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Independence of events

Learning goals

  • Tests independence using conditional probabilities or P(A∩B)=P(A)P(B).
  • Explains independence as one event not changing the probability of the other.
  • Distinguishes mutually exclusive events from independent events.
  • Uses independence in probability calculations when justified.

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Independence of events 1
Events \(A\) and \(B\) satisfy \(P(A)>0\), \(P(B)=0.40\), and \(P(B\mid A)=0.40\). Are \(A\) and \(B\) independent? Explain using the meaning of independence.

Hints

- Ask whether conditioning on \(A\) changes the probability of \(B\). - Compare the conditional probability with the corresponding marginal probability.

Solution

1. Independence means that knowing \(A\) occurred does not change the probability of \(B\). 2. Here, \(P(B\mid A)=0.40=P(B)\). 3. Therefore, \(A\) and \(B\) are independent.

Answer

Yes. They are independent because \(P(B\mid A)=P(B)\).
Independence of events 2
A fair coin is tossed and a fair die is rolled. Event \(H\) is getting heads, and event \(S\) is rolling a six. If the coin lands heads, does that change the chance that the die shows six? Are \(H\) and \(S\) independent?

Hints

- Does the die's set of possible outcomes change when the coin lands? - Compare the chance of a six before the toss with the chance after you know the coin result.

Solution

1. The coin and die are separate random devices. 2. The die still shows six with probability \(\frac{1}{6}\), regardless of the coin result. 3. Thus H and S are independent.

Answer

No. The chance of a six stays \(\frac{1}{6}\); the events are independent.
Independence of events 3
A bag contains two red balls and one blue ball. One ball is drawn and not put back before a second draw. If the first ball is red, does the chance of drawing red next stay the same? Are the colors of the two draws independent?

Hints

- Compare the number of red balls before and after the first red draw. - What happens to the total number of balls when one is not replaced?

Solution

1. Before a draw, two of the three balls are red. 2. After one red ball is removed, only one of the two remaining balls is red. 3. The chance changes from \(\frac{2}{3}\) to \(\frac{1}{2}\), so the draws are dependent.

Answer

No. The red probability changes from \(\frac{2}{3}\) to \(\frac{1}{2}\); the draws are dependent.
Independence of events 4
A fair die is rolled once. Event \(E\) is “even” and event \(O\) is “odd.” Can these events happen together? If you learn that E happened, what is the probability of O? Are they independent?

Hints

- Which outcomes could be even and odd at the same time? - What does knowing an even result tell you about an odd result?

Solution

1. The die cannot show an odd and an even result on the same roll. 2. Given E, the chance of O is \(0\), though without that information \(P(O)=\frac{1}{2}\). 3. Their probabilities change with the condition, so the events are not independent.

Answer

No; \(P(O\mid E)=0\), and the events are not independent.
Independence of events 5
Maia completes her homework on \(80\%\) of school days. She wears her hair in a ponytail on \(70\%\) of school days. Assume these events are independent. What is the probability that she comes to school without a ponytail and without completed homework?

Hints

- Find the complement of each given percentage. - For independent events, multiply the probabilities to find the probability that both occur. - The complements of independent events are also independent.

Solution

1. The probability that Maia has not completed her homework is \(1-0.80=0.20\). 2. The probability that she does not wear a ponytail is \(1-0.70=0.30\). 3. The complements of independent events are also independent, so \(P(\text{no homework and no ponytail})=0.20\cdot 0.30=0.06\).

Answer

\(0.06\), or \(6\%\).
Independence of events 6
Lucas flips a fair coin five times, and every flip lands heads. He says, “Tails must be more likely than heads on the sixth flip to make up for the streak.” Is Lucas correct? Explain using independence.

Hints

- Which earlier outcomes can affect the result of a new fair-coin flip? - What does independence say about the next flip when past flips are known? - Does a short streak force any particular result on the next trial?

Solution

1. Coin flips are independent: knowing the outcomes of earlier flips does not change the probability on the next flip. 2. On the sixth flip, the probability of tails is still \(\frac{1}{2}=50\%\). 3. There is no rule requiring a streak of heads to be balanced by tails on the next flip. Lucas is incorrect.

Answer

No. Each fair-coin flip is independent, so the probability of tails on the sixth flip is still \(\frac{1}{2}=50\%\).
Independence of events 7
A streaming service analyzes \(500\) customers. Customers are classified as under age \(25\) (event \(U\)) or age \(25\) and older (event \(A\)), and by whether they have a premium subscription (event \(S\)). <table><tr><th></th><th>\(U\)</th><th>\(A\)</th><th>Total</th></tr><tr><th>\(S\)</th><td>\(120\)</td><td>\(80\)</td><td>\(200\)</td></tr><tr><th>\(S^c\)</th><td>\(180\)</td><td>\(120\)</td><td>\(300\)</td></tr><tr><th>Total</th><td>\(300\)</td><td>\(200\)</td><td>\(500\)</td></tr></table> Determine whether having a premium subscription is independent of age group by comparing \(P(S\mid U)\) with \(P(S)\).

Hints

- What fraction of all customers have premium subscriptions? - What fraction of the younger age group have premium subscriptions? - Independence means that knowing the age group does not change the subscription probability.

Solution

1. The overall probability of a premium subscription is \(P(S)=\frac{200}{500}=0.40\). 2. Among customers under age \(25\), \(P(S\mid U)=\frac{120}{300}=0.40\). 3. Knowing that the customer is under \(25\) does not change the premium-subscription probability. Thus, the two events are independent in this group.

Answer

\(P(S)=0.40\) and \(P(S\mid U)=0.40\). The events are independent because these probabilities are equal.
Independence of events 8
A streaming service finds that \(60\%\) of its subscribers are under age \(30\) (event \(U\)). Overall, \(30\%\) of subscribers prefer science-fiction movies (event \(S\)). In addition, \(12\%\) of subscribers are age \(30\) or older and prefer science-fiction movies. Determine whether science-fiction preference is independent of age group.

Hints

- Express each percentage as a probability. - Find the probability that a subscriber is age \(30\) or older. - Test whether the joint probability equals the product of the marginal probabilities. - Independence with an event also implies independence with its complement.

Solution

1. The probability that a subscriber is age \(30\) or older is \(P(U^c)=1-0.60=0.40\). 2. If \(U^c\) and \(S\) are independent, then \(P(U^c\cap S)=P(U^c)P(S)\). 3. Compute the product: \(P(U^c)P(S)=0.40\cdot0.30=0.12\). 4. This equals the given value \(P(U^c\cap S)=0.12\), so \(U^c\) and \(S\) are independent. Independence with an event also implies independence with its complement, so \(U\) and \(S\) are independent.

Answer

Science-fiction preference and age group are independent because \(P(U^c\cap S)=0.12=P(U^c)P(S)\).
Independence of events 9
A recreation club has \(120\) members. Of these, \(60\) play tennis (event \(T\)), \(40\) play squash (event \(S\)), and \(20\) play both sports. a) Determine whether membership in the tennis and squash groups is independent. b) Of the \(120\) members, \(40\) have belonged to the club for more than \(10\) years (event \(L\)). Of those long-term members, \(15\) play tennis. Determine whether \(L\) and \(T\) are independent.

Hints

- Convert each count to a probability using \(120\) as the total. - Independence requires \(P(A\cap B)=P(A)P(B)\). - In part b, the \(15\) members belong to both groups. - Compare the joint probability with the product in each part.

Solution

1. For part a, \(P(T)=\frac{60}{120}=\frac{1}{2}\), \(P(S)=\frac{40}{120}=\frac{1}{3}\), and \(P(T\cap S)=\frac{20}{120}=\frac{1}{6}\). 2. Since \(P(T)P(S)=\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}=P(T\cap S)\), the events are independent. 3. For part b, \(P(L)=\frac{40}{120}=\frac{1}{3}\) and \(P(L\cap T)=\frac{15}{120}=\frac{1}{8}\). 4. Since \(P(L)P(T)=\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}\neq\frac{1}{8}=P(L\cap T)\), the events are dependent.

