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Multiplication rule and tree diagrams

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Difficulty: 1 – 5

53750010

A student labels the two branches from one node of a spinner tree diagram with probabilities \(0.6\) and \(0.5\). Explain the error and give the probability that should be paired with \(0.6\).

Figure for problem 537500

Hints

- Add the probabilities leaving the same node. - The total probability distributed from one node must be \(1\).

Solution

1. The two branch probabilities add to \(0.6 + 0.5 = 1.1\). 2. Probabilities on all branches from one node must add to \(1\), so the diagram is invalid. 3. The correct complementary probability is \(1 - 0.6 = 0.4\).

Answer

The diagram is invalid; \(0.5\) should be replaced by \(0.4\).

51007110

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

51009510

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

51362110

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) \(\frac{2}{5} = 40\%\)

51363310

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

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

51363910

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) \(\frac{61}{125} = 48.8\%\)

51364210

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) \(0.128 = 12.8\%\)

51364510

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) \(\frac{13}{20} = 65\%\)

51516910

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

\(32.768\%\)

52382010

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 \(1\).

53749510

A bag contains \(3\) red balls and \(2\) blue balls. Two draws are made with replacement. Find the probability that both balls have the same color.

Figure for problem 537495

Hints

- Identify the complete paths that produce the same color twice. - Because the ball is replaced, the composition of the bag does not change.

Solution

1. The probability of the red-red path is \(\frac{3}{5} \cdot \frac{3}{5} = \frac{9}{25}\). 2. The probability of the blue-blue path is \(\frac{2}{5} \cdot \frac{2}{5} = \frac{4}{25}\). 3. Adding the favorable paths gives \(P(\text{same color}) = \frac{9}{25} + \frac{4}{25} = \frac{13}{25}\).

Answer

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

53749610

A bag contains \(3\) red balls and \(2\) blue balls. Two balls are drawn without replacement. What is the probability that the balls have different colors?

Figure for problem 537496

Hints

- Include both possible color orders. - Update the contents of the bag after the first draw.

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. Therefore, \(P(\text{different colors}) = \frac{3}{10} + \frac{3}{10} = \frac{3}{5}\).

Answer

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

53749810

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

53749910

A ball is drawn twice with replacement. Let \(R\) mean “a red ball is drawn,” and suppose \(P(R) = 0.3\). Describe the complement of “red at least once” and find its probability.

Figure for problem 537499

Hints

- Which outcome sequences are excluded when at least one red result is required? - Can you describe the complement with a single path in the tree diagram?

Solution

1. The complement of “red at least once” is “no red on either draw.” 2. The probability of not drawing red on one draw is \(1 - 0.3 = 0.7\). 3. Because the draws are independent, the probability of no red on both draws is \(0.7 \cdot 0.7 = 0.49\).

Answer

Complement: No red ball is drawn on either draw. The probability is \(0.49\).

53750110

A fair coin is tossed three times. Use the tree diagram to find the probability of getting exactly two heads.

Figure for problem 537501

Hints

- Mark every complete path containing exactly two heads. - Determine whether those complete paths have the same probability.

Solution

1. The favorable outcomes are \(HHT\), \(HTH\), and \(THH\). 2. Each complete path has probability \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\). 3. Therefore, the probability of exactly two heads is \(3 \cdot \frac{1}{8} = \frac{3}{8}\).

Answer

\(\frac{3}{8} = 37.5\%\)

53750310

On a given day, the probability of dry weather is \(0.7\). If the weather is dry, Lio takes the bus with probability \(0.2\). If it rains, Lio takes the bus with probability \(0.6\). Find the probability that Lio takes the bus.

Figure for problem 537503

Hints

- Identify every complete path that ends with taking the bus. - Include both possible weather conditions.

Solution

1. The probability of the dry-and-bus path is \(0.7 \cdot 0.2 = 0.14\). 2. The probability of the rain-and-bus path is \(0.3 \cdot 0.6 = 0.18\). 3. Adding the two bus paths gives \(0.14 + 0.18 = 0.32\).

Answer

\(0.32 = 32\%\)

53750510

A fair spinner labeled with the numbers \(1\), \(2\), \(3\), and \(4\) is spun twice. Find the probability that the product of the two numbers is even.

Figure for problem 537505

Hints

- When is a product of two whole numbers odd? - Is the complement of the desired event easier to calculate?

Solution

1. Use the complement. A product is odd only when both factors are odd. 2. On each spin, \(2\) of the \(4\) outcomes are odd, so the probability of two odd results is \(\frac{2}{4} \cdot \frac{2}{4} = \frac{1}{4}\). 3. Therefore, the probability of an even product is \(1 - \frac{1}{4} = \frac{3}{4}\).

Answer

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

53751410

In the tree diagram, the probability of \(B\) after event \(A\) is \(0.65\). Complete the missing branch and find \(P(A \cap B^c)\) when \(P(A) = 0.4\).

Figure for problem 537514

Hints

- Complete the two branches after \(A\) so they add to \(1\). - Multiply along the single path for \(A \cap B^c\).

Solution

1. The missing probability after \(A\) is \(1 - 0.65 = 0.35\). 2. The requested path has probability \(P(A \cap B^c) = 0.4 \cdot 0.35 = 0.14\).

Answer

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

53751510

A fair six-sided die is rolled twice. Find the probability that at least one roll is a \(6\).

Figure for problem 537515

Hints

- The desired event includes several paths in the tree diagram. - Can you describe its complement with one path?

