\) positions. Each corresponding sequence has probability \(0.1 \cdot 0.9^3 = 0.0729\). 4. Therefore, \(P(\text{exactly one defective}) = 4 \cdot 0.0729 = 0.2916\).

Answer

a) \(0.6561 = 65.61\%\) b) \(0.2916 = 29.16\%\)

Multiplication rule and tree diagrams 46

A fair six-sided number cube is rolled four times. Consider these events: \(E\): At least one \(6\) is rolled. \(F\): Exactly one \(6\) is rolled. Find \(P(E)\) and \(P(F)\). Which event is more likely? Justify your answer without drawing the entire tree diagram.

Hints

- Use the complement of no sixes for event \(E\). - Count the possible positions of the single \(6\) for event \(F\). - Compare the events by considering whether one is contained in the other.

Solution

1. The probability of no \(6\) on one roll is \(\frac{5}{6}\). 2. \(P(E) = 1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.5177\). 3. Exactly one \(6\) can occur in \(4\) positions. Each such sequence has probability \(\frac{1}{6}\left(\frac{5}{6}\right)^3 = \frac{125}{1296}\). Therefore, \(P(F) = 4 \cdot \frac{125}{1296} = \frac{125}{324} \approx 0.3858\). 4. Event \(F\) is contained in event \(E\), while \(E\) also includes two, three, or four sixes. Thus, \(E\) is more likely.

Answer

\(P(E) = \frac{671}{1296} \approx 51.77\%\) \(P(F) = \frac{125}{324} \approx 38.58\%\) Event \(E\) is more likely.

Multiplication rule and tree diagrams 47

A class has \(20\) students: \(12\) girls and \(8\) boys. Two students are selected at random for a project team. a) Find the probability that two boys are selected. b) Find the probability that exactly one girl and one boy are selected. c) Find the probability that at least one girl is selected. Use a complement.

Hints

- Without replacement, update the denominator after the first selection. - For exactly one student of each sex, include both possible orders. - For part c, identify the single complementary event.

Solution

1. For part a, \(P(\text{two boys})=\frac{8}{20}\cdot\frac{7}{19}=\frac{14}{95}\approx0.1474\). 2. For part b, the two possible orders are girl-boy and boy-girl. Thus, \(P(\text{one of each})=\frac{12}{20}\cdot\frac{8}{19}+\frac{8}{20}\cdot\frac{12}{19}=\frac{48}{95}\approx0.5053\). 3. The complement of “at least one girl” is “two boys.” Therefore, \(P(\text{at least one girl})=1-\frac{14}{95}=\frac{81}{95}\approx0.8526\).

Answer

a) \(\frac{14}{95}\approx14.74\%\) b) \(\frac{48}{95}\approx50.53\%\) c) \(\frac{81}{95}\approx85.26\%\)

Multiplication rule and tree diagrams 48

Ten cards show the letters in the word “STATISTICS.” The cards are mixed and placed face down. Two cards are turned over at the same time. a) Find the probability that both cards show T. b) Find the probability that neither card shows S. c) Find the probability that at least one card shows I.

Hints

- Count how many times each relevant letter appears. - Turning over two cards at once is equivalent to drawing two cards without replacement. - Use a complement for “at least one I.”

Solution

1. In STATISTICS, S appears \(3\) times, T appears \(3\) times, I appears \(2\) times, and A and C each appear once. 2. a) \(P(\text{two Ts}) = \frac{3}{10} \cdot \frac{2}{9} = \frac{1}{15} \approx 0.0667\). 3. b) There are \(7\) cards that are not S. Thus, \(P(\text{no S}) = \frac{7}{10} \cdot \frac{6}{9} = \frac{7}{15} \approx 0.4667\). 4. c) There are \(8\) cards that are not I. Using the complement, \(P(\text{at least one I}) = 1 - \frac{8}{10} \cdot \frac{7}{9} = 1 - \frac{28}{45} = \frac{17}{45} \approx 0.3778\).

Answer

a) \(\frac{1}{15} \approx 6.67\%\) b) \(\frac{7}{15} \approx 46.67\%\) c) \(\frac{17}{45} \approx 37.78\%\)

Multiplication rule and tree diagrams 49

A container holds \(5\) red fruit candies and \(3\) green fruit candies. Leon selects two candies one at a time with replacement. a) State the branch probabilities for a two-stage tree diagram representing the selections with replacement. b) Find the probability that the two candies have different colors. c) Now suppose Leon eats the first candy before selecting the second, so the selections are made without replacement. Find the probability of two different colors and compare it with your answer to part b.

Hints

- Identify the two paths that produce different colors. - With replacement, the second-stage probabilities remain unchanged. - Without replacement, update the counts after the first selection.

Solution

1. a) With replacement, the first branches are red with probability \(\frac{5}{8}\) and green with probability \(\frac{3}{8}\). The same two probabilities appear after each first branch. 2. b) The different-color paths are red-green and green-red. Therefore, \(P(\text{different, with replacement}) = \frac{5}{8} \cdot \frac{3}{8} + \frac{3}{8} \cdot \frac{5}{8} = \frac{15}{32} = 0.46875\). 3. c) Without replacement, \(P(\text{red, green}) = \frac{5}{8} \cdot \frac{3}{7} = \frac{15}{56}\), and \(P(\text{green, red}) = \frac{3}{8} \cdot \frac{5}{7} = \frac{15}{56}\). 4. Thus, \(P(\text{different, without replacement}) = \frac{30}{56} = \frac{15}{28} \approx 0.5357\), which is greater than \(\frac{15}{32} \approx 0.4688\).

Answer

a) First draw: red \(\frac{5}{8}\), green \(\frac{3}{8}\); after each branch, the same probabilities repeat. b) \(\frac{15}{32} = 0.46875\) c) \(\frac{15}{28} \approx 0.5357\); the probability is greater without replacement.

Multiplication rule and tree diagrams 50

An LED bulb factory produces bulbs with a \(2\%\) defect rate. Ten bulbs are selected independently from a very large production run. Find the probability that at least one of the ten bulbs is defective. Give your answer as a percent, rounded to the nearest hundredth of a percent.

Hints

- What event would make the statement “at least one defective” false? - How do you find the probability that every one of ten independent selections is not defective? - Use the complement and convert the result to the requested percent.

Solution

1. The probability that one bulb is not defective is \(0.98\). 2. The complement of at least one defective bulb is that all ten bulbs are not defective. 3. The probability is \(1-(0.98)^{10}\approx0.182927\), or approximately \(18.29\%\).

Answer

\(18.29\%\).

Multiplication rule and tree diagrams 51

Two games are available. Game A: Roll a fair six-sided die three times. You win if all three rolls are \(6\). Game B: Draw two cards with replacement from a well-shuffled standard \(52\)-card deck. You win if both cards are the ace of hearts. Find the probability of winning each game and decide which game offers the greater chance of winning.

Hints

- Find the probability of the required result on one trial in each game. - Determine how many consecutive times that result must occur. - Multiply the independent probabilities, then compare the results.

Solution

1. For game A, \(P(A)=\left(\frac{1}{6}\right)^3=\frac{1}{216}\approx 0.00463\). 2. For game B, replacement keeps the draw probability at \(\frac{1}{52}\), so \(P(B)=\left(\frac{1}{52}\right)^2=\frac{1}{2704}\approx 0.00037\). 3. Since \(\frac{1}{216}>\frac{1}{2704}\), game A has the greater winning probability.

Answer

Game A: \(\frac{1}{216}\approx 0.46\%\) Game B: \(\frac{1}{2704}\approx 0.04\%\) Game A offers the greater chance of winning.

Multiplication rule and tree diagrams 52

A bowl contains \(20\) gummy bears: \(10\) red, \(6\) yellow, and \(4\) green. Max selects a gummy bear, records its color, replaces it, and mixes the bowl. He repeats this process three times. Find the probability that all three selected gummy bears are the same color.

Hints

- Replacement keeps the color probabilities constant. - Identify the three possible same-color events. - Multiply along each same-color path. - Add the probabilities of the disjoint same-color events.

Solution

1. The color probabilities are \(P(\text{red})=\frac{10}{20}=0.5\), \(P(\text{yellow})=\frac{6}{20}=0.3\), and \(P(\text{green})=\frac{4}{20}=0.2\). 2. The probability of three red selections is \((0.5)^3=0.125\). The probability of three yellow selections is \((0.3)^3=0.027\). The probability of three green selections is \((0.2)^3=0.008\). 3. These events are disjoint, so the total probability is \(0.125+0.027+0.008=0.16\).

Answer

\(0.16\), or \(16\%\)

Multiplication rule and tree diagrams 53

A factory component is defective with probability \(0.05\). Two components are tested independently. Let \(E\) be the event “At least one of the two components is defective.” State the complement \(E^c\), then use it to find \(P(E)\).

Hints

- State the logical opposite of “at least one.” - Multiply the probabilities for two independent acceptable components. - Subtract the complement probability from \(1\).