Answer

a) \(T\) and \(S\) are independent because \(P(T\cap S)=\frac{1}{6}=P(T)P(S)\). b) \(L\) and \(T\) are dependent because \(P(L\cap T)=\frac{1}{8}\neq\frac{1}{6}=P(L)P(T)\).
Independence of events 10
During a \(30\)-day month, it rained on \(12\) days. A runner went jogging on \(15\) days, including \(3\) rainy days. One day from the month is selected at random. Let \(R\) be the event that it rained, and let \(J\) be the event that the runner went jogging. Determine whether \(R\) and \(J\) are independent, and interpret the result in context.

Hints

- Convert each count to a probability using \(30\) as the total. - Independence requires the joint probability to equal the product of the marginal probabilities. - Compare \(P(J\mid R)\) with \(P(J)\) to interpret the result.

Solution

1. The probabilities are \(P(R)=\frac{12}{30}=0.40\), \(P(J)=\frac{15}{30}=0.50\), and \(P(R\cap J)=\frac{3}{30}=0.10\). 2. If the events were independent, the joint probability would be \(P(R)P(J)=0.40\cdot 0.50=0.20\). 3. Since \(0.10\neq 0.20\), the events are dependent. 4. Also, \(P(J\mid R)=\frac{3}{12}=0.25\), which is lower than the overall jogging rate of \(0.50\). In this month's data, the runner jogged less often on rainy days.

Answer

The events are dependent because \(P(R\cap J)=0.10\neq 0.20=P(R)P(J)\). The runner's observed jogging rate was lower on rainy days than over the month as a whole.
Independence of events 11
A bag contains the \(12\) labeled balls shown. One ball is selected at random. Let \(A\) be the event that the ball is blue, and let \(E\) be the event that its number is even. Determine whether \(A\) and \(E\) are independent.
Figure for problem 532075

Hints

- Count all the balls, the blue balls, and the even-numbered balls. - Identify the balls that satisfy both conditions. - Independence requires the joint probability to equal the product of the marginal probabilities.

Solution

1. There are \(4\) blue balls, so \(P(A)=\frac{4}{12}=\frac{1}{3}\). 2. The even-numbered balls are numbered \(2,4,6\), so \(P(E)=\frac{3}{12}=\frac{1}{4}\). 3. Only the blue ball numbered \(2\) belongs to both events, so \(P(A\cap E)=\frac{1}{12}\). 4. Since \(P(A)P(E)=\frac{1}{3}\cdot\frac{1}{4}=\frac{1}{12}=P(A\cap E)\), the events are independent.

Answer

The events \(A\) and \(E\) are independent because \(P(A\cap E)=\frac{1}{12}=P(A)P(E)\).
Independence of events 12
Components are checked for a defect at Station A (event \(A\)) and a defect at Station B (event \(B\)). What feature of the probability tree shows immediately that the two defect events are independent?
Figure for problem 537488

Hints

- Compare the two branches leading to \(B\) in the second stage. - Ask whether the probability of \(B\) changes depending on the first-stage outcome. - Equal conditional probabilities indicate independence.

Solution

1. Read the second-stage conditional probabilities from the tree. 2. The probability of a defect at Station B is \(P(B\mid A)=0.25\) after a defect at Station A. 3. It is also \(P(B\mid A^c)=0.25\) after no defect at Station A. 4. Because the probability of \(B\) is unchanged by whether \(A\) occurred, the events are independent. The complementary branch probabilities are also equal: both are \(0.75\).

Answer

The second-stage branch probabilities are the same after either first-stage outcome: \(P(B\mid A)=P(B\mid A^c)=0.25\). Therefore, the events are independent.
Independence of events 13
Suppose \(P(A)=0.4\), \(P(B)=0.5\), and \(P(A\cap B)=0.2\). Determine whether \(A\) and \(B\) are independent.

Hints

- Compare the given intersection probability with the product of the individual probabilities. - Independence requires exact equality.

Solution

1. For independent events, \(P(A\cap B)=P(A)P(B)\). 2. Here, \(P(A)P(B)=0.4\cdot0.5=0.2\). 3. This equals the given value \(P(A\cap B)=0.2\), so the events are independent.

Answer

\(A\) and \(B\) are independent.
Independence of events 14
Find the missing branch probability \(x\) that would make events \(A\) and \(B\) independent.
Figure for problem 537568

Hints

- Compare the two second-stage branches that lead to \(B\). - What would independence mean for those two conditional probabilities? - You can check your result by adding the probabilities of both paths ending in \(B\).

Solution

1. For independent events, the probability of \(B\) does not change based on whether \(A\) occurred. 2. The tree gives \(P(B\mid A^c)=0.30\), so independence requires \(P(B\mid A)=x=0.30\). 3. Check: \(P(B)=0.40\cdot0.30+0.60\cdot0.30=0.30\), equal to \(P(B\mid A)\).

Answer

\(x=0.30\).
Independence of events 15
Among patients who have a particular disease, a laboratory uses two tests, A and B. Use the probability tree to explain why the laboratory cannot calculate every two-test path by using one fixed probability for a positive result on Test B.
Figure for problem 537577

Hints

- Compare the two second-stage branches that lead to \(B^+\). - Under independence, would the result of Test A change the probability of a positive Test B result?

Solution

1. The probability of a positive result on Test B depends on the result of Test A. 2. After \(A^+\), the tree gives \(P(B^+\mid A^+)=0.95\), but after \(A^-\), it gives \(P(B^+\mid A^-)=0.60\). 3. Because these conditional probabilities differ, the test results are conditionally dependent within the diseased group. Each path must use the second-stage probability that matches its first-stage outcome.

Answer

A single fixed probability is not valid because \(P(B^+\mid A^+)=0.95\neq0.60=P(B^+\mid A^-)\). The two test results are conditionally dependent in this group.
Independence of events 16
Events \(A\) and \(B\) are disjoint, with \(P(A)=0.30\) and \(P(B)=0.20\). Are \(A\) and \(B\) independent? Explain by comparing \(P(A\cap B)\) with \(P(A)P(B)\).

Hints

- What does disjointness tell you about the intersection? - Independence has a separate product criterion involving the intersection. - Compare those two required intersection probabilities.

Solution

1. Because the events are disjoint, \(P(A\cap B)=0\). 2. If they were independent, the product criterion would require \(P(A\cap B)=P(A)P(B)=0.30\cdot 0.20=0.06\). 3. Since \(0\ne 0.06\), the events are not independent. 4. More generally, two disjoint events with positive probabilities cannot be independent.

Answer

No. \(P(A\cap B)=0\), but \(P(A)P(B)=0.06\), so the independence criterion fails.
Independence of events 17
A number is chosen at random from \(\{1,2,3,4,5,6,7,8\}\). Let \(A\) mean “even” and \(B\) mean “greater than four.” Compare \(P(A)\) with \(P(A\mid B)\). Are \(A\) and \(B\) independent?

Hints

- List the even numbers in the full sample space. - Repeat that count using only numbers greater than four.

Solution

1. The even results are \(2,4,6,8\), so \(P(A)=\frac{4}{8}=\frac{1}{2}\). 2. Restricting to \(B=\{5,6,7,8\}\), two of four results are even; thus \(P(A\mid B)=\frac{2}{4}=\frac{1}{2}\). 3. The probability is unchanged, so A and B are independent.

Answer

\(P(A)=P(A\mid B)=\frac{1}{2}\); independent.
Independence of events 18
For events \(W\) (“wears glasses”) and \(R\) (“rides the bus”), suppose \(P(W)=0.35\) and \(P(W\mid R)=0.50\). Are W and R independent? Explain without doing any further calculations.

Hints

- Which probability describes glasses-wearing before you learn anything about the bus ride? - What changes in the information available after you learn the student rides the bus?