Solution

1. Use the complement: neither roll is a \(6\). 2. The probability of not rolling a \(6\) on one roll is \(\frac{5}{6}\), so the probability of no \(6\) on both rolls is \(\frac{5}{6} \cdot \frac{5}{6} = \frac{25}{36}\). 3. Therefore, the probability of at least one \(6\) is \(1 - \frac{25}{36} = \frac{11}{36}\).

Answer

\(\frac{11}{36} \approx 30.6\%\)

53751610

A soccer player scores a penalty kick with probability \(0.8\). She takes two penalty kicks under the same independent conditions. What is the probability of scoring exactly once?

Figure for problem 537516

Hints

- List the two orders in which exactly one goal can occur. - Determine whether the two paths have the same probability.

Solution

1. The probability of the goal-miss path is \(0.8 \cdot 0.2 = 0.16\). 2. The probability of the miss-goal path is \(0.2 \cdot 0.8 = 0.16\). 3. Adding the two favorable paths gives \(0.16 + 0.16 = 0.32\).

Answer

\(0.32 = 32\%\)

53751710

A raffle drum contains \(3\) winning tickets and \(7\) losing tickets. Two tickets are drawn without replacement. Find the probability of drawing at least one winning ticket.

Figure for problem 537517

Hints

- Use the complement of the event. - Remember that the first ticket is not replaced.

Solution

1. Use the complement, drawing two losing tickets. Its probability is \(\frac{7}{10} \cdot \frac{6}{9} = \frac{7}{15}\). 2. Therefore, the probability of at least one winning ticket is \(1 - \frac{7}{15} = \frac{8}{15}\).

Answer

\(\frac{8}{15} \approx 53.3\%\)

53751910

The tree diagram represents two draws without replacement from a bag. Reconstruct the bag’s original composition. How many red balls and blue balls were in the bag?

Figure for problem 537519

Hints

- Use the first fraction’s numerator and denominator to identify the original red count and total count. - Check that the second-stage probability agrees after one red ball is removed.

Solution

1. The first red probability is \(\frac{5}{8}\), so the bag originally had \(8\) balls, \(5\) of them red. 2. After a red ball is drawn, \(4\) red balls remain among \(7\) total balls, which matches the second probability \(\frac{4}{7}\). 3. Therefore, the original bag contained \(5\) red balls and \(3\) blue balls.

Answer

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

53753710

Two production lines make equal numbers of parts. Line A has a defective-part rate of \(0.03\), and line B has a defective-part rate of \(0.01\). One part is selected at random from the combined production. Find the probability that the selected part is defective.

Figure for problem 537537

Hints

- A defective part can come from either production line. - Use the equal production amounts to find the probability of selecting each line. - Multiply along each path and add the two path probabilities.

Solution

1. Because the lines make equal numbers of parts, \(P(A)=P(B)=0.5\). 2. The probability of selecting a defective part from line A is \(0.5\cdot 0.03=0.015\). 3. The probability of selecting a defective part from line B is \(0.5\cdot 0.01=0.005\). 4. Therefore, \(P(\text{defective})=0.015+0.005=0.02\).

Answer

\(0.02=2\%\)

53755810

In the tree diagram, the branch probabilities along the path A then B are \(0.4\) and \(0.7\). A student adds them and claims that the path probability is \(0.4+0.7=1.1\). Explain the error and find the correct probability.

Figure for problem 537558

Hints

- Decide whether the two branches represent sequential conditions or alternative cases. - Check whether the student’s result is even within the possible range for a probability.

Solution

1. The two branches occur in sequence along the same path, so their probabilities must be multiplied, not added. 2. The correct path probability is \(P(A\cap B)=0.4\cdot 0.7=0.28\). 3. The value \(1.1\) also cannot be a probability because probabilities cannot exceed \(1\).

Answer

The student should multiply the branch probabilities. \(P(A\cap B)=0.28\).

51361910

A multiple-choice test has \(6\) questions. Each question has \(3\) answer choices, exactly one of which is correct. A student guesses randomly on every question. a) Find the probability that the student answers every question incorrectly. b) Find the probability that the student answers at least one question correctly. c) A second student can eliminate one incorrect choice on each question and then guesses between the two remaining choices. Find the probability that this student 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. a) The probability of an incorrect answer on one question is \(\frac{2}{3}\). Because the questions are independent, \(P(\text{all incorrect}) = \left(\frac{2}{3}\right)^6 = \frac{64}{729} \approx 0.0878\). 2. b) Use the complement: \(P(\text{at least one correct}) = 1 - \frac{64}{729} = \frac{665}{729} \approx 0.9122\). 3. c) 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, \(P(\text{all correct}) = \left(\frac{1}{3}\right)^6 = \frac{1}{729} \approx 0.00137\). Eliminating one choice makes a perfect score about \(\frac{729}{64} \approx 11.4\) times as likely.

Answer

a) \(\frac{64}{729} \approx 0.0878\) b) \(\frac{665}{729} \approx 0.9122\) c) \(\frac{1}{64} = 0.015625\), compared with \(\frac{1}{729} \approx 0.00137\) for completely random guessing

51362010

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 \(\frac{1}{216}\).

51362210

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) \(\frac{11}{32} = 0.34375\)

51362510

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) \(\frac{4}{7}\)

51363510

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

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

51364110

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) \(0.3439 = 34.39\%\)

51364310

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) \(0.2916 = 29.16\%\)

51364410

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.

51364710

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) \(\frac{17}{45} \approx 37.78\%\)

51366110

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.

51517110

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.

51517610

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 \(16\%\)

52184810

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) \(0.936\)

52196810

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 \(30.77\%\).