Solution

1. The complement is \(E^c\): Neither component is defective, so both components are acceptable. 2. The probability that one component is acceptable is \(1-0.05=0.95\). 3. By independence, \(P(E^c)=(0.95)^2=0.9025\). 4. Therefore, \(P(E)=1-P(E^c)=1-0.9025=0.0975\).

Answer

\(E^c\): Both components are acceptable. \(P(E)=0.0975\), or \(9.75\%\)

Multiplication rule and tree diagrams 54

An archer hits the center of a target with probability \(0.60\) on each of three independent shots. Find the probability that the archer: 1) Misses the center on all three shots. 2) Hits the center exactly twice. 3) Hits the center at least once.

Hints

- The hit and miss probabilities remain constant. - Count the orders containing exactly two hits. - The complement of at least one hit is no hits. - Multiply along each path and add favorable paths.

Solution

1. The miss probability is \(0.40\), so \(P(\text{three misses})=(0.40)^3=0.064\). 2. The favorable paths are hit-hit-miss, hit-miss-hit, and miss-hit-hit. Each has probability \((0.60)^2\cdot 0.40=0.144\), so the total is \(3\cdot 0.144=0.432\). 3. Use the complement of three misses: \(P(\text{at least one hit})=1-0.064=0.936\).

Answer

1) \(0.064\) 2) \(0.432\) 3) \(0.936\)

Multiplication rule and tree diagrams 55

A fruit basket contains \(15\) pieces of fruit: \(9\) apples and \(6\) pears. Three pieces are selected one at a time without replacement. a) What is the probability of selecting pear, then apple, then pear? b) Find the probability that at least one pear is selected. c) Given that the first piece selected was an apple, what is the probability that the second and third pieces are also apples?

Hints

- In part a, keep the required order and update both the numerator and denominator after each selection. - For “at least one,” consider using the complement. - In part c, begin with the basket after the first apple has already been removed.

Solution

1. For the specified order, multiply the changing probabilities: \(P(\text{pear, apple, pear})=\frac{6}{15}\cdot\frac{9}{14}\cdot\frac{5}{13}=\frac{9}{91}\). 2. The complement of “at least one pear” is “three apples.” Its probability is \(\frac{9}{15}\cdot\frac{8}{14}\cdot\frac{7}{13}=\frac{12}{65}\). Therefore, \(P(\text{at least one pear})=1-\frac{12}{65}=\frac{53}{65}\). 3. After one apple has been selected, \(8\) apples remain among \(14\) pieces. Thus, \(P(\text{apple, apple after first apple})=\frac{8}{14}\cdot\frac{7}{13}=\frac{4}{13}\).

Answer

a) \(P(\text{pear, apple, pear})=\frac{9}{91}\approx 0.0989\), or about \(9.89\%\). b) \(P(\text{at least one pear})=\frac{53}{65}\approx 0.8154\), or about \(81.54\%\). c) \(\frac{4}{13}\approx 0.3077\), or about \(30.77\%\).

Multiplication rule and tree diagrams 56

A drawer contains \(6\) gray socks and \(4\) blue socks. Two socks are selected one at a time without replacement. a) How many terminal paths are in the corresponding tree diagram? b) Find the probability that the two selected socks are the same color. c) Find the probability that at least one of the socks is blue.

Hints

- Identify the two possible colors on each draw. - The total number of socks and the number of socks of a selected color both change after the first draw. - Add the probabilities of the two same-color paths. - For part c, use the complement of selecting no blue socks.

Solution

1. Each draw can result in gray or blue, so the four color paths are gray-gray, gray-blue, blue-gray, and blue-blue. 2. The probability of two gray socks is \(\frac{6}{10}\cdot\frac{5}{9}=\frac{1}{3}\). The probability of two blue socks is \(\frac{4}{10}\cdot\frac{3}{9}=\frac{2}{15}\). Therefore, \(P(\text{same color})=\frac{1}{3}+\frac{2}{15}=\frac{7}{15}\). 3. The complement of at least one blue sock is two gray socks. Thus, \(P(\text{at least one blue})=1-\frac{1}{3}=\frac{2}{3}\).

Answer

a) \(4\) paths b) \(\frac{7}{15}\approx 46.7\%\) c) \(\frac{2}{3}\approx 66.7\%\)

Multiplication rule and tree diagrams 57

A factory produces electronic components. Each component is acceptable \((A)\) with probability \(0.90\) and defective \((D)\) with probability \(0.10\). Three components are selected independently from a very large production run. a) How many terminal nodes are in a complete tree diagram for this three-stage experiment? b) Find the probability that all three components are acceptable. c) Find the probability that exactly one of the three components is defective.

Hints

- Count the number of branches created at each stage. - Multiply probabilities along one path. - Identify every path containing exactly one defective component. - Check whether those paths have equal probabilities.

Solution

1. Each of the three stages has two possible outcomes, so the tree has \(2^3=8\) terminal nodes. 2. The probability of the path \((A, A, A)\) is \((0.90)^3=0.729\). 3. Exactly one defective component occurs on the paths \((D, A, A)\), \((A, D, A)\), and \((A, A, D)\). Each path has probability \(0.10\cdot (0.90)^2=0.081\), so the total probability is \(3\cdot 0.081=0.243\).

Answer

a) \(8\) terminal nodes b) \(0.729\), or \(72.9\%\) c) \(0.243\), or \(24.3\%\)

Multiplication rule and tree diagrams 58

A factory produces components that are acceptable \((A)\) with probability \(0.95\) and defective \((D)\) with probability \(0.05\). Four components are tested independently. Consider these events: \(A_1\): Only the second component is defective. \(B\): Exactly three components are acceptable. \(C\): The first three components are acceptable. \(D_1\): Only the fourth component is defective. a) List all outcomes in each event as ordered quadruples. b) Which events have the same probability? Explain without calculating. c) Find \(P(A_1)\), \(P(B)\), and \(P(C)\).

Hints

- Determine how many ordered outcomes belong to each event. - Pay attention to the difference between “only” and “exactly.” - In event \(C\), the fourth component is not specified. - Outcomes with the same numbers of each result have equal probabilities in this independent setting.

Solution

1. \(A_1=\{(A, D, A, A)\}\). 2. \(B=\{(D, A, A, A),(A, D, A, A),(A, A, D, A),(A, A, A, D)\}\). 3. \(C=\{(A, A, A, A),(A, A, A, D)\}\). 4. \(D_1=\{(A, A, A, D)\}\). 5. Events \(A_1\) and \(D_1\) each contain one outcome with three acceptable components and one defective component, so they have the same probability. 6. \(P(A_1)=(0.95)^3\cdot0.05=0.04286875\). 7. \(P(B)=4\cdot(0.95)^3\cdot0.05=0.171475\). 8. The fourth component is unrestricted in event \(C\), so \(P(C)=(0.95)^3=0.857375\).

Answer

a) \(A_1=\{(A, D, A, A)\}\); \(B=\{(D, A, A, A),(A, D, A, A),(A, A, D, A),(A, A, A, D)\}\); \(C=\{(A, A, A, A),(A, A, A, D)\}\); \(D_1=\{(A, A, A, D)\}\) b) \(P(A_1)=P(D_1)\) because each event contains one outcome with the same numbers of acceptable and defective components. c) \(P(A_1)\approx 0.0429\); \(P(B)\approx 0.1715\); \(P(C)\approx 0.8574\)

Multiplication rule and tree diagrams 59

A basketball player makes each free throw with probability \(0.25\). The player takes three independent free throws. Let \(M\) represent a make and \(X\) represent a miss. Consider these events: \(A\): Only the second shot is made. \(B\): The first made shot occurs on the second attempt. \(C\): The second shot is made. \(D\): The second shot is missed. a) List all ordered outcomes in events \(A\), \(B\), and \(C\). b) Find \(P(A)\), \(P(B)\), \(P(C)\), and \(P(D)\).

Hints

- A tree diagram can help organize the ordered outcomes. - Interpret words such as “only” and “first” carefully. - Multiply probabilities along one ordered outcome. - If a shot is unrestricted, its possible outcomes have total probability \(1\).

Solution

1. \(A=\{(X, M, X)\}\). 2. \(B=\{(X, M, X),(X, M, M)\}\). 3. \(C=\{(M, M, M),(M, M, X),(X, M, M),(X, M, X)\}\). 4. \(P(A)=0.75\cdot0.25\cdot0.75=0.140625=\frac{9}{64}\). 5. In event \(B\), the third shot is unrestricted, so \(P(B)=0.75\cdot0.25=0.1875=\frac{3}{16}\). 6. Event \(C\) depends only on the second shot, so \(P(C)=0.25=\frac{1}{4}\). 7. Event \(D\) is the complement of \(C\), so \(P(D)=0.75=\frac{3}{4}\).