Solution

1. Independence would require \(P(W\mid R)=P(W)\), provided \(P(R)>0\). 2. Because \(0.50\ne0.35\), the probabilities differ and the events are dependent.

Answer

No. Knowing R changes the probability of W from \(0.35\) to \(0.50\).
Independence of events 19
Events \(A\) and \(B\) have \(P(A)=0.30\), \(P(B)=0.40\), and \(P(A\cap B)=0.12\). Use the multiplication test to decide whether the events are independent.

Hints

- What equation must hold if two events are independent? - Compare the product with the given intersection probability.

Solution

1. Compute \(P(A)P(B)=0.30\cdot0.40=0.12\). 2. This equals the given \(P(A\cap B)\), so the events are independent.

Answer

Independent, because \(P(A)P(B)=P(A\cap B)=0.12\).
Independence of events 20
A fair die is rolled once. Let \(A=\{1,2,3\}\) and \(B=\{1,4\}\). Are A and B independent? Show your reasoning using the intersection and the product of the event probabilities.

Hints

- Write down the outcomes common to A and B. - Compare the common fraction with the product of the two event fractions.

Solution

1. The intersection is \(A\cap B=\{1\}\), so \(P(A\cap B)=\frac{1}{6}\). 2. Also \(P(A)=\frac{3}{6}=\frac{1}{2}\) and \(P(B)=\frac{2}{6}=\frac{1}{3}\). 3. The product is \(\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}\). Therefore, A and B are independent even though they overlap.

Answer

Yes. \(P(A\cap B)=P(A)P(B)=\frac{1}{6}\).
Independence of events 21
The table shows participation in a volunteer project and in a school sport. <table><tr><th></th><th>Plays sport</th><th>Does not play</th><th>Total</th></tr><tr><th>Volunteers</th><td>6</td><td>4</td><td>10</td></tr><tr><th>Does not volunteer</th><td>12</td><td>8</td><td>20</td></tr><tr><th>Total</th><td>18</td><td>12</td><td>30</td></tr></table> Are “volunteers” and “plays sport” independent in this table?

Hints

- Find the overall sport fraction. - Check whether the sport fraction changes when you restrict to volunteers.

Solution

1. Among volunteers, the fraction who play sport is \(\frac{6}{10}=0.60\). 2. Among all students, the fraction who play sport is \(\frac{18}{30}=0.60\). 3. The proportions agree, so the events are independent in this probability model.

Answer

Yes. Both the overall and volunteer-conditioned sport probabilities equal \(0.60\).
Independence of events 22
Events \(F\) and \(G\) are independent. \(P(F)=0.60\) and \(P(F\cap G)=0.18\). Find \(P(G)\).

Hints

- Which multiplication relationship holds because F and G are independent? - Which factor in that relationship is unknown?

Solution

1. Independence gives \(P(F\cap G)=P(F)P(G)\). 2. Thus \(P(G)=\frac{0.18}{0.60}=0.30\).

Answer

\(P(G)=0.30\)
Independence of events 23
Two balls are drawn from the bag. The first ball is red. Compare the probability that the second ball is red if the first ball is replaced and if it is not replaced.
Figure for problem 560989

Hints

- Count the balls remaining when the first ball is put back. - Count again if the first red ball stays out of the bag.

Solution

1. With replacement, the bag still has two red balls among four, so the second-red probability is \(\frac{2}{4}=\frac{1}{2}\). 2. Without replacement, one red ball is removed, leaving one red among three, so the probability is \(\frac{1}{3}\). 3. Replacing the ball keeps the original chances; not replacing it changes them.

Answer

With replacement: \(\frac{1}{2}\). Without replacement: \(\frac{1}{3}\).
Independence of events 24
Events \(A\) and \(B\) cannot happen together, but each has a probability greater than zero. Could A and B be independent? Explain using the multiplication test, without assigning particular probabilities.

Hints

- What is the probability of the intersection when two events are disjoint? - Is the product of two positive numbers ever zero?

Solution

1. Since A and B cannot occur together, \(P(A\cap B)=0\). 2. Since both probabilities are positive, their product \(P(A)P(B)>0\). 3. The product is not the intersection probability, so the events cannot be independent.

Answer

No. \(P(A\cap B)=0\) but \(P(A)P(B)>0\).
Independence of events 25
A fair six-sided die is rolled \(600\) times, and the result \(6\) occurs \(85\) times. a) Find the observed relative frequency of rolling a \(6\) as a fraction in simplest form. b) Njeri says, “Since the probability of a \(6\) is \(\frac{1}{6}\), the next \(600\) rolls must include more than \(100\) sixes to make up for the earlier results.” Explain why the claim is incorrect.

Hints

- Relative frequency compares the number of times an outcome occurred with the number of trials. - Can the results from earlier rolls change the chance on the next roll of a fair die? - Does a known probability guarantee an exact number of successes in a new series of trials?

Solution

1. The observed relative frequency is \(\frac{85}{600}=\frac{17}{120}\). 2. Each future roll is independent of the previous \(600\) rolls. The probability of a \(6\) on each new roll remains \(\frac{1}{6}\). 3. The number of sixes in a new set of \(600\) rolls is not guaranteed to be above \(100\). Earlier results do not force later ones to make up a difference.

Answer

a) \(\frac{17}{120}\). b) The claim is incorrect. Each new roll is independent and has probability \(\frac{1}{6}\) of showing a \(6\). No catch-up count is guaranteed.
Independence of events 26
A bag contains \(5\) red balls and \(5\) blue balls. In each of two experiments, three red balls have already been drawn in a row. Experiment A: Each ball is replaced after it is drawn. Experiment B: Balls are not replaced. For each experiment, determine how the first three draws affect the probability of drawing a blue ball next. Compare the results.

Hints

- Determine the contents of the bag after the three draws in each experiment. - Decide whether the total and color counts change. - Connect replacement with independence.

Solution

1. In Experiment A, replacement keeps the bag at \(5\) red and \(5\) blue balls after every draw. Therefore, the probability of blue on the next draw remains \(\frac{5}{10} = \frac{1}{2}\). The draws are independent. 2. In Experiment B, after three red balls have been removed, \(7\) balls remain: \(2\) red and \(5\) blue. Therefore, the probability of blue next is \(\frac{5}{7} \approx 0.7143\). 3. With replacement, earlier results do not change the next-draw probability. Without replacement, the draws are dependent, and removing red balls increases the probability of blue.

Answer

In Experiment A, the probability remains \(\frac{1}{2}\). In Experiment B, it increases to \(\frac{5}{7} \approx 0.7143\) because the earlier red balls were not replaced.
Independence of events 27
A smartphone manufacturer tracks two types of defects: display defects \((D)\) and battery defects \((B)\). Of all phones, \(5\%\) have a display defect, \(3\%\) have a battery defect, and \(1\%\) have both defects. a) Create a two-way probability table. b) A phone is known to have a battery defect. Find the probability that it also has a display defect. c) Compare the result from part b with the overall probability of a display defect. What does the comparison show about the two defects?

Hints

- Subtract the overlap from each marginal probability to fill two inner cells. - For part b, restrict the sample space to phones with a battery defect. - Independence would require the conditional display-defect probability to equal the overall display-defect probability.

Solution

1. The remaining inner cells are \(P(D\cap B^c)=0.05-0.01=0.04\), \(P(D^c\cap B)=0.03-0.01=0.02\), and \(P(D^c\cap B^c)=1-(0.01+0.04+0.02)=0.93\). 2. \(P(D\mid B)=\frac{P(D\cap B)}{P(B)}=\frac{0.01}{0.03}=\frac{1}{3}\). 3. Since \(P(D\mid B)=\frac{1}{3}\neq0.05=P(D)\), the events are dependent. In these data, a display defect is much more common among phones with a battery defect than among all phones.