52312310

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) \(\frac{2}{3}\approx 66.7\%\)

52312410

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 \(24.3\%\)

52335110

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) \(P(A_1)\approx 0.0429\); \(P(B)\approx 0.1715\); \(P(C)\approx 0.8574\)

52337210

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

52380310

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) \(\frac{98}{125}=0.784\)

52380910

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 of the four bulbs is defective.

Hints

- Find the probability that one bulb works. - Multiply probabilities along the specified path in part b. - The complement of at least one defective bulb is that all bulbs work.

Solution

1. The probability that a bulb is defective is \(0.05\), and the probability that it works is \(0.95\). 2. \(P(\text{all work})=(0.95)^4\approx 0.8145\). 3. \(P(\text{only first defective})=0.05\cdot (0.95)^3\approx 0.0429\). 4. Use the complement of all four working: \(P(\text{at least one defective})=1-(0.95)^4\approx 0.1855\).

Answer

a) About \(0.8145\), or \(81.45\%\) b) About \(0.0429\), or \(4.29\%\) c) About \(0.1855\), or \(18.55\%\)

52381010

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 \(44.1\%\)

52381910

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. Statement a is true. A path probability is the 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. 2. Statement b is true for a complete tree diagram. The complete paths are mutually exclusive and together represent every possible outcome, so their probabilities add to \(1\).

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 \(1\).

53088110

A company tests electronic components for defects. For a component that is actually defective, an automated test gives a positive result with probability \(92\%\). Two independent automated tests are performed. - If both tests are positive, the component is immediately rejected. - If exactly one test is positive, an expert performs a manual inspection. - If both tests are negative, the component is classified as acceptable. For a defective component, the manual inspection detects the defect with probability \(96\%\). For a component that is actually defective, find the probability that: (1) it is immediately rejected; (2) neither automated test detects the defect; (3) a manual inspection is required; (4) the defect is first detected during the manual inspection; (5) the component is ultimately identified as defective.

Hints

- Represent the process as a multistage experiment. - Multiply probabilities along a path and add probabilities for different paths leading to the same outcome. - “Exactly one positive” includes two distinct test orders. - For the final part, combine immediate rejection with successful manual detection.

Solution

1. Let \(p=0.92\) be the probability of a positive automated result and \(q=0.08\) the probability of a negative result. 2. Both tests are positive with probability \(p^2=0.92^2=0.8464\). 3. Both tests are negative with probability \(q^2=0.08^2=0.0064\). 4. Exactly one test is positive with probability \(2pq=2\cdot 0.92\cdot 0.08=0.1472\). 5. The defect is first detected manually with probability \(0.1472\cdot 0.96=0.141312\). 6. The component is ultimately identified as defective with probability \(0.8464+0.141312=0.987712\).

Answer

(1) \(0.8464\), or \(84.64\%\). (2) \(0.0064\), or \(0.64\%\). (3) \(0.1472\), or \(14.72\%\). (4) \(0.141312\), or about \(14.13\%\). (5) \(0.987712\), or about \(98.77\%\).

53088210

A security system uses two identical sensors that independently detect an intruder with probability \(0.88\) each. - If both sensors respond, the system triggers a main alarm immediately. - If exactly one sensor responds, a security center performs a video review. - If neither sensor responds, the system remains silent. The video review confirms an intrusion with probability \(0.95\). Given that an intrusion has occurred, find the probability that: 1) The main alarm is triggered immediately. 2) The system gives no response. 3) A video review is initiated. 4) The intrusion is confirmed only through video review. 5) The intrusion is detected either immediately or through video review.

Hints

- Write the probabilities that a sensor responds and does not respond. - Account for both orders in which exactly one sensor responds. - Multiply along the video-review path. - Add the disjoint paths that lead to successful detection.

Solution

1. Both sensors respond with probability \((0.88)^2=0.7744\). 2. Neither sensor responds with probability \((0.12)^2=0.0144\). 3. Exactly one sensor responds with probability \(2\cdot 0.88\cdot 0.12=0.2112\). 4. Confirmation only through video has probability \(0.2112\cdot 0.95=0.20064\). 5. The total detection probability is \(0.7744+0.20064=0.97504\).

Answer

1) \(0.7744\), or \(77.44\%\) 2) \(0.0144\), or \(1.44\%\) 3) \(0.2112\), or \(21.12\%\) 4) \(0.20064\), or about \(20.06\%\) 5) \(0.97504\), or about \(97.50\%\)

53093310

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 \(94\%\)

53093410

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.

53093510

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 \(P(B\mid N)=\frac{4}{7}\).

53200410

A spinner has \(8\) equal sections: \(3\) red, \(3\) blue, and \(2\) green. 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

- Find the probability of each color on one spin. - Multiply probabilities along a single two-spin path. - Add the probabilities of the red-red, blue-blue, and green-green paths.

Solution

1. The one-spin probabilities are \(P(\text{red})=\frac{3}{8}\), \(P(\text{blue})=\frac{3}{8}\), 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) \(P(E_2)=\frac{11}{32}=34.375\%\)

53204110

A fair spinner has \(8\) equal sections: \(5\) green sections marked G, \(2\) yellow sections marked Y, and \(1\) red section marked R. The spinner is spun twice. Find the probability of event \(E\): “The spinner lands on green at least once.” First describe the complement \(\overline{E}\) 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? - How many sections are not green? - Find the probability of a non-green result on one spin. - How do you find the probability of two independent results both occurring? - 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. There are \(3\) 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 \(P(\overline{E}) = \frac{3}{8} \cdot \frac{3}{8} = \frac{9}{64}\). 4. Using the complement rule, \(P(E) = 1 - P(\overline{E}) = 1 - \frac{9}{64} = \frac{55}{64}\).