Answer

a) \(A=\{(X, M, X)\}\); \(B=\{(X, M, X),(X, M, M)\}\); \(C=\{(M, M, M),(M, M, X),(X, M, M),(X, M, X)\}\) b) \(P(A)=\frac{9}{64}\approx 14.1\%\); \(P(B)=\frac{3}{16}=18.75\%\); \(P(C)=\frac{1}{4}=25\%\); \(P(D)=\frac{3}{4}=75\%\)

Multiplication rule and tree diagrams 60

A bag contains five balls: two red and three blue. A ball is drawn, its color is recorded, and it is replaced. This process is repeated three times. Find the probability that: a) All three draws are red. b) Exactly two draws are blue. c) At least one draw is red.

Hints

- Find the red and blue probabilities for one draw. - Replacement keeps the probabilities constant. - List the paths containing exactly two blue results. - For “at least one red,” use the complement.

Solution

1. Because each ball is replaced, \(P(\text{red})=\frac{2}{5}=0.4\) and \(P(\text{blue})=\frac{3}{5}=0.6\) on every draw. 2. \(P(\text{three red})=\left(\frac{2}{5}\right)^3=\frac{8}{125}=0.064\). 3. Exactly two blue draws can occur as blue-blue-red, blue-red-blue, or red-blue-blue. Each path has probability \(\left(\frac{3}{5}\right)^2\cdot\frac{2}{5}=\frac{18}{125}\), so the total is \(3\cdot\frac{18}{125}=\frac{54}{125}=0.432\). 4. Use the complement of three blue draws: \(P(\text{at least one red})=1-\left(\frac{3}{5}\right)^3=\frac{98}{125}=0.784\).

Answer

a) \(\frac{8}{125}=0.064\) b) \(\frac{54}{125}=0.432\) c) \(\frac{98}{125}=0.784\)

Multiplication rule and tree diagrams 61

A manufacturer reports that \(5\%\) of its LED bulbs are defective. Four bulbs are selected and tested independently. a) Find the probability that all four bulbs work properly. b) Find the probability that only the first bulb is defective. c) Find the probability that at least one bulb is defective. Give each answer as a percent rounded to the nearest hundredth of a percent.

Hints

- What is the chance of one bulb working, rather than being defective? - In part b), which positions have fixed outcomes, and what is the probability of that exact sequence? - What is the opposite of at least one defective bulb? - Convert your probabilities to percents only after completing the calculations.

Solution

1. One bulb works with probability \(0.95\) and is defective with probability \(0.05\). 2. All four working has probability \((0.95)^4=0.81450625\approx81.45\%\). 3. Only the first being defective has probability \(0.05\cdot(0.95)^3=0.04286875\approx4.29\%\). 4. At least one defective is the complement of all four working, giving \(1-(0.95)^4=0.18549375\approx18.55\%\).

Answer

a) \(81.45\%\). b) \(4.29\%\). c) \(18.55\%\).

Multiplication rule and tree diagrams 62

An archer hits the center of a target with probability \(0.70\) on each of three independent shots. a) Find the probability of the sequence hit, miss, hit. b) Find the probability of exactly two hits in the three shots.

Hints

- Multiply probabilities along the specified sequence. - List all orders containing exactly two hits. - Add the probabilities of the favorable paths.

Solution

1. The hit probability is \(0.70\), and the miss probability is \(0.30\). 2. The probability of hit-miss-hit is \(0.70\cdot 0.30\cdot 0.70=0.147\). 3. Exactly two hits can occur as hit-hit-miss, hit-miss-hit, or miss-hit-hit. Each path has probability \((0.70)^2\cdot 0.30=0.147\), so the total is \(3\cdot 0.147=0.441\).

Answer

a) \(0.147\), or \(14.7\%\) b) \(0.441\), or \(44.1\%\)

Multiplication rule and tree diagrams 63

Decide whether each statement about tree diagrams for multistage probability experiments is true or false. Justify each answer. a) The probability of a complete path can never be greater than the probability on the first branch of that path. b) The sum of the probabilities of all complete paths in a tree diagram is always \(1\).

Hints

- Recall the possible values of a probability. - Consider what happens when a number is multiplied by a value between \(0\) and \(1\). - Identify what all complete paths represent together. - Recall the probability of the entire sample space.

Solution

1. A complete path probability is a product \(p_1p_2\cdots p_n\). Because every branch probability is between \(0\) and \(1\), multiplying \(p_1\) by the remaining factors cannot increase it. Thus, the first statement is true. 2. In a complete tree diagram, the complete paths are mutually exclusive and together represent every possible outcome, so their probabilities add to \(1\). Thus, the second statement is true.

Answer

a) True. Multiplying the first branch probability by additional factors between \(0\) and \(1\) cannot make the result larger. b) True. In a complete tree, all terminal paths form the entire sample space, so their probabilities sum to \(1\).

Multiplication rule and tree diagrams 64

Two fair eight-sided dice labeled \(1\) through \(8\) are rolled. Find the probability that at least one die: (1) shows \(8\); (2) shows a prime number; (3) shows a number less than \(3\).

Hints

- Use the complement that neither die meets the condition. - Multiply probabilities because the dice are independent. - List the prime numbers and the numbers less than \(3\).

Solution

1. For each part, use the complement that neither die meets the condition. 2. For an \(8\), \(P(\text{at least one }8)=1-\left(\frac{7}{8}\right)^2=\frac{15}{64}\). 3. The primes are \(2,3,5,7\), so one die is not prime with probability \(\frac{1}{2}\). Thus, \(P(\text{at least one prime})=1-\left(\frac{1}{2}\right)^2=\frac{3}{4}\). 4. The numbers less than \(3\) are \(1\) and \(2\), so one die fails the condition with probability \(\frac{6}{8}=\frac{3}{4}\). Thus, \(P(\text{at least one result below }3)=1-\left(\frac{3}{4}\right)^2=\frac{7}{16}\).

Answer

(1) \(\frac{15}{64}\) (2) \(\frac{3}{4}\) (3) \(\frac{7}{16}\)

Multiplication rule and tree diagrams 65

A fair eight-sided die labeled \(1\) through \(8\) is rolled three times. Find the probability that: 1. the sum of the three results is less than \(6\); 2. an \(8\) occurs at least once; 3. all three results are different.

Hints

- Count ordered triples for the small possible sums. - Use the complement of no \(8\). - For all different values, count the remaining choices at each roll.

Solution

1. There are \(8^3=512\) equally likely ordered outcomes. Sums less than \(6\) are \(3\), \(4\), and \(5\), occurring in \(1\), \(3\), and \(6\) ordered outcomes, respectively. Thus, \(P(S<6)=\frac{10}{512}=\frac{5}{256}\). 2. The probability of no \(8\) in three rolls is \(\left(\frac{7}{8}\right)^3\). Therefore, \(P(\text{at least one }8)=1-\frac{343}{512}=\frac{169}{512}\). 3. The first result can be any of \(8\) values, the second any of the remaining \(7\), and the third any of the remaining \(6\). Thus, \(P(\text{all different})=\frac{8\cdot7\cdot6}{8^3}=\frac{21}{32}\).

Answer

1. \(\frac{5}{256}\approx 0.01953\) 2. \(\frac{169}{512}\approx 0.33008\) 3. \(\frac{21}{32}=0.65625\)

Multiplication rule and tree diagrams 66

A factory checks electronic components for defects. For a component that is defective, each of two independent automated tests gives a positive result with probability \(92\%\). If both tests are positive, the component is rejected at once. If exactly one is positive, the component is sent for manual inspection. If neither is positive, the component is classified as acceptable. Among defective components sent for manual inspection, the manual inspection detects the defect with probability \(96\%\). For a defective component, find the probability that: (1) it is immediately rejected; (2) neither automated test detects the defect; (3) manual inspection is needed; (4) the defect is first detected during manual inspection; (5) the component is ultimately identified as defective.

Hints

- First list the three possible patterns: two positives, one positive, and no positives. - Which of those patterns leads to manual inspection? - Use the detection percentage within the manual-inspection group, not within all components. - The final outcome can happen by immediate rejection or by manual detection.

Solution

1. Each automated test has positive probability \(0.92\) and negative probability \(0.08\) for a defective component. 2. Both tests are positive with probability \(0.92^2=0.8464\). 3. Both are negative with probability \(0.08^2=0.0064\). 4. Exactly one is positive with probability \(2\cdot0.92\cdot0.08=0.1472\). 5. Of the components requiring manual inspection, \(96\%\) are detected. Thus the probability of first detecting the defect manually is \(0.1472\cdot0.96=0.141312\). 6. The component is ultimately identified as defective if both automated tests are positive or the manual inspection detects it: \(0.8464+0.141312=0.987712\).

Answer

(1) \(0.8464\) (2) \(0.0064\) (3) \(0.1472\) (4) \(0.141312\) (5) \(0.987712\)

Multiplication rule and tree diagrams 67

A seed-cleaning machine removes weed seeds in three successive stages. The first stage removes \(75\%\) of the weed seeds. The second stage removes \(60\%\) of the weed seeds that remain, and the third stage removes \(40\%\) of those still remaining. Find the probability that a weed seed passes through all three stages without being removed. Then find the probability that it is removed during the process.

Hints

- Find the fraction that remains after each stage. - Multiply the probabilities along the path that survives every stage. - Use the complement to find the probability of removal.