Answer

a) <table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(D\)</td><td>\(0.01\)</td><td>\(0.04\)</td><td>\(0.05\)</td></tr> <tr><td>\(D^c\)</td><td>\(0.02\)</td><td>\(0.93\)</td><td>\(0.95\)</td></tr> <tr><td>Total</td><td>\(0.03\)</td><td>\(0.97\)</td><td>\(1\)</td></tr> </table> b) \(P(D\mid B)=\frac{1}{3}\approx0.3333\), or about \(33.33\%\). c) Because \(P(D\mid B)\neq P(D)\), display and battery defects are dependent in these production data.
Independence of events 28
A survey of \(200\) households records dog ownership \((D)\) and yard ownership \((Y)\). <table> <tr><td></td><td>\(Y\)</td><td>\(Y^c\)</td><td>Total</td></tr> <tr><td>\(D\)</td><td>\(48\)</td><td>\(12\)</td><td>\(60\)</td></tr> <tr><td>\(D^c\)</td><td>\(56\)</td><td>\(84\)</td><td>\(140\)</td></tr> <tr><td>Total</td><td>\(104\)</td><td>\(96\)</td><td>\(200\)</td></tr> </table> a) Find \(P(Y\mid D)\), the probability that a household has a yard given that it has a dog. b) Find \(P(Y\mid D^c)\). Compare the two probabilities and describe the association between dog ownership and yard ownership in this sample.

Hints

- The event after the vertical bar determines the relevant row. - Divide the appropriate inner count by that row's total. - Compare the two conditional proportions to assess dependence.

Solution

1. Among the \(60\) households with a dog, \(48\) have a yard. Thus, \(P(Y\mid D)=\frac{48}{60}=0.80\). 2. Among the \(140\) households without a dog, \(56\) have a yard. Thus, \(P(Y\mid D^c)=\frac{56}{140}=0.40\). 3. The yard-ownership rate is twice as high among households with a dog, so the two variables are associated in this sample and are not independent.

Answer

a) \(P(Y\mid D)=0.80\), or \(80\%\). b) \(P(Y\mid D^c)=0.40\), or \(40\%\). Yard ownership is twice as common among dog-owning households in this sample, indicating dependence.
Independence of events 29
A sports organization studies whether participation in a mental-skills training program \((M)\) is associated with passing a coaching-certification exam \((E)\). The results for \(500\) participants are shown. <table> <tr><td></td><td>\(M\) (training)</td><td>\(M^c\) (no training)</td><td>Total</td></tr> <tr><td>\(E\) (passed)</td><td>\(150\)</td><td>\(200\)</td><td>\(350\)</td></tr> <tr><td>\(E^c\) (did not pass)</td><td>\(50\)</td><td>\(100\)</td><td>\(150\)</td></tr> <tr><td>Total</td><td>\(200\)</td><td>\(300\)</td><td>\(500\)</td></tr> </table> Determine whether \(M\) and \(E\) are independent, and interpret the result in context.

Hints

- Independence requires \(P(M\cap E)=P(M)P(E)\). - Compute the marginal and joint probabilities from the counts. - A statistical association does not automatically establish a causal effect.

Solution

1. From the table, \(P(M)=\frac{200}{500}=0.40\), \(P(E)=\frac{350}{500}=0.70\), and \(P(M\cap E)=\frac{150}{500}=0.30\). 2. If the events were independent, \(P(M\cap E)\) would equal \(P(M)P(E)=0.40\cdot0.70=0.28\). 3. Since \(0.30\neq0.28\), the events are dependent. 4. Equivalently, the pass rate among training participants is \(P(E\mid M)=\frac{150}{200}=0.75\), which is higher than the overall pass rate \(0.70\). The data show a positive association, but the table alone does not establish that the training caused the difference.

Answer

The events are dependent because \(P(M\cap E)=0.30\neq0.28=P(M)P(E)\). In this sample, training participation is positively associated with passing the exam, but this observational comparison does not by itself prove causation.
Independence of events 30
A study examines whether using a vocabulary app (event \(A\)) is associated with passing a language exam (event \(E\)). The results for \(800\) participants are shown. <table><tr><th></th><th>\(E\)</th><th>\(E^c\)</th><th>Total</th></tr><tr><th>\(A\)</th><td>\(240\)</td><td>\(80\)</td><td>\(320\)</td></tr><tr><th>\(A^c\)</th><td>\(120\)</td><td>\(360\)</td><td>\(480\)</td></tr><tr><th>Total</th><td>\(360\)</td><td>\(440\)</td><td>\(800\)</td></tr></table> a) Use the multiplication test to decide whether app use and exam success are independent. b) Evaluate the claim, “Using the app causes students to pass the exam.”

Hints

- Use the row and column totals to find the two marginal probabilities. - What value would the joint probability have if the events were independent? - Could app users and nonusers differ in ways not measured by the table?

Solution

1. \(P(A)=\frac{320}{800}=0.40\) and \(P(E)=\frac{360}{800}=0.45\). 2. Independence would require \(P(A\cap E)=P(A)P(E)=0.40\cdot0.45=0.18\). 3. But \(P(A\cap E)=\frac{240}{800}=0.30\), so the events are dependent. 4. Students' motivation, prior skill, and study time may affect both app use and passing. The survey does not establish that using the app caused exam success.

Answer

a) Dependent, because \(P(A\cap E)=0.30\ne0.18=P(A)P(E)\). b) An association does not prove that using the app causes success; other differences between students may explain it.
Independence of events 31
A random experiment has sample space \(S=\{z_1,z_2,z_3,z_4,z_5\}\) with the probability distribution shown. <table> <tr><td>Outcome</td><td>\(z_1\)</td><td>\(z_2\)</td><td>\(z_3\)</td><td>\(z_4\)</td><td>\(z_5\)</td></tr> <tr><td>Probability</td><td>\(0.12\)</td><td>\(0.28\)</td><td>\(0.18\)</td><td>\(0.22\)</td><td>\(0.20\)</td></tr> </table> Determine whether the events \(E=\{z_1,z_2\}\) and \(F=\{z_1,z_3\}\) are independent.

Hints

- Add the probabilities of the outcomes contained in each event. - Identify the outcomes that belong to both events. - Independence requires \(P(E\cap F)=P(E)P(F)\). - Compare the joint probability with the product of the marginal probabilities.

Solution

1. Add the probabilities of the outcomes in each event: \(P(E)=0.12+0.28=0.40\) and \(P(F)=0.12+0.18=0.30\). 2. The intersection is \(E\cap F=\{z_1\}\), so \(P(E\cap F)=0.12\). 3. Compute \(P(E)P(F)=0.40\cdot 0.30=0.12\). 4. Since \(P(E\cap F)=P(E)P(F)\), the events are independent.

Answer

The events \(E\) and \(F\) are independent because \(P(E\cap F)=0.12=P(E)P(F)\).
Independence of events 32
A survey asked \(200\) people whether they prefer the beach or the mountains for a vacation. The results are shown by age group. <table> <tr><td></td><td>Beach</td><td>Mountains</td><td>Total</td></tr> <tr><td>Under age \(40\)</td><td>\(72\)</td><td>\(48\)</td><td>\(120\)</td></tr> <tr><td>Age \(40\) or older</td><td>\(48\)</td><td>\(32\)</td><td>\(80\)</td></tr> <tr><td>Total</td><td>\(120\)</td><td>\(80\)</td><td>\(200\)</td></tr> </table> One person is selected at random. Let \(A\) be the event that the person is under age \(40\), and let \(B\) be the event that the person prefers the beach. Determine whether \(A\) and \(B\) are independent.

Hints

- Use the row and column totals to find the marginal probabilities. - Identify the table cell that represents both events occurring. - Independence requires the joint probability to equal the product of the marginal probabilities. - Compare the two values.

Solution

1. From the row and column totals, \(P(A)=\frac{120}{200}=0.60\) and \(P(B)=\frac{120}{200}=0.60\). 2. The joint probability is \(P(A\cap B)=\frac{72}{200}=0.36\). 3. Compute \(P(A)P(B)=0.60\cdot 0.60=0.36\). 4. Since \(P(A\cap B)=P(A)P(B)\), the events are independent.