Answer

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

53215310

The spinner shown has \(5\) equal sections: \(4\) blue sections marked B and \(1\) red section marked R. It is spun three times. Let \(A\) be the event “The spinner lands on red at least once.” a) Describe the complement \(\overline{A}\) in words. b) Find \(P(\overline{A})\) 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? - Describe the complement of “red at least once” in words. - What is the probability of blue on one spin? - How do you find the probability of blue on all three independent spins? - How are the probabilities of an event and its complement related?

Solution

1. a) 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. b) The probability of blue on one spin is \(\frac{4}{5}\). Because the spins are independent, \(P(\overline{A}) = \left(\frac{4}{5}\right)^3 = \frac{64}{125}\). 3. c) By the complement rule, \(P(A) = 1 - P(\overline{A}) = 1 - \frac{64}{125} = \frac{61}{125}\).

Answer

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

53591310

Spinner 1 and Spinner 2, each divided into equal sections, are spun once. On Spinner 1, the blue sections are numbered \(2, 4, 6, 8\). On Spinner 2, the blue sections are numbered \(1\) and \(4\). Find each probability and briefly explain your method. Write each result 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

- Count the total number of possible ordered spinner results. - Find the probability of the desired result on each spinner separately. - The two spins are independent, so multiply the two probabilities. - Remember that \(1\) is not prime. - You may also count favorable ordered pairs and divide by the total.

Solution

1. For part a, Spinner 1 has \(4\) blue sections out of \(8\), and Spinner 2 has \(2\) blue sections out of \(6\). Because the spins are independent, \(P(\text{both blue}) = \frac{4}{8} \cdot \frac{2}{6} = \frac{1}{6} \approx 16.7\%\). 2. For part b, the prime-numbered sections on Spinner 1 are \(2, 3, 5, 7\), giving probability \(\frac{4}{8}\). On Spinner 2, the prime-numbered sections are \(2, 3, 5\), giving probability \(\frac{3}{6}\). Therefore, \(P(\text{both prime}) = \frac{4}{8} \cdot \frac{3}{6} = \frac{1}{4} = 25.0\%\).

Answer

a) \(\frac{1}{6} \approx 16.7\%\) b) \(\frac{1}{4} = 25.0\%\)

53593510

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

53594910

A fair spinner has \(10\) equal sections marked by color initials: \(5\) orange sections marked O, \(3\) blue sections marked B, and \(2\) yellow sections marked Y. The spinner is spun twice. Find the probability of event \(E\): “The two spins show different colors.” Use the complement method: 1. Describe the complement \(\overline{E}\) in words. 2. Find \(P(\overline{E})\). 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? - Find the probability of each color on one spin. - How do you find the probability of one same-color pair, such as blue followed by blue? - After finding the probability of matching colors, 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. On one spin, \(P(\text{orange}) = \frac{5}{10} = 0.5\), \(P(\text{blue}) = \frac{3}{10} = 0.3\), and \(P(\text{yellow}) = \frac{2}{10} = 0.2\). 3. Because the spins are independent and the three same-color cases do not overlap, \(P(\overline{E}) = 0.5 \cdot 0.5 + 0.3 \cdot 0.3 + 0.2 \cdot 0.2 = 0.25 + 0.09 + 0.04 = 0.38\). 4. Therefore, \(P(E) = 1 - P(\overline{E}) = 1 - 0.38 = 0.62\), or \(62\%\).

Answer

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

53608710

A spinner has \(8\) equal sections: \(2\) blue, \(3\) yellow, and \(3\) green. The spinner 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

- Find the probability of each color on one spin. - 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. On one spin, \(P(\text{blue}) = \frac{2}{8} = \frac{1}{4}\), \(P(\text{yellow}) = \frac{3}{8}\), and \(P(\text{not yellow}) = P(\text{not green}) = \frac{5}{8}\). 2. For part a, \(P(\text{three blue}) = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \approx 1.56\%\). 3. For part b, the one yellow result can occur on any of the three spins. Therefore, \(P(\text{exactly one yellow}) = 3 \cdot \frac{3}{8} \cdot \left(\frac{5}{8}\right)^2 = \frac{225}{512} \approx 43.95\%\). 4. For part c, every spin must be not green. Thus, \(P(\text{no green}) = \left(\frac{5}{8}\right)^3 = \frac{125}{512} \approx 24.41\%\).

Answer

a) \(\frac{1}{64} \approx 1.56\%\) b) \(\frac{225}{512} \approx 43.95\%\) c) \(\frac{125}{512} \approx 24.41\%\)

53612310

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 \(50\%\)?

Figure for problem 536123

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 \(P(\text{same color}) = \frac{12}{30} + \frac{2}{30} = \frac{7}{15} \approx 46.7\%\). 5. Since \(46.7\% < 50\%\), the probability is less than \(50\%\).

Answer

a) \(\frac{7}{15} \approx 46.7\%\) b) Less than \(50\%\)

53744610

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 \(n(E)=500\cdot 0.65=325\) and \(n(\overline{E})=500\cdot 0.35=175\). 2. Follow each second-stage branch: \(n(E\cap M)=325\cdot 0.20=65\), \(n(E\cap\overline{M})=325\cdot 0.80=260\), \(n(\overline{E}\cap M)=175\cdot 0.40=70\), and \(n(\overline{E}\cap\overline{M})=175\cdot 0.60=105\). 3. The column totals are \(n(M)=65+70=135\) and \(n(\overline{M})=260+105=365\).