Solution

1. The probabilities of surviving the three stages are \(0.25\), \(0.40\), and \(0.60\), respectively. 2. Therefore, \(P(\text{not removed})=0.25\cdot 0.40\cdot 0.60=0.06\). 3. Use the complement: \(P(\text{removed})=1-0.06=0.94\).

Answer

Not removed: \(0.06\), or \(6\%\) Removed: \(0.94\), or \(94\%\)

Multiplication rule and tree diagrams 68

A data center uses three successive security systems to detect unauthorized access. Their detection rates are \(90\%\), \(85\%\), and \(70\%\), respectively, for threats that reach each system. a) Find the probability that an attack passes all three systems undetected. b) Find the probability that at least one system detects the attack. c) A fourth system will be added. What detection rate must it have so that the overall detection probability is greater than \(99.9\%\)?

Hints

- An attack remains undetected only if every system misses it. - Use the complement for at least one detection. - In part c, write an inequality for the new probability of passing every system undetected.

Solution

1. The probabilities of passing the three systems undetected are \(0.10\), \(0.15\), and \(0.30\). 2. Thus, \(P(\text{undetected})=0.10\cdot 0.15\cdot 0.30=0.0045\). 3. Therefore, \(P(\text{detected})=1-0.0045=0.9955\). 4. Let \(p\) be the fourth system's detection rate. Require \(0.0045(1-p)<0.001\). 5. Solving gives \(1-p<\frac{0.001}{0.0045}=\frac{2}{9}\), so \(p>\frac{7}{9}\approx 0.7778\).

Answer

a) \(0.0045\), or \(0.45\%\) b) \(0.9955\), or \(99.55\%\) c) The fourth system's detection rate must be greater than \(\frac{7}{9}\approx 77.78\%\). If rates are stated to the nearest tenth of a percent, \(77.8\%\) is the smallest acceptable rate.

Multiplication rule and tree diagrams 69

A market research agency surveyed \(2000\) people about a streaming service. Respondents were classified as under age \(30\) \((A)\) or age \(30\) and older \((B)\), and as regular users \((R)\) or rare/nonusers \((N)\). <table> <tr><th colspan="2">Use by age group</th><th colspan="2">Use</th><th rowspan="2">Total</th></tr> <tr><th colspan="2"></th><th>Regular \((R)\)</th><th>Rare/none \((N)\)</th></tr> <tr><td rowspan="2">Age group</td><td>Under \(30\) \((A)\)</td><td>\(900\)</td><td></td><td>\(1200\)</td></tr> <tr><td>Age \(30\) or older \((B)\)</td><td></td><td></td><td></td></tr> <tr><td colspan="2">Total</td><td>\(1300\)</td><td></td><td>\(2000\)</td></tr> </table> 1. Complete the table. 2. Find the branch probabilities for a tree diagram that branches first by age group. 3. Find the branch probabilities for a tree diagram that branches first by usage category.

Hints

- Use the row and column totals to complete the table. - First-stage branch probabilities are the marginal relative frequencies. - Second-stage branch probabilities are conditional relative frequencies within the first-stage group. - Probabilities leaving each node must sum to \(1\).

Solution

1. Complete the counts: \(n(A\cap N)=1200-900=300\), \(n(B\cap R)=1300-900=400\), \(n(B)=2000-1200=800\), \(n(B\cap N)=800-400=400\), and \(n(N)=300+400=700\). 2. For the age-first tree, \(P(A)=\frac{1200}{2000}=0.60\) and \(P(B)=0.40\). The second-stage probabilities are \(P(R\mid A)=\frac{900}{1200}=0.75\), \(P(N\mid A)=0.25\), \(P(R\mid B)=\frac{400}{800}=0.50\), and \(P(N\mid B)=0.50\). 3. For the usage-first tree, \(P(R)=\frac{1300}{2000}=0.65\) and \(P(N)=0.35\). The second-stage probabilities are \(P(A\mid R)=\frac{900}{1300}=\frac{9}{13}\), \(P(B\mid R)=\frac{4}{13}\), \(P(A\mid N)=\frac{300}{700}=\frac{3}{7}\), and \(P(B\mid N)=\frac{4}{7}\).

Answer

1. <table> <tr><th colspan="2">Use by age group</th><th colspan="2">Use</th><th rowspan="2">Total</th></tr> <tr><th colspan="2"></th><th>Regular \((R)\)</th><th>Rare/none \((N)\)</th></tr> <tr><td rowspan="2">Age group</td><td>Under \(30\) \((A)\)</td><td>\(900\)</td><td>\(300\)</td><td>\(1200\)</td></tr> <tr><td>Age \(30\) or older \((B)\)</td><td>\(400\)</td><td>\(400\)</td><td>\(800\)</td></tr> <tr><td colspan="2">Total</td><td>\(1300\)</td><td>\(700\)</td><td>\(2000\)</td></tr> </table> 2. Age-first tree: \(P(A)=0.60\), \(P(B)=0.40\), \(P(R\mid A)=0.75\), \(P(N\mid A)=0.25\), \(P(R\mid B)=0.50\), and \(P(N\mid B)=0.50\). 3. Usage-first tree: \(P(R)=0.65\), \(P(N)=0.35\), \(P(A\mid R)=\frac{9}{13}\), \(P(B\mid R)=\frac{4}{13}\), \(P(A\mid N)=\frac{3}{7}\), and \(P(B\mid N)=\frac{4}{7}\).

Multiplication rule and tree diagrams 70

The spinner shown has \(8\) equal sections. It is spun twice. Find each probability as a simplified fraction and a percent. a) Event \(E_1\): Both spins are green. b) Event \(E_2\): Both spins are the same color.

Figure for problem 532004

Hints

- Read the probability of each color from the equal-section spinner. - Multiply probabilities along a single two-spin path. - Add the probabilities of the three same-color paths.

Solution

1. From the spinner, \(P(\text{red})=\frac{3}{8}\), \(P(\text{blue})=\frac{3}{8}\), and \(P(\text{green})=\frac{2}{8}=\frac{1}{4}\). 2. \(P(E_1)=\left(\frac{2}{8}\right)^2=\frac{1}{16}=6.25\%\). 3. Event \(E_2\) consists of red-red, blue-blue, and green-green. Therefore, \(P(E_2)=\left(\frac{3}{8}\right)^2+\left(\frac{3}{8}\right)^2+\left(\frac{2}{8}\right)^2=\frac{22}{64}=\frac{11}{32}=34.375\%\).

Answer

a) \(P(E_1)=\frac{1}{16}=6.25\%\) b) \(P(E_2)=\frac{11}{32}=34.375\%\)

Multiplication rule and tree diagrams 71

The fair spinner shown has \(8\) equal labeled sections. It is spun twice. Find the probability of event \(E\): “The spinner lands on green at least once.” First describe the complement \(E^c\) in words. Then find \(P(E)\) as a fraction.

Figure for problem 532041

Hints

- What is the opposite of landing on green at least once? - Count the sections that are not green in the spinner. - Find the probability of two independent non-green results. - Use the relationship between an event and its complement.

Solution

1. The complement of “green at least once” is “no green on either spin,” or equivalently, “a non-green color on both spins.” 2. The spinner shows \(3\) non-green sections out of \(8\), so the probability of a non-green result on one spin is \(\frac{3}{8}\). 3. The two spins are independent, so \(P(E^c)=\frac{3}{8}\cdot\frac{3}{8}=\frac{9}{64}\). 4. Using the complement rule, \(P(E)=1-P(E^c)=1-\frac{9}{64}=\frac{55}{64}\).

Answer

The complement \(E^c\) is: The spinner does not land on green on either spin. \(P(E)=\frac{55}{64}\)

Multiplication rule and tree diagrams 72

The spinner shown has \(5\) equal labeled sections. It is spun three times. Let \(A\) be the event “The spinner lands on red at least once.” a) Describe the complement \(A^c\) in words. b) Find \(P(A^c)\) as a fraction in simplest form. c) Use the complement to find \(P(A)\) as a fraction in simplest form.

Figure for problem 532153

Hints

- What must happen on all three spins for “red at least once” to fail? - Read the one-spin blue probability from the spinner. - Multiply along the three independent spins, then use the complement rule.

Solution

1. The complement of “red at least once” over three spins is “no red on any spin,” which is the same as “blue on all three spins.” 2. From the spinner, the probability of blue on one spin is \(\frac{4}{5}\). Because the spins are independent, \(P(A^c)=\left(\frac{4}{5}\right)^3=\frac{64}{125}\). 3. By the complement rule, \(P(A)=1-P(A^c)=1-\frac{64}{125}=\frac{61}{125}\).

Answer

a) \(A^c\): The spinner lands on blue on all three spins. b) \(P(A^c)=\frac{64}{125}\) c) \(P(A)=\frac{61}{125}\)

Multiplication rule and tree diagrams 73

Spin each spinner once. The spins are independent. Find each probability as a fraction in simplest form and as a percent rounded to one decimal place. a) Both spinners land on blue. b) Both spinners land on a prime number.