Answer

The events \(A\) and \(B\) are independent because \(P(A\cap B)=0.36=P(A)P(B)\).
Independence of events 33
A fitness center has \(1500\) members. Of these members, \(900\) regularly use the sauna, \(600\) regularly attend group exercise classes, and \(400\) do both. One member is selected at random. Let \(G\) be the event that the member attends group classes, and let \(S\) be the event that the member uses the sauna. Determine whether \(G\) and \(S\) are independent.

Hints

- Convert each count to a probability using the total number of members. - Find the probability that both events occur. - Independence requires \(P(G\cap S)=P(G)P(S)\). - Compare the joint probability with the product.

Solution

1. The marginal probabilities are \(P(G)=\frac{600}{1500}=0.40\) and \(P(S)=\frac{900}{1500}=0.60\). 2. The joint probability is \(P(G\cap S)=\frac{400}{1500}=\frac{4}{15}\approx 0.2667\). 3. If the events were independent, the joint probability would be \(P(G)P(S)=0.40\cdot 0.60=0.24\). 4. Since \(\frac{4}{15}\neq 0.24\), the events are dependent.

Answer

The events \(G\) and \(S\) are dependent because \(P(G\cap S)=\frac{4}{15}\approx 0.2667\), while \(P(G)P(S)=0.24\).
Independence of events 34
The following two-way probability table for independent events \(E\) and \(F\) is incomplete. <table> <tr><td></td><td>\(F\)</td><td>\(F^c\)</td><td>Total</td></tr> <tr><td>\(E\)</td><td></td><td>\(0.12\)</td><td></td></tr> <tr><td>\(E^c\)</td><td></td><td>\(0.18\)</td><td></td></tr> <tr><td>Total</td><td></td><td></td><td>\(1\)</td></tr> </table> Find all missing probabilities and complete the table.

Hints

- First add the two given entries in the \(F^c\) column. - For independent events, a joint probability equals the product of the corresponding marginal probabilities. - Each pair of complementary marginal probabilities sums to \(1\).

Solution

1. Add the entries in the \(F^c\) column: \(P(F^c)=0.12+0.18=0.30\), so \(P(F)=1-0.30=0.70\). 2. Since \(E\) and \(F\) are independent, \(E\) and \(F^c\) are also independent. Thus, \(P(E\cap F^c)=P(E)P(F^c)\), so \(0.12=P(E)(0.30)\) and \(P(E)=0.40\). 3. Therefore, \(P(E^c)=0.60\). 4. Use independence to find the remaining joint probabilities: \(P(E\cap F)=0.40\cdot0.70=0.28\) and \(P(E^c\cap F)=0.60\cdot0.70=0.42\).

Answer

<table> <tr><td></td><td>\(F\)</td><td>\(F^c\)</td><td>Total</td></tr> <tr><td>\(E\)</td><td>\(0.28\)</td><td>\(0.12\)</td><td>\(0.40\)</td></tr> <tr><td>\(E^c\)</td><td>\(0.42\)</td><td>\(0.18\)</td><td>\(0.60\)</td></tr> <tr><td>Total</td><td>\(0.70\)</td><td>\(0.30\)</td><td>\(1\)</td></tr> </table>
Independence of events 35
Events \(A\) and \(B\) are independent. Suppose \(P(A)=0.60\) and \(P(A\cap B)=0.12\). a) Use independence to find \(P(B)\). b) Make a two-way probability table for \(A\) and \(B\). c) Find \(P(A\cup B)\).

Hints

- For independent events, how is the intersection related to the two individual probabilities? - Find the second marginal probability before completing the table. - Each row and column total is the sum of its two inner cells. - Use the addition rule for the final union.

Solution

1. Independence gives \(P(A\cap B)=P(A)P(B)\). Therefore, \(0.12=0.60P(B)\), so \(P(B)=0.20\). 2. The remaining inner cells are \(P(A\cap B^c)=0.60-0.12=0.48\), \(P(A^c\cap B)=0.20-0.12=0.08\), and \(P(A^c\cap B^c)=1-(0.12+0.48+0.08)=0.32\). 3. The completed table is <table><tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr><tr><th>\(A\)</th><td>\(0.12\)</td><td>\(0.48\)</td><td>\(0.60\)</td></tr><tr><th>\(A^c\)</th><td>\(0.08\)</td><td>\(0.32\)</td><td>\(0.40\)</td></tr><tr><th>Total</th><td>\(0.20\)</td><td>\(0.80\)</td><td>\(1\)</td></tr></table> 4. The addition rule gives \(P(A\cup B)=0.60+0.20-0.12=0.68\).

Answer

a) \(P(B)=0.20\). b) <table><tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr><tr><th>\(A\)</th><td>\(0.12\)</td><td>\(0.48\)</td><td>\(0.60\)</td></tr><tr><th>\(A^c\)</th><td>\(0.08\)</td><td>\(0.32\)</td><td>\(0.40\)</td></tr><tr><th>Total</th><td>\(0.20\)</td><td>\(0.80\)</td><td>\(1\)</td></tr></table> c) \(P(A\cup B)=0.68\).
Independence of events 36
A school surveys \(200\) students about whether they regularly eat breakfast (event \(B\)) and whether they show high concentration in first period (event \(C\)). <table><tr><th></th><th>\(C\) (high)</th><th>\(C^c\) (lower)</th><th>Total</th></tr><tr><th>\(B\) (breakfast)</th><td>\(72\)</td><td>\(48\)</td><td>\(120\)</td></tr><tr><th>\(B^c\) (no breakfast)</th><td>\(18\)</td><td>\(62\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(90\)</td><td>\(110\)</td><td>\(200\)</td></tr></table> a) Show that \(B\) and \(C\) are dependent. b) Find \(P(B\cap C)\) and \(P(C\mid B)\). Describe their association and give a possible factor that could affect both habits without claiming that breakfast causes higher concentration.

Hints

- Use row and column totals to find the two individual probabilities. - Under independence, what should their product equal? - Among students who eat breakfast, which count measures high concentration? - Could another daily habit affect both outcomes?

Solution

1. \(P(B)=\frac{120}{200}=0.60\) and \(P(C)=\frac{90}{200}=0.45\). 2. The joint probability is \(P(B\cap C)=\frac{72}{200}=0.36\), but independence would require \(P(B)P(C)=0.60\cdot0.45=0.27\). Thus the events are dependent. 3. Among students who eat breakfast, \(P(C\mid B)=\frac{72}{120}=0.60\), greater than the overall \(0.45\). The sample shows a positive association. 4. Sleep, household routines, or access to food could affect both habits; this survey does not establish causation.

Answer

a) Dependent, because \(P(B\cap C)=0.36\ne0.27=P(B)P(C)\). b) \(P(B\cap C)=0.36\) and \(P(C\mid B)=0.60\). The events are associated; for example, sleep habits may influence both breakfast and concentration.
Independence of events 37
Students prepare for a quiz using one of two methods: flashcards (event \(T\)) or practice questions (event \(T^c\)). Event \(R\) means a student passed. <table><tr><th></th><th>Passed</th><th>Did not pass</th></tr><tr><th>Flashcards</th><td>\(120\)</td><td>\(80\)</td></tr><tr><th>Practice questions</th><td>\(45\)</td><td>\(y\)</td></tr></table> 1. Find \(y\) so that passing is independent of the study method. 2. Explain what your result means.

Hints

- What should be true of the two groups' pass rates if the events are independent? - Which total belongs in the denominator for each method? - How can the second group's pass rate be expressed using \(y\)?

Solution

1. Passing is independent of the study method when the two groups have the same pass rate. 2. The flashcard group's pass rate is \(\frac{120}{120+80}=0.60\). The practice-question group's pass rate is \(\frac{45}{45+y}\). 3. Solve \(\frac{45}{45+y}=0.60\). Then \(45=27+0.60y\), so \(y=30\). 4. Both study groups then have a \(60\%\) pass rate. In this table, whether a student passed does not depend on which method the student used.