Answer

<table> <tr><td></td><td>\(M\)</td><td>\(\overline{M}\)</td><td>Total</td></tr> <tr><td>\(E\)</td><td>\(65\)</td><td>\(260\)</td><td>\(325\)</td></tr> <tr><td>\(\overline{E}\)</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>

53744710

A bakery uses two ovens to bake dinner rolls. Oven 1 produces \(60\%\) of the rolls, and Oven 2 produces the other \(40\%\). Each roll is classified as perfect, too dark, or unusable. The tree diagram shows the category rates for each oven. Of \(2000\) rolls produced, how many are expected to be classified as perfect?

Figure for problem 537447

Hints

- Find how many rolls come from each oven. - For each oven, multiply its production count by its perfect-roll rate. - Add the two perfect-roll counts. - Alternatively, find the overall probability of a perfect roll first, then multiply by \(2000\).

Solution

1. Oven 1 produces \(2000 \cdot 0.60 = 1200\) rolls, and Oven 2 produces \(2000 \cdot 0.40 = 800\) rolls. 2. The expected number of perfect rolls from Oven 1 is \(1200 \cdot 0.95 = 1140\). 3. The expected number of perfect rolls from Oven 2 is \(800 \cdot 0.85 = 680\). 4. The total expected number is \(1140 + 680 = 1820\) perfect rolls.

Answer

About \(1820\) rolls are expected to be classified as perfect.

53744810

An electronics store sells rechargeable batteries from two manufacturers. The tree diagram shows each manufacturer's share of the inventory and the probability that a battery is defective \((D)\) or working properly \((W)\) during the warranty period. a) Write a brief quality-control report that summarizes the main information in the tree diagram. b) Create the corresponding two-way probability table.

Figure for problem 537448

Hints

- Multiply the probabilities along each complete path. - Each inner cell of the table represents a joint event. - Add the inner cells by row and column to check the marginal probabilities.

Solution

1. The inventory is \(45\%\) from manufacturer \(M_1\) and \(55\%\) from manufacturer \(M_2\). The defect rates are \(4\%\) for \(M_1\) and \(6\%\) for \(M_2\). 2. Multiply along each path: \(P(M_1\cap D)=0.45\cdot 0.04=0.018\), \(P(M_1\cap W)=0.45\cdot 0.96=0.432\), \(P(M_2\cap D)=0.55\cdot 0.06=0.033\), and \(P(M_2\cap W)=0.55\cdot 0.94=0.517\). 3. The column totals are \(P(D)=0.018+0.033=0.051\) and \(P(W)=0.432+0.517=0.949\).

Answer

a) Sample report: “Manufacturer \(M_1\) supplies \(45\%\) of the batteries and has a \(4\%\) defect rate. Manufacturer \(M_2\) supplies \(55\%\) of the batteries and has a slightly higher defect rate of \(6\%\).” 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>

53745010

A bowl contains \(20\) candies, exactly one of which is lemon. Three children each take one candy without replacement. In the tree diagram, let \(L\) mean lemon and \(L^c\) mean not lemon. 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

- Count the candies remaining after each selection. - Branch probabilities from one node must add to \(1\). - Multiply the probabilities along the required path. - Simplify the product of the fractions after multiplying.

Solution

1. a) Initially, \(19\) of the \(20\) candies are not lemon, so \(a = \frac{19}{20}\). 2. After the first child takes a non-lemon candy, \(19\) candies remain, including the lemon candy. Thus, \(b = \frac{1}{19}\) and \(c = \frac{18}{19}\). 3. b) The second child gets lemon along the path \(L^c, L\), so \(P = \frac{19}{20} \cdot \frac{1}{19} = \frac{1}{20}\). 4. c) The third child gets lemon along the path \(L^c, L^c, L\), whose probability is \(\frac{19}{20} \cdot \frac{18}{19} \cdot \frac{1}{18} = \frac{1}{20}\).

Answer

a) \(a = \frac{19}{20}\), \(b = \frac{1}{19}\), \(c = \frac{18}{19}\) b) \(\frac{1}{20}\) c) \(\frac{19}{20} \cdot \frac{18}{19} \cdot \frac{1}{18} = \frac{1}{20}\)

53745110

A lockbox can be opened by exactly one of \(15\) keys. A person tries the keys one at a time without retrying a key. The tree diagram records each attempt as success \((S)\) or failure \((F)\). a) Find the probability that the lockbox opens on exactly the fourth attempt. b) Find the probability that the lockbox opens by the fourth attempt. c) Compare the probability of success on the first attempt with the probability of success on the fourth attempt.

Figure for problem 537451

Hints

- Track how many keys remain available after each failed attempt. - Follow the failure-failure-failure-success path for part a. - Multiply probabilities along one path. - For “by the fourth attempt,” add the probabilities of success on attempts \(1\) through \(4\).

Solution

1. Success on exactly the fourth attempt follows the path \(F, F, F, S\). Its probability is \(\frac{14}{15} \cdot \frac{13}{14} \cdot \frac{12}{13} \cdot \frac{1}{12} = \frac{1}{15}\). 2. Opening by the fourth attempt means success on attempt \(1\), \(2\), \(3\), or \(4\). Each exact-attempt probability is \(\frac{1}{15}\), so the total is \(4 \cdot \frac{1}{15} = \frac{4}{15}\). 3. The probability of success on the first attempt is \(\frac{1}{15}\), and the probability of success on exactly the fourth attempt is also \(\frac{1}{15}\). They are equal.

Answer

a) \(\frac{1}{15}\) b) \(\frac{4}{15}\) c) The probabilities are equal; each is \(\frac{1}{15}\).