Figure for problem 535913

Hints

- Use the sectors of each spinner to count the favorable outcomes and the total outcomes. - For the prime-number event, check which numbers on each spinner are prime. - What operation combines the probabilities from two independent spins? - Convert your final fractions to percents only after multiplying.

Solution

1. Spinner 1 has \(4\) blue sections out of \(8\), and Spinner 2 has \(2\) out of \(6\). Multiply the one-spin probabilities: \(P(\text{both blue})=\frac{4}{8}\cdot\frac{2}{6}=\frac{1}{6}\approx16.7\%\). 2. The prime numbers on Spinner 1 are \(2,3,5,7\), so its prime probability is \(\frac{4}{8}\). Those on Spinner 2 are \(2,3,5\), so its prime probability is \(\frac{3}{6}\). 3. Thus, \(P(\text{both prime})=\frac{4}{8}\cdot\frac{3}{6}=\frac{1}{4}=25.0\%\).

Answer

a) \(\frac{1}{6}\approx16.7\%\). b) \(\frac{1}{4}=25.0\%\).

Multiplication rule and tree diagrams 74

The spinner shown has \(6\) equal sections labeled \(1\), \(2\), and \(3\). It is spun twice, and the two results are added. a) Find the probability that the sum is exactly \(4\). b) Use the complement to find the probability that the sum is at least \(3\).

Figure for problem 535935

Hints

- Find the probability of each number on one spin. - For part a, list the ordered pairs whose sum is \(4\). - Multiply along each path and add the favorable path probabilities. - For part b, identify the only outcome outside the event.

Solution

1. On one spin, \(P(1) = \frac{3}{6} = \frac{1}{2}\), \(P(2) = \frac{2}{6} = \frac{1}{3}\), and \(P(3) = \frac{1}{6}\). 2. A sum of \(4\) occurs for \((1, 3)\), \((3, 1)\), or \((2, 2)\). Therefore, \(P(\text{sum }4) = \frac{1}{2} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{3} = \frac{5}{18}\). 3. For part b, the complement of a sum of at least \(3\) is a sum less than \(3\). The only such outcome is \((1, 1)\), with probability \(\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\). 4. Thus, \(P(\text{sum at least }3) = 1 - \frac{1}{4} = \frac{3}{4}\).

Answer

a) \(\frac{5}{18} \approx 27.8\%\) b) \(\frac{3}{4} = 75\%\)

Multiplication rule and tree diagrams 75

The fair spinner shown has \(10\) equal labeled sections. It is spun twice. Find the probability of event \(E\): “The two spins show different colors.” Use the complement method: 1. Describe the complement \(E^c\) in words. 2. Find \(P(E^c)\). 3. Use that result to find \(P(E)\).

Figure for problem 535949

Hints

- What must happen for the two colors to fail to be different? - Read each one-spin color probability from the spinner. - Add the probabilities of the same-color paths, then subtract from \(1\).

Solution

1. The complement of “different colors” is “the same color on both spins.” The possible same-color pairs are orange-orange, blue-blue, and yellow-yellow. 2. From the spinner, \(P(\text{orange})=\frac{5}{10}=0.5\), \(P(\text{blue})=\frac{3}{10}=0.3\), and \(P(\text{yellow})=\frac{2}{10}=0.2\). 3. Because the spins are independent and the three same-color cases do not overlap, \(P(E^c)=0.5\cdot0.5+0.3\cdot0.3+0.2\cdot0.2=0.38\). 4. Therefore, \(P(E)=1-P(E^c)=1-0.38=0.62\), or \(62\%\).

Answer

1. \(E^c\): The same color appears on both spins. 2. \(P(E^c)=0.38\) 3. \(P(E)=0.62=62\%\)

Multiplication rule and tree diagrams 76

The spinner shown has \(8\) equal sections and is spun three times. Find the probability of each event. a) Blue appears three times. b) Yellow appears exactly once. c) Green does not appear.

Figure for problem 536087

Hints

- Read the probability of each color from the equal-section spinner. - Multiply probabilities along a specific three-spin path. - For exactly one yellow, count the possible positions of the yellow result. - “No green” means a non-green result on all three spins.

Solution

1. From the spinner, \(P(\text{blue})=\frac{2}{8}=\frac{1}{4}\), \(P(\text{yellow})=\frac{3}{8}\), and \(P(\text{not yellow})=P(\text{not green})=\frac{5}{8}\). 2. For part a, \(P(\text{three blue})=\left(\frac{1}{4}\right)^3=\frac{1}{64}\approx1.56\%\). 3. For part b, the one yellow result can occur on any of the three spins. Therefore, \(P(\text{exactly one yellow})=3\cdot\frac{3}{8}\cdot\left(\frac{5}{8}\right)^2=\frac{225}{512}\approx43.95\%\). 4. For part c, every spin must be not green. Thus, \(P(\text{no green})=\left(\frac{5}{8}\right)^3=\frac{125}{512}\approx24.41\%\).

Answer

a) \(\frac{1}{64}\approx1.56\%\) b) \(\frac{225}{512}\approx43.95\%\) c) \(\frac{125}{512}\approx24.41\%\)

Multiplication rule and tree diagrams 77

An urn contains \(4\) orange balls and \(2\) purple balls. Two balls are drawn in order without replacement. a) Find the probability that both balls have the same color. b) Is this probability greater than, less than, or equal to \(50\%\)?

Hints

- Use the orange-orange and purple-purple paths. - Without replacement, both the favorable count and the total count change after the first draw. - Add the favorable path probabilities, then compare the result with \(\frac{1}{2}\).

Solution

1. There are \(6\) balls total. 2. The probability of drawing two orange balls is \(\frac{4}{6}\cdot\frac{3}{5}=\frac{12}{30}\). 3. The probability of drawing two purple balls is \(\frac{2}{6}\cdot\frac{1}{5}=\frac{2}{30}\). 4. Adding the two favorable paths gives \(P(\text{same color})=\frac{12}{30}+\frac{2}{30}=\frac{7}{15}\approx46.7\%\). 5. Since \(46.7\%<50\%\), the probability is less than \(50\%\).

Answer

a) \(\frac{7}{15}\approx46.7\%\) b) Less than \(50\%\)

Multiplication rule and tree diagrams 78

A survey of \(500\) teenagers records whether they exercise regularly \((E)\) and whether they play a musical instrument \((M)\). The tree diagram shows the relative frequencies. Use the diagram to create a two-way table showing the counts of teenagers in each category.

Figure for problem 537446

Hints

- Use the first-stage probabilities to find the numbers who do and do not exercise regularly. - Multiply each first-stage count by its second-stage conditional probability. - Add the inner cells to find the row and column totals. - Check that all four inner counts sum to \(500\).

Solution

1. The first-stage counts are \(n(E)=500\cdot0.65=325\) and \(n(E^c)=500\cdot0.35=175\). 2. Follow each second-stage branch: \(n(E\cap M)=325\cdot0.20=65\), \(n(E\cap M^c)=325\cdot0.80=260\), \(n(E^c\cap M)=175\cdot0.40=70\), and \(n(E^c\cap M^c)=175\cdot0.60=105\). 3. The column totals are \(n(M)=65+70=135\) and \(n(M^c)=260+105=365\).

Answer

<table> <tr><td></td><td>\(M\)</td><td>\(M^c\)</td><td>Total</td></tr> <tr><td>\(E\)</td><td>\(65\)</td><td>\(260\)</td><td>\(325\)</td></tr> <tr><td>\(E^c\)</td><td>\(70\)</td><td>\(105\)</td><td>\(175\)</td></tr> <tr><td>Total</td><td>\(135\)</td><td>\(365\)</td><td>\(500\)</td></tr> </table>

Multiplication rule and tree diagrams 79

A bakery uses two ovens to bake dinner rolls. The probability tree gives the share of rolls from each oven and the rates at which they are classified as perfect, too dark, or unusable. Of \(2000\) rolls produced, how many are expected to be classified as perfect?

Figure for problem 537447

Hints

- Read the first-stage rates to find how many rolls each oven produces. - For each oven, what fraction of its rolls are classified as perfect? - Add the expected perfect-roll counts from the two ovens.

Solution

1. The first oven produces \(2000\cdot0.60=1200\) rolls, and the second produces \(2000\cdot0.40=800\) rolls. 2. Of these, the expected number of perfect rolls from Oven 1 is \(1200\cdot0.95=1140\). 3. From Oven 2, the expected number is \(800\cdot0.85=680\). 4. Together, the ovens are expected to produce \(1140+680=1820\) perfect rolls.

Answer

\(1820\) perfect rolls, on average.

Multiplication rule and tree diagrams 80

A battery factory uses two production lines, \(M_1\) and \(M_2\). The tree shows each line's share of batteries and the chance a battery is defective (\(D\)) or works properly (\(W\)). a) Which line makes more batteries? Which has the higher defect rate? Give each difference in percentage points. b) Make a two-way probability table using the tree. Use the two production lines as rows and \(D\), \(W\), and totals as columns.