Answer

1. \(y=30\). 2. Both groups have a \(60\%\) pass rate, so passing and study method are independent in this table.
Independence of events 38
A bag contains \(10\) balls: \(2\) blue, \(3\) red, and \(5\) yellow. Two balls are drawn in sequence. Let \(B\) be the event that the first ball is blue, and let \(R\) be the event that the second ball is red. a) Without replacement, find \(P(R)\) and \(P(R\mid B)\). Are \(B\) and \(R\) independent? b) With replacement after the first draw, determine whether \(B\) and \(R\) are independent.

Hints

- After a blue ball is removed, identify the new total number of balls and the number of red balls. - A tree diagram can organize the possible first-draw colors that lead to red on the second draw. - Compare \(P(R\mid B)\) with \(P(R)\) in each situation.

Solution

1. Without replacement, split according to the color of the first ball: \(P(R)=\frac{2}{10}\cdot\frac{3}{9}+\frac{3}{10}\cdot\frac{2}{9}+\frac{5}{10}\cdot\frac{3}{9}=\frac{3}{10}=0.30\). 2. Given that the first ball is blue, \(3\) of the remaining \(9\) balls are red, so \(P(R\mid B)=\frac{3}{9}=\frac{1}{3}\). 3. Since \(P(R\mid B)\neq P(R)\), the events are dependent without replacement. 4. With replacement, the bag again contains \(3\) red balls out of \(10\) before the second draw. Thus, \(P(R)=\frac{3}{10}\) and \(P(R\mid B)=\frac{3}{10}\). 5. Since these probabilities are equal, the events are independent with replacement.

Answer

a) \(P(R)=0.30\) and \(P(R\mid B)=\frac{1}{3}\). The events are dependent. b) \(P(R)=P(R\mid B)=0.30\). The events are independent.
Independence of events 39
Two jars sit on a table. The left jar has \(5\) red chips and \(15\) blue chips. The right jar has \(10\) red chips and \(30\) blue chips. First choose a jar at random, with equal chances, then draw one chip. Let \(L\) mean choosing the left jar and \(R\) mean drawing red. a) Show that \(L\) and \(R\) are independent at first. b) You can add \(15\) red chips. Plan 1 adds \(5\) to the left jar and \(10\) to the right jar. Plan 2 adds \(10\) to the left jar and \(5\) to the right jar. Which plan keeps \(L\) and \(R\) independent? c) Instead, add \(5\) red chips to the left jar and \(10\) blue chips to the right jar. Are \(L\) and \(R\) independent now?

Hints

- Compare red chips as a fraction of all chips in each jar. - When chips are added, which counts change in the numerator and denominator? - Would knowing which jar was chosen change your chance of drawing red?

Solution

1. Initially \(P(R\mid L)=\frac{5}{20}=0.25\) and \(P(R\mid L^c)=\frac{10}{40}=0.25\), so drawing red does not depend on the jar chosen. 2. Under Plan 1, the left jar has \(10\) red out of \(25\), and the right jar has \(20\) red out of \(50\). Both fractions equal \(0.40\). 3. Under Plan 2, the red proportions are \(\frac{15}{30}=\frac{1}{2}\) and \(\frac{15}{45}=\frac{1}{3}\), so these events are dependent. 4. In part c), the proportions are \(\frac{10}{25}=0.40\) and \(\frac{10}{50}=0.20\). They differ, so the events are dependent.

Answer

a) Independent; the two red proportions both equal \(0.25\). b) Plan 1 keeps independence: both red proportions are \(0.40\). c) Dependent: \(0.40\ne0.20\).
Independence of events 40
A market-research survey includes \(200\) people: \(120\) teenagers and \(80\) adults. Overall, \(60\) respondents say they would buy a new soft drink. Let \(T\) be the event that a selected respondent is a teenager, and let \(B\) be the event that the respondent would buy the drink. a) How many teenagers and how many adults would need to be interested in buying the drink for \(T\) and \(B\) to be independent? b) Suppose \(40\) teenagers are actually interested in buying the drink. Show that age group and purchase interest are dependent. c) Under the condition in part b, determine without further calculation whether \(T^c\) and \(B^c\) are independent or dependent. Explain briefly.

Hints

- For independence, the purchase-interest rate in each age group must equal the overall rate. - In part b, use the total of \(60\) interested respondents to find the number of interested adults. - Recall how independence behaves when events are replaced by their complements.

Solution

1. The overall purchase-interest rate is \(P(B)=\frac{60}{200}=0.30\). For independence, each age group must have this same rate. 2. The required counts are \(120\cdot0.30=36\) teenagers and \(80\cdot0.30=24\) adults. 3. If \(40\) teenagers are interested, then \(P(B\mid T)=\frac{40}{120}=\frac{1}{3}\). The remaining \(20\) interested respondents are adults, so \(P(B\mid T^c)=\frac{20}{80}=0.25\). 4. Since the conditional probabilities differ, \(T\) and \(B\) are dependent. 5. Dependence is preserved when both events are replaced by their complements. Therefore, \(T^c\) and \(B^c\) are also dependent.

Answer

a) \(36\) teenagers and \(24\) adults. b) \(P(B\mid T)=\frac{1}{3}\neq0.25=P(B\mid T^c)\), so the events are dependent. c) \(T^c\) and \(B^c\) are dependent because replacing both events with their complements preserves independence or dependence.
Independence of events 41
A fair six-sided die is rolled twice. Consider these events: \(A\): The first roll is a prime number. \(B\): The second roll is greater than \(4\). \(C\): The sum of the two rolls is exactly \(6\). a) Show that \(A\) and \(B\) are independent. b) Determine whether \(A\) and \(C\) are independent.

Hints

- Independence requires the joint probability to equal the product of the individual probabilities. - There are \(36\) equally likely ordered outcomes for two rolls. - List the ordered pairs that satisfy each event. - Compare each joint probability with the corresponding product.

Solution

1. The prime faces are \(2,3,5\), so \(P(A)=\frac{3}{6}=\frac{1}{2}\). The faces greater than \(4\) are \(5,6\), so \(P(B)=\frac{2}{6}=\frac{1}{3}\). 2. There are \(3\cdot 2=6\) ordered outcomes in \(A\cap B\), so \(P(A\cap B)=\frac{6}{36}=\frac{1}{6}\). 3. Since \(P(A)P(B)=\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}=P(A\cap B)\), the events are independent. 4. The outcomes whose sum is \(6\) are \((1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\), so \(P(C)=\frac{5}{36}\). 5. The outcomes in both \(A\) and \(C\) are \((2, 4), (3, 3), (5, 1)\), so \(P(A\cap C)=\frac{3}{36}=\frac{1}{12}\). 6. Since \(P(A)P(C)=\frac{1}{2}\cdot\frac{5}{36}=\frac{5}{72}\neq\frac{1}{12}\), the events are dependent.

Answer

a) \(A\) and \(B\) are independent because \(P(A\cap B)=\frac{1}{6}=P(A)P(B)\). b) \(A\) and \(C\) are dependent because \(P(A\cap C)=\frac{1}{12}\neq\frac{5}{72}=P(A)P(C)\).
Independence of events 42
A survey of smartphone users found that \(40\%\) also own a laptop made by the same company (event \(L\)). Overall, \(55\%\) are very satisfied with the company's customer service (event \(S\)). Both statements are true for \(25\%\) of respondents. a) Complete the two-way probability table. <table> <tr><td></td><td>\(L\)</td><td>\(L^c\)</td><td>Total</td></tr> <tr><td>\(S\)</td><td>\(25\%\)</td><td></td><td>\(55\%\)</td></tr> <tr><td>\(S^c\)</td><td></td><td></td><td></td></tr> <tr><td>Total</td><td>\(40\%\)</td><td></td><td>\(100\%\)</td></tr> </table> b) Determine whether \(L\) and \(S\) are independent. c) Find \(P(S\mid L)\) and \(P(S\mid L^c)\). Interpret each probability in context. d) Describe the association between owning the company's laptop and satisfaction with customer service.