53746410

An archer hits the center of a target with probability \(0.80\). The archer takes three independent shots. Let \(H\) represent a hit and \(M\) represent a miss. a) How many terminal nodes are in the complete tree diagram for all three shots? b) Find the probability of event \(A\): The archer gets exactly two hits. c) Find the probability of event \(B\): At least one shot is a hit.

Figure for problem 537464

Hints

- The experiment has one stage for each shot. - The branch probabilities remain the same from shot to shot. - List every path containing exactly two hits. - For “at least one,” consider the complementary event.

Solution

1. At every stage of the tree, the hit branch has probability \(0.80\), and the miss branch has probability \(0.20\). A complete three-stage tree has \(2^3=8\) terminal nodes. 2. Event \(A\) contains \(HHM\), \(HMH\), and \(MHH\). Each path has probability \((0.80)^2\cdot 0.20=0.128\), so \(P(A)=3\cdot 0.128=0.384\). 3. The complement of event \(B\) is three misses. Therefore, \(P(B)=1-(0.20)^3=0.992\).

Answer

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

53746510

A factory produces electronic components. Historically, \(10\%\) are defective \((D)\), and the rest are good \((G)\). Three components are selected independently from a very large production run and tested. a) Explain why this is a binomial experiment. Give \(n\) and the success probability \(p\) for “defective.” b) Use the tree diagram and path rules to find the probability that at most one selected component is defective.

Figure for problem 537465

Hints

- What conditions define a binomial experiment? - Which defect counts are included in “at most one”? - Multiply probabilities along each path and add the probabilities of the qualifying paths.

Solution

1. Each selection has two relevant outcomes, defective or good. The selections are independent, and the defect probability remains \(0.10\). Therefore, \(n=3\) and \(p=0.10\). 2. “At most one defective” includes \(GGG\), \(DGG\), \(GDG\), and \(GGD\). 3. The all-good path has probability \((0.90)^3=0.729\). 4. Each path with exactly one defective component has probability \(0.10\cdot(0.90)^2=0.081\). There are three such paths, so their total probability is \(3\cdot 0.081=0.243\). 5. Add the disjoint path probabilities: \(0.729+0.243=0.972\).

Answer

a) The trials have two outcomes, constant probability, and independence. \(n=3\), \(p=0.10\). b) \(P(X\le 1)=0.972\)

53746610

At a school, students either participate in the lunch program \((M)\) or do not \((\overline{M})\). The school also records whether they ride a bicycle to school \((B)\) or not \((\overline{B})\). The tree diagram gives the relevant probabilities. a) Describe in words what the probability \(0.20\) means. b) Find \(P(M\cap B)\) and \(P(\overline{M}\cap B)\). c) Find \(P(B)\) and \(P(\overline{B})\). d) Find \(P(\overline{B}\mid M)\) and \(P(M\cap\overline{B})\). 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: \(P(M\cap B)=0.30\cdot 0.20=0.06\) and \(P(\overline{M}\cap B)=0.70\cdot 0.40=0.28\). 3. Add the paths ending at \(B\): \(P(B)=0.06+0.28=0.34\). Therefore, \(P(\overline{B})=1-0.34=0.66\). 4. The diagram gives \(P(\overline{B}\mid M)=0.80\). Thus, \(P(M\cap\overline{B})=0.30\cdot 0.80=0.24\). 5. The requested conditional probability is \(P(B\mid\overline{M})=0.40\).

Answer

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

53746710

A candy company packages boxes either by hand \((H)\) or by machine \((\overline{H})\). 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 \((\overline{D})\). c) Find \(P(H\mid D)\). Interpret the result in context. d) Determine whether packaging method and defects are independent. Justify your answer.

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. - To reverse the condition, divide the appropriate joint probability by the probability of the new condition. - For independence, ask whether knowing the packaging method changes the defect probability.

Solution

1. Add the two paths that end in \(D\): \(P(D)=0.20\cdot 0.05+0.80\cdot 0.10=0.01+0.08=0.09\). 2. The hand-packaged, nondefective path has probability \(P(H\cap\overline{D})=0.20\cdot 0.95=0.19\). 3. The hand-packaged defective path has probability \(P(H\cap D)=0.20\cdot 0.05=0.01\). Therefore, \(P(H\mid D)=\frac{0.01}{0.09}=\frac{1}{9}\approx 0.1111\). About \(11.1\%\) of defective boxes were packaged by hand. 4. Packaging method and defects are not independent because \(P(D\mid H)=0.05\neq P(D)=0.09\).

Answer

a) \(P(D)=0.09\) b) \(P(H\cap\overline{D})=0.19\) c) \(P(H\mid D)=\frac{1}{9}\approx 0.1111\). About \(11.1\%\) of defective boxes were packaged by hand. d) The events are dependent because \(P(D\mid H)\neq P(D)\).

53752010

A bag contains \(3\) red balls and \(2\) blue balls. Two balls are drawn without replacement. Mia claims, “Red then blue is more likely than blue then red because red is more likely on the first draw.” Evaluate her claim.

Figure for problem 537520

Hints

- Calculate the probability of each complete path separately. - A larger first probability can be offset by the probability on the second branch.

Solution

1. The probability of red then blue is \(\frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}\). 2. The probability of blue then red is \(\frac{2}{5} \cdot \frac{3}{4} = \frac{3}{10}\). 3. The two orders have the same probability, so Mia’s claim is false.

Answer

Both orders have probability \(\frac{3}{10}\), so Mia’s claim is false.

53752410

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

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

53752710

A component is defective with probability \(0.02\). Three independently selected components are inspected. Find the probability that at least one component is defective.