Figure for problem 537448

Hints

- What does the first split tell you about each production line? - Compare the defective branches leaving the two lines, not just the first branches. - Each inner table cell is the probability of following one complete path through the tree. - Add the inner cells across rows and down columns to check the totals.

Solution

1. Line \(M_2\) makes \(55\%\) of the batteries, compared with \(45\%\) from \(M_1\). The difference is \(10\) percentage points. The defect rates are \(6\%\) and \(4\%\), so \(M_2\) has the higher defect rate by \(2\) percentage points. 2. Multiply along each tree path: \(P(M_1\cap D)=0.45\cdot0.04=0.018\), \(P(M_1\cap W)=0.45\cdot0.96=0.432\), \(P(M_2\cap D)=0.55\cdot0.06=0.033\), and \(P(M_2\cap W)=0.55\cdot0.94=0.517\). 3. Add the cells to check the margins: \(P(D)=0.051\), \(P(W)=0.949\), and the grand total is \(1\).

Answer

a) \(M_2\) supplies more by \(10\) percentage points. Its defect rate is higher by \(2\) percentage points. b) <table><tr><th></th><th>\(D\)</th><th>\(W\)</th><th>Total</th></tr><tr><th>\(M_1\)</th><td>\(0.018\)</td><td>\(0.432\)</td><td>\(0.45\)</td></tr><tr><th>\(M_2\)</th><td>\(0.033\)</td><td>\(0.517\)</td><td>\(0.55\)</td></tr><tr><th>Total</th><td>\(0.051\)</td><td>\(0.949\)</td><td>\(1\)</td></tr></table>

Multiplication rule and tree diagrams 81

A bowl contains \(20\) candies, exactly one of which is lemon. Three children each take one candy without replacement. In the tree, \(L\) means lemon and \(L^c\) means not lemon. The tree tracks who gets the lemon candy; after someone gets it, that path ends because the lemon cannot be drawn again. a) Find the missing branch probabilities \(a\), \(b\), and \(c\). b) Find the probability that the second child gets the lemon candy. c) Explain why the probability that the third child gets the lemon candy is also \(\frac{1}{20}\).

Figure for problem 537450

Hints

- Follow the branch where the lemon has not yet been taken. How many candies remain? - Each pair of branches leaving the same node must have probabilities totaling \(1\). - Which path ends with the second child taking lemon? Which path ends with the third child taking it?

Solution

1. Initially, \(19\) of the \(20\) candies are not lemon, so \(a=\frac{19}{20}\). 2. After a non-lemon candy is taken, \(19\) remain, including the lemon. Therefore, \(b=\frac{1}{19}\) and \(c=\frac{18}{19}\). 3. For the second child to get lemon, the first must not: \(P(L^c,L)=\frac{19}{20}\cdot\frac{1}{19}=\frac{1}{20}\). 4. For the third child to get lemon, the first two must not: \(P(L^c,L^c,L)=\frac{19}{20}\cdot\frac{18}{19}\cdot\frac{1}{18}=\frac{1}{20}\).

Answer

a) \(a=\frac{19}{20}\), \(b=\frac{1}{19}\), \(c=\frac{18}{19}\) b) \(\frac{1}{20}\) c) \(\frac{19}{20}\cdot\frac{18}{19}\cdot\frac{1}{18}=\frac{1}{20}\). The three positions are equally likely to receive the only lemon candy.

Multiplication rule and tree diagrams 82

An automated maintenance system can make up to four repair attempts. It stops as soon as an attempt succeeds. In the tree diagram, \(S_k\) means success on attempt \(k\), and \(F_k\) means failure on attempt \(k\). Each edge label is the conditional probability for that outcome given that the process reached that attempt. a) Find the probability that the first success occurs on exactly the fourth attempt. b) Find the probability that the system succeeds by the end of the fourth attempt. c) Compare the probability of success on the first attempt with the probability that the first success occurs on the fourth attempt.

Figure for problem 537451

Hints

- For part a), trace the one path with three failures followed by a success. - Multiply the edge probabilities along a single path. - For “succeeds by the end of the fourth attempt,” consider the complement event shown by the all-failure path.

Solution

1. First success on the fourth attempt follows the path \(F_1,F_2,F_3,S_4\). Its probability is \(0.90\cdot0.85\cdot0.75\cdot0.40=0.2295\). 2. Failure on all four attempts has probability \(0.90\cdot0.85\cdot0.75\cdot0.60=0.34425\). 3. Therefore, the probability of succeeding by the end of the fourth attempt is \(1-0.34425=0.65575\). 4. The first-attempt success probability is \(0.10\). Since \(0.2295>0.10\), first success on the fourth attempt is more likely than success on the first attempt.

Answer

a) \(0.2295\), or \(22.95\%\). b) \(0.65575\), or \(65.575\%\). c) First success on the fourth attempt is more likely: \(0.2295>0.10\).

Multiplication rule and tree diagrams 83

The tree shows three independent shots at a target. \(H\) means a hit and \(M\) means a miss. a) How many terminal nodes does this complete three-shot tree have? b) Find the probability of exactly two hits. c) Find the probability of at least one hit.

Figure for problem 537464

Hints

- Follow each branch through all three shots, even if a hit occurs early. - For exactly two hits, identify every possible position of the one miss. - For at least one hit, is there a single complementary path that is easier to calculate?

Solution

1. Each shot has two possible outcomes, so the complete tree has \(2^3=8\) terminal nodes. 2. Exactly two hits occur on the paths \(HHM\), \(HMH\), and \(MHH\). Each path has probability \(0.8\cdot0.8\cdot0.2=0.128\), giving \(3\cdot0.128=0.384\). 3. The complement of at least one hit is three misses. Therefore, the probability is \(1-0.2^3=0.992\).

Answer

a) \(8\) terminal nodes b) \(0.384\), or \(38.4\%\) c) \(0.992\), or \(99.2\%\)

Multiplication rule and tree diagrams 84

At Mesa Vista High School, students either participate in the lunch program \((M)\) or do not \((M^c)\). The school also records whether they ride a bicycle to school \((B)\) or not \((B^c)\). The tree diagram gives the relevant probabilities. a) Describe in words what the probability \(0.20\) means. b) Find \(P(M\cap B)\) and \(P(M^c\cap B)\). c) Find \(P(B)\) and \(P(B^c)\). d) Find \(P(B^c\mid M)\) and \(P(M\cap B^c)\). e) Write the notation and value for the probability that a student rides a bicycle, given that the student does not participate in the lunch program.

Figure for problem 537466

Hints

- Identify what each stage of the tree represents. - Multiply along a path and add different paths that end in the same event. - Distinguish a conditional probability on one branch from a joint probability for a complete path. - In \(P(B\mid A)\), the event after the vertical bar is the condition.

Solution

1. The value \(0.20\) is \(P(B\mid M)\). It means that \(20\%\) of students who participate in the lunch program ride a bicycle to school. 2. Multiply along each path: \(P(M\cap B)=0.30\cdot0.20=0.06\) and \(P(M^c\cap B)=0.70\cdot0.40=0.28\). 3. Add the paths ending at \(B\): \(P(B)=0.06+0.28=0.34\). Therefore, \(P(B^c)=1-0.34=0.66\). 4. The diagram gives \(P(B^c\mid M)=0.80\). Thus, \(P(M\cap B^c)=0.30\cdot0.80=0.24\). 5. The requested conditional probability is \(P(B\mid M^c)=0.40\).

Answer

a) Among students who participate in the lunch program, \(20\%\) ride a bicycle to school. b) \(P(M\cap B)=0.06\); \(P(M^c\cap B)=0.28\). c) \(P(B)=0.34\); \(P(B^c)=0.66\). d) \(P(B^c\mid M)=0.80\); \(P(M\cap B^c)=0.24\). e) \(P(B\mid M^c)=0.40\).

Multiplication rule and tree diagrams 85

A candy company packages boxes either by hand \((H)\) or by machine \((H^c)\). Some boxes have a defect \((D)\). The probabilities are shown in the tree diagram. a) Find the overall probability that a randomly selected box has a defect. b) Find the probability that a box was packaged by hand and has no defect \((D^c)\). c) Find the overall probability of no defect by adding the two complete tree paths that end in \(D^c\). Verify the result using a complement.

Figure for problem 537467

Hints

- Add the probabilities of all complete paths that end at the event of interest. - Multiply probabilities along one path to find an intersection. - For part c, identify both paths ending in no defect, then compare their sum with the complement of part a.

Solution

1. Add the two paths that end in \(D\): \(P(D)=0.20\cdot0.05+0.80\cdot0.10=0.01+0.08=0.09\). 2. The hand-packaged, nondefective path has probability \(P(H\cap D^c)=0.20\cdot0.95=0.19\). 3. The machine-packaged, nondefective path has probability \(P(H^c\cap D^c)=0.80\cdot0.90=0.72\). 4. Add the two disjoint no-defect paths: \(P(D^c)=0.19+0.72=0.91\). 5. This agrees with the complement calculation \(1-P(D)=1-0.09=0.91\).