Hints

- Use each row or column total to find its missing inner cell. - Independence requires \(P(L\cap S)=P(L)P(S)\). - For a conditional probability, restrict the denominator to the group named after the vertical bar. - Compare the two conditional satisfaction rates.

Solution

1. Use the row and column totals: \(P(L\cap S^c)=0.40-0.25=0.15\), \(P(L^c\cap S)=0.55-0.25=0.30\), and \(P(L^c\cap S^c)=1-(0.25+0.15+0.30)=0.30\). The remaining margins are \(P(L^c)=0.60\) and \(P(S^c)=0.45\). 2. For independence, \(P(L\cap S)\) must equal \(P(L)P(S)\). Here, \(P(L)P(S)=0.40\cdot0.55=0.22\), while \(P(L\cap S)=0.25\). Therefore, the events are dependent. 3. The conditional probabilities are \(P(S\mid L)=\frac{0.25}{0.40}=0.625\) and \(P(S\mid L^c)=\frac{0.30}{0.60}=0.50\). 4. Satisfaction is more common among respondents who also own the company's laptop. This is an association in the survey data and does not by itself establish causation.

Answer

a) <table> <tr><td></td><td>\(L\)</td><td>\(L^c\)</td><td>Total</td></tr> <tr><td>\(S\)</td><td>\(25\%\)</td><td>\(30\%\)</td><td>\(55\%\)</td></tr> <tr><td>\(S^c\)</td><td>\(15\%\)</td><td>\(30\%\)</td><td>\(45\%\)</td></tr> <tr><td>Total</td><td>\(40\%\)</td><td>\(60\%\)</td><td>\(100\%\)</td></tr> </table> b) The events are dependent because \(P(L\cap S)=0.25\neq0.22=P(L)P(S)\). c) \(P(S\mid L)=0.625\), or \(62.5\%\), is the satisfaction rate among respondents who own the company's laptop. \(P(S\mid L^c)=0.50\), or \(50\%\), is the satisfaction rate among those who do not. d) Laptop ownership and service satisfaction are positively associated in this sample.
Independence of events 43
A distribution center receives packages from two shipping providers. Provider A supplies \(60\%\) of the packages (event \(A\)); Provider B supplies the rest. Overall, \(5\%\) of packages arrive damaged (event \(D\)), and \(3\%\) of all packages are both from Provider A and damaged. a) Create a complete two-way probability table. b) Determine whether package damage is independent of the provider. c) Find \(P(D\mid A)\) and \(P(D\mid B)\), and interpret them in context. d) Compare the providers with respect to the observed damage rate.

Hints

- Use the given marginal probabilities and joint probability to fill the table. - Test whether the joint probability equals the product of the marginal probabilities. - For each provider's damage rate, use only packages from that provider as the denominator. - Base the comparison only on the damage data provided.

Solution

1. The marginal probabilities are \(P(A)=0.60\), \(P(B)=0.40\), \(P(D)=0.05\), and \(P(D^c)=0.95\). 2. Complete the inner cells: \(P(A\cap D^c)=0.60-0.03=0.57\), \(P(B\cap D)=0.05-0.03=0.02\), and \(P(B\cap D^c)=0.40-0.02=0.38\). 3. Since \(P(A)P(D)=0.60\cdot0.05=0.03=P(A\cap D)\), the events are independent. 4. The conditional damage rates are \(P(D\mid A)=\frac{0.03}{0.60}=0.05\) and \(P(D\mid B)=\frac{0.02}{0.40}=0.05\). 5. Both providers have an observed damage rate of \(5\%\). These data show no difference in this measure, but they do not compare other aspects of service quality.

Answer

a) <table> <tr><td></td><td>\(D\)</td><td>\(D^c\)</td><td>Total</td></tr> <tr><td>\(A\)</td><td>\(3\%\)</td><td>\(57\%\)</td><td>\(60\%\)</td></tr> <tr><td>\(B\)</td><td>\(2\%\)</td><td>\(38\%\)</td><td>\(40\%\)</td></tr> <tr><td>Total</td><td>\(5\%\)</td><td>\(95\%\)</td><td>\(100\%\)</td></tr> </table> b) Damage and provider are independent because \(P(A\cap D)=0.03=P(A)P(D)\). c) \(P(D\mid A)=5\%\) and \(P(D\mid B)=5\%\). Each is the damage rate among packages from that provider. d) The providers have the same observed damage rate, so neither has an advantage on this measure.
Independence of events 44
Bag A has \(6\) red balls and \(4\) blue balls. Bag B has \(2\) red balls and \(4\) blue balls. A ball is selected at random from Bag A and moved to Bag B. Then a ball is selected at random from Bag B and moved to Bag A. Let \(R\) be the event that the first ball moved is red. Let \(G\) be the event that Bag A ends with its original numbers of red and blue balls. a) Find \(P(G)\). b) Find \(P(G\mid R)\). c) Are \(R\) and \(G\) independent? Explain.

Hints

- What color must move back to Bag A for its original composition to be restored? - Consider separately the cases in which the first ball is red and blue. - After either first transfer, how many balls are in Bag B? - Compare the chance of restoring Bag A after a red first transfer with the overall chance of restoring it.

Solution

1. Bag A returns to its original colors if both transfers move red balls or both move blue balls. 2. If red moves first, Bag B has \(3\) red balls out of \(7\). This path has probability \(\frac{6}{10}\cdot\frac{3}{7}=\frac{9}{35}\). 3. If blue moves first, Bag B has \(5\) blue balls out of \(7\). This path has probability \(\frac{4}{10}\cdot\frac{5}{7}=\frac{2}{7}\). 4. Add the disjoint paths: \(P(G)=\frac{9}{35}+\frac{2}{7}=\frac{19}{35}\). 5. Given that the first ball moved is red, Bag A is restored only if a red ball moves back, so \(P(G\mid R)=\frac{3}{7}\). 6. Because \(\frac{3}{7}\neq\frac{19}{35}\), the events are not independent.

Answer

a) \(P(G)=\frac{19}{35}\). b) \(P(G\mid R)=\frac{3}{7}\). c) No. \(P(G\mid R)\neq P(G)\), so the events are dependent.
Independence of events 45
A data center records whether Server \(S_1\) and Server \(S_2\) experience any downtime on each of \(20\) days. An “X” means that the server had at least one outage that day. <table> <tr><td>Day</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td></tr> <tr><td>\(D_1\)</td><td></td><td>X</td><td></td><td></td><td>X</td><td></td><td>X</td><td></td><td></td><td>X</td></tr> <tr><td>\(D_2\)</td><td></td><td>X</td><td></td><td></td><td></td><td></td><td>X</td><td></td><td></td><td></td></tr> </table> <br> <table> <tr><td>Day</td><td>\(11\)</td><td>\(12\)</td><td>\(13\)</td><td>\(14\)</td><td>\(15\)</td><td>\(16\)</td><td>\(17\)</td><td>\(18\)</td><td>\(19\)</td><td>\(20\)</td></tr> <tr><td>\(D_1\)</td><td></td><td>X</td><td></td><td></td><td>X</td><td></td><td>X</td><td></td><td></td><td>X</td></tr> <tr><td>\(D_2\)</td><td></td><td>X</td><td></td><td></td><td></td><td></td><td>X</td><td>X</td><td>X</td><td></td></tr> </table> Network maintenance (event \(M\)) occurred on days \(2,3,7,8,12\). a) Use relative frequencies to determine whether \(D_1\) and \(D_2\) are independent. b) Determine whether outages on Server \(S_1\) are independent of network maintenance. Interpret the result in context.

Hints

- Count the days on which each event occurs, then divide by \(20\). - Count the days on which both events occur. - Independence requires the joint relative frequency to equal the product of the two marginal relative frequencies. - For part b, compare the outage rate on maintenance days with the overall outage rate.