Figure for problem 537527

Hints

- The event includes several possible paths. - Describe its complement with one path. - Subtract the complement probability from \(1\).

Solution

1. Use the complement that no component is defective. A component is not defective with probability \(0.98\). 2. Therefore, \(P(\text{at least one defective})=1-(0.98)^3=1-0.941192=0.058808\).

Answer

\(0.058808\approx5.88\%\)

53752810

A trip has three legs. The probabilities of a delay on the three legs are \(0.10\), \(0.20\), and \(0.15\), respectively. Model the delays as independent. Find the probability that exactly one leg is delayed.

Figure for problem 537528

Hints

- List the three cases in which exactly one leg is delayed. - Use the appropriate delay or on-time probability for each leg. - Add the probabilities of the three cases.

Solution

1. If only the first leg is delayed, the probability is \(0.10\cdot 0.80\cdot 0.85=0.068\). 2. If only the second leg is delayed, the probability is \(0.90\cdot 0.20\cdot 0.85=0.153\). 3. If only the third leg is delayed, the probability is \(0.90\cdot 0.80\cdot 0.15=0.108\). 4. Add the three disjoint cases: \(P(\text{exactly one delay})=0.068+0.153+0.108=0.329\).

Answer

\(0.329=32.9\%\)

53753410

A six-sided die has three red faces, two blue faces, and one green face. It is rolled twice. Find the probability that the two rolls show different colors.

Figure for problem 537534

Hints

- Consider the complement: both rolls show the same color. - Find the probabilities of red-red, blue-blue, and green-green. - Add those probabilities and subtract from \(1\).

Solution

1. Use the complement that both rolls show the same color. 2. The probability of the same color is \(\left(\frac{3}{6}\right)^2+\left(\frac{2}{6}\right)^2+\left(\frac{1}{6}\right)^2=\frac{14}{36}\). 3. Therefore, \(P(\text{different colors})=1-\frac{14}{36}=\frac{22}{36}=\frac{11}{18}\).

Answer

\(\frac{11}{18}\approx 61.1\%\)

53754310

A team wins each game of a best-of-three series with probability \(0.60\). The series ends as soon as one team has two wins. Let \(W\) represent a win and \(L\) represent a loss. Find the probability that the team wins the series.

Figure for problem 537543

Hints

- Some paths end after only two games. - Identify every terminal path on which the team reaches two wins first.

Solution

1. The team can win in two games with the sequence \(WW\), whose probability is \((0.60)^2=0.36\). 2. The team can win in three games with \(WLW\), whose probability is \(0.60\cdot 0.40\cdot 0.60=0.144\). 3. It can also win with \(LWW\), whose probability is \(0.40\cdot (0.60)^2=0.144\). 4. Adding the disjoint winning paths gives \(0.36+0.144+0.144=0.648\).

Answer

\(0.648\), or \(64.8\%\)

53754410

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) \(0.81\)

53754810

Three consecutive traffic lights are green independently with probabilities \(0.7\), \(0.6\), and \(0.8\), respectively. In the tree diagram, G means green and R means red. Find the probability that exactly two lights are green.

Figure for problem 537548

Hints

- Exactly one of the three lights must be red. - List the three possible positions of the red light. - Multiply along each path and add the three probabilities.

Solution

1. The probability of green-green-red is \(0.7\cdot 0.6\cdot 0.2=0.084\). 2. The probability of green-red-green is \(0.7\cdot 0.4\cdot 0.8=0.224\). 3. The probability of red-green-green is \(0.3\cdot 0.6\cdot 0.8=0.144\). 4. Add the three disjoint cases: \(P(\text{exactly two green})=0.084+0.224+0.144=0.452\).

Answer

\(0.452=45.2\%\)

53755310

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

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

53755410

A search procedure succeeds on each attempt with probability \(0.25\). The procedure stops after the first success or after the third attempt, whichever comes first. Find the probability that it ends: a) After the first attempt. b) After the second attempt. c) After the third attempt.

Figure for problem 537554

Hints

- The final stopping time includes either result on the third attempt. - Describe the outcomes that must occur before each stopping time.

Solution

1. The procedure ends after attempt \(1\) when the first attempt succeeds, so the probability is \(0.25\). 2. It ends after attempt \(2\) when the first attempt fails and the second succeeds: \(0.75\cdot 0.25=0.1875\). 3. It reaches and ends after attempt \(3\) whenever the first two attempts fail, regardless of the third result. The probability is \((0.75)^2=0.5625\). 4. The probabilities check: \(0.25+0.1875+0.5625=1\).

Answer

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

53756310

A tree diagram represents a survey of \(1000\) people with these four terminal frequencies: \(A\cap B=240\), \(A\cap\overline{B}=60\), \(\overline{A}\cap B=140\), and \(\overline{A}\cap\overline{B}=560\). Find every branch probability when \(A\) is the first stage of the tree.

Figure for problem 537563

Hints

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

Solution

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

Answer

\(P(A)=0.3\), \(P(\overline{A})=0.7\); after \(A\): \(0.8\), \(0.2\); after \(\overline{A}\): \(0.2\), \(0.8\).

51364610

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) One boy, Lucas, knows that he was not selected. From his perspective, what is the probability that two particular friends of his are the students selected?

Hints

- For part b, include both possible selection orders. - In part c, remove Lucas from the possible students. - Count the number of possible pairs among the remaining \(19\) students.