Answer

a) \(P(D)=0.09\) b) \(P(H\cap D^c)=0.19\) c) \(P(D^c)=0.91\), both by adding the two no-defect paths and by \(1-P(D)\).

Multiplication rule and tree diagrams 86

A three-stage experiment is shown in the tree. At each stage the outcome is success \(S\) or failure \(F\); the chance of success can depend on earlier results. Find the probability of exactly two successes.

Figure for problem 537501

Hints

- List the complete paths containing exactly two \(S\) outcomes. - Follow the conditional probabilities on each selected path instead of assuming every success has the same chance. - How do you combine probabilities from different, nonoverlapping paths?

Solution

1. Exactly two successes occur on three paths: \(SSF\), \(SFS\), and \(FSS\). 2. Multiply along each path: \(P(SSF)=0.60\cdot0.70\cdot0.20=0.084\), \(P(SFS)=0.60\cdot0.30\cdot0.40=0.072\), and \(P(FSS)=0.40\cdot0.50\cdot0.60=0.120\). 3. Add the disjoint path probabilities: \(0.084+0.072+0.120=0.276\).

Answer

\(0.276=27.6\%\)

Multiplication rule and tree diagrams 87

Lio sometimes takes the bus to school. a) Find the three missing branch probabilities in the tree. b) What is the probability that Lio takes the bus on a randomly selected day?

Figure for problem 537503

Hints

- What must the probabilities leaving a single branching point total? - Which two different routes through the tree lead to Bus? - How could you find the chance of each route before combining them?

Solution

1. The rain probability is \(1-0.7=0.3\). The no-bus probabilities are \(1-0.2=0.8\) on dry days and \(1-0.6=0.4\) on rainy days. 2. The dry-and-bus path has probability \(0.7\cdot0.2=0.14\), and the rain-and-bus path has probability \(0.3\cdot0.6=0.18\). 3. Add those two paths: \(0.14+0.18=0.32\).

Answer

a) Rain: \(0.3\); No bus after Dry: \(0.8\); No bus after Rain: \(0.4\). b) \(0.32=32\%\).

Multiplication rule and tree diagrams 88

The partial tree shows probabilities for two draws without replacement from a bag. Use the first-draw and second-draw probabilities together to find the original numbers of red and blue balls.

Figure for problem 537519

Hints

- The first red probability describes a ratio, not necessarily the exact number of balls. - How could you express the red count and total count using one positive whole-number multiplier? - After a red ball is removed, what happens to both counts? Use the second red probability too.

Solution

1. The first-draw probability of red is \(\frac{5}{8}\). This means the initial red count \(r\) and total count \(n\) have the form \(r=5k\), \(n=8k\) for some positive integer \(k\); it does not, by itself, show that there are exactly \(8\) balls. 2. After one red ball is removed, the red probability is \(\frac{5k-1}{8k-1}\). The tree gives \(\frac{5k-1}{8k-1}=\frac{4}{7}\). 3. Cross-multiply: \(7(5k-1)=4(8k-1)\), so \(35k-7=32k-4\), \(3k=3\), and \(k=1\). 4. Therefore, the bag originally had \(5\) red balls and \(8-5=3\) blue balls. The second-draw probability checks: \(\frac{4}{7}\).

Answer

\(5\) red balls and \(3\) blue balls

Multiplication rule and tree diagrams 89

A bag contains \(3\) red balls and \(2\) blue balls. Two balls are drawn without replacement. Mia says, “Red then blue is more likely than blue then red because drawing red first is more likely.” Is she right? Use the probabilities of both draws to explain.

Hints

- What is the composition of the bag after a red ball is removed? After a blue ball is removed? - Compare the products for the two full draw sequences, not just the first branches. - How can a smaller chance on the first draw be offset on the second?

Solution

1. Red then blue has probability \(\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{10}\). 2. Blue then red has probability \(\frac{2}{5}\cdot\frac{3}{4}=\frac{3}{10}\). 3. The orders are equally likely. Mia looks only at the first draw; she misses that the second conditional probability changes with the first color.

Answer

Mia is incorrect. Both orders have probability \(\frac{3}{10}\) because the second-draw probabilities differ in opposite ways.

Multiplication rule and tree diagrams 90

A bag contains two red balls and one blue ball. A ball is drawn, replaced, and then the process is repeated until three draws have been made. Find the probability of drawing red exactly twice.

Figure for problem 537524

Hints

- List the sequences containing exactly two red results. - Replacement keeps the probabilities the same on every draw. - Multiply along each favorable path and add the path probabilities.

Solution

1. The favorable color sequences are RRB, RBR, and BRR. 2. Each sequence has probability \(\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}=\frac{4}{27}\). 3. Therefore, \(P(\text{exactly two red})=3\cdot\frac{4}{27}=\frac{4}{9}\).

Answer

\(\frac{4}{9}\approx 44.4\%\)

Multiplication rule and tree diagrams 91

The tree shows the chances of a delay or an on-time arrival on each of three independent legs of a trip. Find the probability that exactly one leg is delayed.

Figure for problem 537528

Hints

- Locate the three complete paths with exactly one delay. - The probability of a delay is different on each leg. Read the appropriate edge at every stage. - Combine the probabilities of these disjoint paths.

Solution

1. Delay only on leg 1: \(0.10\cdot0.80\cdot0.85=0.068\). 2. Delay only on leg 2: \(0.90\cdot0.20\cdot0.85=0.153\). 3. Delay only on leg 3: \(0.90\cdot0.80\cdot0.15=0.108\). 4. Add the three disjoint paths: \(0.068+0.153+0.108=0.329\).

Answer

\(0.329=32.9\%\)

Multiplication rule and tree diagrams 92

A die with red, blue, and green faces is rolled twice. The tree shows the chances of each color on each roll. Find the probability that the rolls show different colors.

Figure for problem 537534

Hints

- Would you rather list all different-color paths or the matching-color paths? - Read the chance of each color from the branches on both stages. - The matching-color cases are disjoint. How does their total help find the answer?

Solution

1. The complement is the same color on both rolls. 2. Add the matching-color paths: \(P(\text{same})=\left(\frac{3}{6}\right)^2+\left(\frac{2}{6}\right)^2+\left(\frac{1}{6}\right)^2=\frac{14}{36}\). 3. Subtract from \(1\): \(P(\text{different})=1-\frac{14}{36}=\frac{11}{18}\).

Answer

\(\frac{11}{18}\approx61.1\%\)

Multiplication rule and tree diagrams 93

The partial tree shows a two-stage experiment. The overall probability of event \(B\) is \(0.52\). Find \(P(B\mid A^c)\) and the other missing branch probability after \(A^c\).

Figure for problem 537538

Hints

- Which two complete paths end in event \(B\)? - Find the contribution from the known \(A,B\) path, then use the overall chance of \(B\). - The two probabilities leaving \(A^c\) must add to \(1\).

Solution

1. The \(A,B\) path has probability \(0.4\cdot0.7=0.28\). 2. The remaining part of \(B\) is \(P(A^c\cap B)=0.52-0.28=0.24\). 3. Since \(P(A^c)=0.6\), \(P(B\mid A^c)=\frac{0.24}{0.6}=0.4\). 4. The other outgoing branch after \(A^c\) has probability \(1-0.4=0.6\).

Answer

\(P(B\mid A^c)=0.4\); \(P(B^c\mid A^c)=0.6\).

Multiplication rule and tree diagrams 94

A team plays a best-of-three series. The series ends when either team has won two games. The tree shows the possible results and their probabilities. Here \(W\) means the team wins a game, and \(L\) means it loses. Find the probability that this team wins the series.

Figure for problem 537543

Hints

- Some branches end after two games because a team already has two wins. - Which terminal paths end in a series win? - Multiply along each winning path, then add the different path probabilities.

Solution

1. The team wins in two games along \(WW\), with probability \(0.6\cdot0.6=0.36\). 2. It can win in three games along \(WLW\), with probability \(0.6\cdot0.4\cdot0.6=0.144\). 3. It can also win along \(LWW\), with probability \(0.4\cdot0.6\cdot0.6=0.144\). 4. Add the mutually exclusive winning paths: \(0.36+0.144+0.144=0.648\).

Answer

\(0.648=64.8\%\)

Multiplication rule and tree diagrams 95

A machine inspects at most three parts and stops when it finds the first defective part. Each part is defective with probability \(0.10\), independently of the others. Find the probability that: a) Exactly two parts are inspected. b) The third inspection position is reached.

Figure for problem 537544

Hints

- Determine the exact sequence that makes the process stop after the second inspection. - To reach the third inspection, only the first two results matter.

Solution

1. Exactly two parts are inspected when the first part is acceptable and the second is defective. The probability is \(0.90\cdot 0.10=0.09\). 2. The third inspection position is reached when the first two parts are acceptable. The third part's condition is not yet relevant, so the probability is \((0.90)^2=0.81\).

Answer

a) \(0.09\) b) \(0.81\)

Multiplication rule and tree diagrams 96

The tree shows the chances that each of three consecutive traffic lights is green \(G\) or red \(R\). Assume the lights act independently. Find the probability that exactly two lights are green.