Solution

1. a) \(P(\text{two boys}) = \frac{8}{20} \cdot \frac{7}{19} = \frac{14}{95} \approx 0.1474\). 2. b) The two possible orders are girl-boy and boy-girl. Thus, \(P(\text{one of each}) = \frac{12}{20} \cdot \frac{8}{19} + \frac{8}{20} \cdot \frac{12}{19} = \frac{48}{95} \approx 0.5053\). 3. c) Given that Lucas was not selected, the team is an equally likely pair chosen from the other \(19\) students. There are \(\binom{19}{2} = 171\) possible pairs and only one pair consisting of the two named friends. Therefore, the probability is \(\frac{1}{171} \approx 0.00585\).

Answer

a) \(\frac{14}{95} \approx 14.74\%\) b) \(\frac{48}{95} \approx 50.53\%\) c) \(\frac{1}{171} \approx 0.585\%\)

53592710

At a carnival game, a player spins each of three spinners once. The spinners have equal sections labeled with digits and colored red or blue. On Spinner 1, section \(1\) is red and sections \(2\) through \(4\) are blue. On Spinner 2, sections \(1\) and \(2\) are red and sections \(3\) through \(5\) are blue. On Spinner 3, sections \(1\) through \(3\) are red and sections \(4\) through \(6\) are blue. 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

- Treat the experiment as three independent stages. - Count the sections with the required digit or color on each spinner. - Multiply probabilities along each favorable path. - For part b, list all paths containing exactly two blue results, then add their probabilities.

Solution

1. The probabilities for Spinner 1 are \(P(1) = \frac{1}{4}\), \(P(\text{blue}) = \frac{3}{4}\), and \(P(\text{red}) = \frac{1}{4}\). 2. The probabilities for Spinner 2 are \(P(1) = \frac{1}{5}\), \(P(\text{blue}) = \frac{3}{5}\), and \(P(\text{red}) = \frac{2}{5}\). 3. The probabilities for Spinner 3 are \(P(1) = \frac{1}{6}\), \(P(\text{blue}) = \frac{1}{2}\), 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) \(\frac{1}{120} \approx 0.83\%\) b) \(\frac{9}{20} = 45\%\)

53746110

A table tennis championship match uses a best-of-five format: the first player to win three games wins the match. The finalists are equally matched, so each player has probability \(0.5\) of winning any game. Use a tree diagram to find the probability that the match lasts all five games.

Figure for problem 537461

Hints

- Determine the score after four games that requires a fifth game. - Count the paths in which each player wins exactly two of the first four games. - Use the multiplication and addition rules for the tree paths.

Solution

1. A fifth game is needed exactly when the score is tied \(2\)–\(2\) after four games. 2. There are \(\binom{4}{2}=6\) sequences in which each player wins two of the first four games. 3. Each sequence has probability \((0.5)^4=0.0625\). 4. Therefore, the probability is \(6\cdot0.0625=0.375=37.5\%\).

Answer

\(0.375\), or \(37.5\%\)

53748510

An online store reports that \(60\%\) of its orders come from returning customers \((R)\). Of those orders, \(80\%\) use express shipping \((E)\). Of the orders from occasional customers \((\overline{R})\), \(15\%\) use express shipping. a) Complete the probability tree. b) What percentage of all orders use express shipping? c) Given that an order does not use express shipping \((\overline{E})\), find the probability that it came from a returning customer. d) Explain the difference between \(P(R\cap E)\) and \(P(E\mid R)\) in this context. e) Interpret the expression \(\frac{0.40\cdot 0.15}{0.60\cdot 0.80+0.40\cdot 0.15}\) in context.

Figure for problem 537485

Hints

- At each node, the outgoing branch probabilities sum to \(1\). - Add all complete paths that end at express shipping. - In part c, orders without express shipping form the new denominator. - In part e, identify the event represented by the numerator and the event represented by the denominator.

Solution

1. The missing branch probabilities are \(P(\overline{R})=0.40\), \(P(\overline{E}\mid R)=0.20\), and \(P(\overline{E}\mid\overline{R})=0.85\). 2. Add the express-shipping paths: \(P(E)=0.60\cdot 0.80+0.40\cdot 0.15=0.54\). 3. Then \(P(\overline{E})=0.46\) and \(P(R\cap\overline{E})=0.60\cdot 0.20=0.12\). Therefore, \(P(R\mid\overline{E})=\frac{0.12}{0.46}=\frac{6}{23}\approx 0.2609\). 4. \(P(R\cap E)\) is the proportion of all orders that are both from returning customers and use express shipping. \(P(E\mid R)\) is the proportion using express shipping within the returning-customer group. 5. The numerator is \(P(\overline{R}\cap E)\), and the denominator is \(P(E)\). The expression equals \(P(\overline{R}\mid E)\), the probability that an express order came from an occasional customer.

Answer

a) \(P(\overline{R})=0.40\), \(P(\overline{E}\mid R)=0.20\), and \(P(\overline{E}\mid\overline{R})=0.85\). b) \(54\%\). c) \(P(R\mid\overline{E})=\frac{6}{23}\approx 0.2609\), or about \(26.09\%\). d) \(P(R\cap E)\) uses all orders as the reference group; \(P(E\mid R)\) uses only returning-customer orders. e) It is \(P(\overline{R}\mid E)\), the probability that an express order came from an occasional customer.

53757010

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 \(P(\overline{A}\cap B)=0.7-0.4=0.3\). 3. Since \(P(\overline{A})=0.2\), the required branch probability would be \(P(B\mid\overline{A})=\frac{0.3}{0.2}=1.5\). 4. A probability cannot exceed \(1\), so the original values are inconsistent.

Answer

The values are impossible because they require \(P(B\mid\overline{A})=1.5\).

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