Figure for problem 537548

Hints

- Exactly one light must be red. Which three complete paths satisfy that condition? - The green probabilities change by stage. Read each branch carefully. - Add the probabilities after multiplying along those three paths.

Solution

1. Green, green, red: \(0.7\cdot0.6\cdot0.2=0.084\). 2. Green, red, green: \(0.7\cdot0.4\cdot0.8=0.224\). 3. Red, green, green: \(0.3\cdot0.6\cdot0.8=0.144\). 4. Add the three paths: \(0.084+0.224+0.144=0.452\).

Answer

\(0.452=45.2\%\)

Multiplication rule and tree diagrams 97

A fair die is rolled. If the result is \(1\), \(2\), \(3\), or \(4\), a fair coin is flipped, and heads counts as a success. If the result is \(5\) or \(6\), the die is rolled again, and only a \(6\) counts as a success. Find the probability of success.

Figure for problem 537553

Hints

- The procedure branches into two different second-stage experiments. - Find the success probability on each branch separately, then add them.

Solution

1. The probability of entering the coin branch and then getting heads is \(\frac{4}{6}\cdot\frac{1}{2}=\frac{1}{3}\). 2. The probability of entering the second-die branch and then rolling a \(6\) is \(\frac{2}{6}\cdot\frac{1}{6}=\frac{1}{18}\). 3. The two success paths are disjoint, so the total probability is \(\frac{1}{3}+\frac{1}{18}=\frac{7}{18}\).

Answer

\(\frac{7}{18}\approx 38.9\%\)

Multiplication rule and tree diagrams 98

The tree shows a search that ends when an attempt succeeds or after three attempts, whichever comes first. Find the probability that the search ends: a) After the first attempt. b) After the second attempt. c) After the third attempt.

Figure for problem 537554

Hints

- Which outcomes cause the search to stop early? - To reach the last attempt, what must happen on the first two attempts? - Does the final result matter when counting the probability of ending after three attempts?

Solution

1. The search ends after attempt one when the first attempt succeeds, with probability \(0.25\). 2. It ends after attempt two if the first attempt fails and the second succeeds: \(0.75\cdot0.25=0.1875\). 3. It ends after attempt three whenever the first two attempts fail, regardless of the third result: \(0.75^2=0.5625\). 4. The probabilities total \(0.25+0.1875+0.5625=1\).

Answer

a) \(0.25\) b) \(0.1875\) c) \(0.5625\)

Multiplication rule and tree diagrams 99

A tree diagram represents a survey of \(1000\) people. Find every branch probability when \(A\) is the first stage of the tree.

Figure for problem 537563

Hints

- Read the four terminal frequencies from the tree. - First total the two terminal frequencies under each first-stage event. - At each split, divide a child frequency by its parent frequency.

Solution

1. From the terminal frequencies in the tree, the first-stage frequencies are \(240+60=300\) for \(A\) and \(140+560=700\) for \(A^c\). 2. Therefore, \(P(A)=\frac{300}{1000}=0.3\) and \(P(A^c)=0.7\). 3. After \(A\), \(P(B\mid A)=\frac{240}{300}=0.8\) and \(P(B^c\mid A)=0.2\). 4. After \(A^c\), \(P(B\mid A^c)=\frac{140}{700}=0.2\) and \(P(B^c\mid A^c)=0.8\).

Answer

\(P(A)=0.3\), \(P(A^c)=0.7\); after \(A\): \(0.8\), \(0.2\); after \(A^c\): \(0.2\), \(0.8\).

Multiplication rule and tree diagrams 100

In a production process, \(5\%\) of components are defective. Six components are tested independently. Analyze the error in each calculation and find the correct probability. a) “The second component tested is defective.” Incorrect calculation: \(0.95\cdot0.05\cdot(0.95)^4\) b) “At least one of the six components is defective.” Incorrect calculation: \(6\cdot0.05\cdot(0.95)^5\)

Hints

- Distinguish a condition on one trial from a condition that specifies an entire ordered outcome. - For “at least one,” consider the complementary event.

Solution

1. In part a, only the second component is specified. The other five components may be either defective or acceptable. The incorrect calculation instead describes the event that exactly the second component is defective. The correct probability is \(0.05\). 2. In part b, “at least one” includes one through six defective components. The incorrect expression calculates only the probability of exactly one defective component. Use the complement: \(P(\text{at least one defective})=1-(0.95)^6\approx 0.2649\).

Answer

a) The calculation incorrectly requires all other components to be acceptable. The correct probability is \(0.05\), or \(5\%\). b) The calculation includes only exactly one defective component. The correct probability is \(1-(0.95)^6\approx 0.2649\), or about \(26.49\%\).

Multiplication rule and tree diagrams 101

At a carnival game, a player spins each of the three equal-section spinners shown once. Find the probability of each event. a) All three spinners show the digit \(1\). b) Exactly two of the three spinners land on blue.

Figure for problem 535927

Hints

- Read each spinner's digit and color probabilities from its equal sections. - Multiply probabilities along each favorable path. - For part b, list all paths containing exactly two blue results, then add their probabilities.

Solution

1. From Spinner 1, \(P(1)=\frac{1}{4}\), \(P(\text{blue})=\frac{3}{4}\), and \(P(\text{red})=\frac{1}{4}\). 2. From Spinner 2, \(P(1)=\frac{1}{5}\), \(P(\text{blue})=\frac{3}{5}\), and \(P(\text{red})=\frac{2}{5}\). 3. From Spinner 3, \(P(1)=\frac{1}{6}\), \(P(\text{blue})=\frac{1}{2}\), and \(P(\text{red})=\frac{1}{2}\). 4. For part a, multiply along the single favorable path: \(P(1,1,1)=\frac{1}{4}\cdot\frac{1}{5}\cdot\frac{1}{6}=\frac{1}{120}\). 5. For part b, exactly two blue results can occur as blue-blue-red, blue-red-blue, or red-blue-blue. Therefore, \(P(\text{exactly two blue})=\frac{3}{4}\cdot\frac{3}{5}\cdot\frac{1}{2}+\frac{3}{4}\cdot\frac{2}{5}\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{3}{5}\cdot\frac{1}{2}=\frac{9}{20}\).

Answer

a) \(\frac{1}{120}\approx0.83\%\) b) \(\frac{9}{20}=45\%\)

Multiplication rule and tree diagrams 102

A table tennis championship match uses a best-of-five format: the first player to win three games wins the match. The tree diagram shows the probability that Player A or Player B wins each game; the game-win probabilities change from one game to the next. Scores in the tree are written as Player A–Player B. Identify the six tree paths that reach a \(2\)–\(2\) score after four games, and use their branch probabilities to find the probability that the match lasts all five games.

Figure for problem 537461

Hints

- A fifth game occurs only if the score after four games is \(2\)–\(2\). - The branch probabilities change by game, so the six qualifying paths do not all have the same probability. - Multiply along each qualifying path, then add the six disjoint path probabilities.

Solution

1. A fifth game is needed exactly when each player has won two of the first four games. 2. The six qualifying winner sequences and their path probabilities are: \(AABB: 0.60\cdot0.50\cdot0.30\cdot0.60=0.054\), \(ABAB: 0.60\cdot0.50\cdot0.70\cdot0.60=0.126\), \(ABBA: 0.60\cdot0.50\cdot0.30\cdot0.40=0.036\), \(BAAB: 0.40\cdot0.50\cdot0.70\cdot0.60=0.084\), \(BABA: 0.40\cdot0.50\cdot0.30\cdot0.40=0.024\), and \(BBAA: 0.40\cdot0.50\cdot0.70\cdot0.40=0.056\). 3. These paths are disjoint, so add their probabilities: \(0.054+0.126+0.036+0.084+0.024+0.056=0.380\).

Answer

The Game-5 paths are \(AABB\), \(ABAB\), \(ABBA\), \(BAAB\), \(BABA\), and \(BBAA\). Their total probability is \(0.38\), or \(38\%\).

Multiplication rule and tree diagrams 103

Suppose \(P(A)=0.8\), \(P(B)=0.7\), and \(P(A\cap B)=0.4\). Use a tree diagram to show that these three values cannot all be valid.

Figure for problem 537570

Hints

- First find the conditional probability on the \(A\)-then-\(B\) branch. - Determine how much of event \(B\) would have to occur outside \(A\). - Check whether the resulting branch probability is between \(0\) and \(1\).

Solution

1. From the \(A\) branch, \(P(B\mid A)=\frac{0.4}{0.8}=0.5\). 2. The portion of \(B\) outside \(A\) would have to be \(P(A^c\cap B)=0.7-0.4=0.3\). 3. Since \(P(A^c)=0.2\), the required branch probability would be \(P(B\mid A^c)=\frac{0.3}{0.2}=1.5\). 4. A probability cannot exceed \(1\), so the original values are inconsistent.

Answer

The values are impossible because they require \(P(B\mid A^c)=1.5\).

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