Solution

1. Server \(S_1\) had outages on \(8\) of the \(20\) days, so \(P(D_1)=\frac{8}{20}=0.40\). Server \(S_2\) had outages on \(6\) days, so \(P(D_2)=\frac{6}{20}=0.30\). 2. Both servers had outages on days \(2,7,12,17\), so \(P(D_1\cap D_2)=\frac{4}{20}=0.20\). 3. Since \(P(D_1)P(D_2)=0.40\cdot 0.30=0.12\neq 0.20\), the outage events are dependent in these observations. 4. Maintenance occurred on \(5\) days, so \(P(M)=\frac{5}{20}=0.25\). Server \(S_1\) had an outage on three maintenance days: \(2,7,12\). Thus, \(P(D_1\cap M)=\frac{3}{20}=0.15\). 5. Since \(P(D_1)P(M)=0.40\cdot 0.25=0.10\neq 0.15\), outages on \(S_1\) and maintenance are dependent in these observations. 6. Also, \(P(D_1\mid M)=\frac{3}{5}=0.60\), which is greater than the overall outage rate of \(0.40\). This shows an association, not necessarily that maintenance caused the outages.

Answer

a) \(P(D_1)=0.40\), \(P(D_2)=0.30\), and \(P(D_1\cap D_2)=0.20\). Because \(0.20\neq 0.40\cdot 0.30\), the events are dependent. b) \(P(M)=0.25\) and \(P(D_1\cap M)=0.15\). Because \(0.15\neq 0.40\cdot 0.25\), the events are dependent. The observed outage rate for \(S_1\) was higher on maintenance days, but this does not by itself prove causation.
Independence of events 46
A data center has \(5000\) servers. A security system \((S)\) is installed on \(85\%\) of them. After a wave of cyberattacks, \(350\) servers were successfully compromised (event \(C\)); \(150\) of those compromised servers had the security system. a) Find \(P(C\mid S)\) and \(P(C\mid S^c)\). b) Show that \(C\) and \(S\) are dependent. c) A technician says, “The security system is not very effective because almost half of the compromised servers had it installed.” Evaluate this statement using your results from part a, and explain the limitation of making a causal conclusion from these data.

Hints

- First find how many servers have the system and how many do not. - Use each security group as the denominator for its conditional compromise rate. - Compare \(P(C\mid S)\) with either \(P(C)\) or \(P(C\mid S^c)\). - Distinguish a statistical association from evidence of causation.

Solution

1. The number of servers with the system is \(0.85\cdot5000=4250\), so \(750\) servers do not have it. 2. Of the \(350\) compromised servers, \(350-150=200\) did not have the system. 3. Therefore, \(P(C\mid S)=\frac{150}{4250}=\frac{3}{85}\approx0.0353\), and \(P(C\mid S^c)=\frac{200}{750}=\frac{4}{15}\approx0.2667\). 4. The overall compromise rate is \(P(C)=\frac{350}{5000}=0.07\). Since \(P(C\mid S)\neq P(C)\), the events are dependent. 5. The technician's comparison ignores the very different group sizes. The observed compromise rate is about \(3.53\%\) with the system and \(26.67\%\) without it. This association is consistent with a protective effect, but without a controlled study it does not prove that the system alone caused the difference.

Answer

a) \(P(C\mid S)=\frac{3}{85}\approx3.53\%\), and \(P(C\mid S^c)=\frac{4}{15}\approx26.67\%\). b) The events are dependent because \(P(C\mid S)\approx0.0353\neq0.07=P(C)\). c) The statement is misleading because it ignores the base rates: most servers had the system. The compromise rate was much lower among protected servers, although these observational data alone do not prove causation.
Independence of events 47
In a survey, \(40\%\) of respondents exercise regularly (event \(E\)), \(55\%\) are satisfied with their fitness (event \(F\)), and \(35\%\) report both. a) Find \(P(F\mid E)\) and \(P(F\mid E^c)\). b) Are \(E\) and \(F\) independent? Explain. c) Does this survey prove that exercise causes greater fitness satisfaction? Explain.

Hints

- Use exercisers, not all respondents, as the reference group for the first conditional probability. - Find the percentage who are satisfied but do not exercise regularly before calculating the second conditional probability. - If satisfaction were independent of exercise, how would the two group-specific rates compare? - Could another factor influence both exercise and satisfaction?

Solution

1. Among regular exercisers, \(P(F\mid E)=\frac{0.35}{0.40}=0.875\). 2. The proportion who are satisfied but do not exercise regularly is \(0.55-0.35=0.20\), and \(P(E^c)=0.60\). Therefore, \(P(F\mid E^c)=\frac{0.20}{0.60}=\frac{1}{3}\). 3. The conditional satisfaction rates differ, so exercise and satisfaction are dependent in these survey results. 4. This is an association, not proof of cause. Health, lifestyle, or other differences may affect both responses.

Answer

a) \(P(F\mid E)=0.875\) and \(P(F\mid E^c)=\frac{1}{3}\). b) No. The satisfaction rates differ between the two groups. c) No. The survey shows an association but cannot establish that exercise caused the difference.
Independence of events 48
On an online learning platform, \(60\%\) of students complete all optional practice materials (event \(M\)). Overall, \(65\%\) pass the final exam (event \(E\)). Another \(10\%\) of all students pass without completing all the practice materials. a) Find \(P(E\mid M)\). b) Are passing and completing the materials independent? Explain. c) Why don't these results prove that the materials caused students to pass?

Hints

- Which students passed but did not complete all the practice? - Remove those students from the overall pass group to find how many did both. - Use the group that completed practice as the denominator of the requested conditional probability. - Does an observed association rule out other possible explanations?

Solution

1. Of all students, \(P(M\cap E)=P(E)-P(M^c\cap E)=0.65-0.10=0.55\). 2. Among students who completed the materials, \(P(E\mid M)=\frac{0.55}{0.60}=\frac{11}{12}\approx0.9167\). 3. For independent events, \(P(M\cap E)\) would equal \(P(M)P(E)=0.60\cdot0.65=0.39\). The actual joint probability is \(0.55\), so the events are dependent. 4. Students may differ in motivation or prior knowledge. The association alone does not establish a cause.

Answer

a) \(P(E\mid M)=\frac{11}{12}\approx0.9167\). b) No. \(P(M\cap E)=0.55\neq0.39=P(M)P(E)\). c) Other factors, such as motivation or prior knowledge, may affect both practice completion and exam success.
Independence of events 49
A factory produces \(500\) parts per day. The evening shift produces \(200\) parts, and the day shift produces the remaining parts. At the end of one day, \(40\) defective parts are found, including \(24\) from the evening shift. One part is selected at random. Let \(E\) be the event that the part was made during the evening shift, and let \(D\) be the event that the part is defective. Determine whether defect status is independent of production shift. Explain one appropriate response for quality control.

Hints

- Find the proportions of all parts that came from the evening shift, were defective, and met both conditions. - Test whether the joint probability equals the product of the marginal probabilities. - Compare the evening-shift defect rate with the overall defect rate. - Recommend investigation without assuming the data establish a cause.

Solution

1. The probabilities are \(P(E)=\frac{200}{500}=0.40\), \(P(D)=\frac{40}{500}=0.08\), and \(P(E\cap D)=\frac{24}{500}=0.048\). 2. If the events were independent, \(P(E\cap D)\) would equal \(P(E)P(D)=0.40\cdot 0.08=0.032\). 3. Since \(0.048\neq 0.032\), defect status and production shift are dependent in this day's data. 4. The evening-shift defect rate is \(P(D\mid E)=\frac{24}{200}=0.12\), compared with an overall defect rate of \(0.08\). Quality control should investigate evening-shift materials, equipment, and procedures rather than assuming a cause from this single day alone.

Answer

The events are dependent because \(P(E\cap D)=0.048\neq 0.032=P(E)P(D)\). The evening-shift defect rate was \(12\%\), above the overall rate of \(8\%\), so quality control should investigate the shift's production conditions.
Independence of events 50
The spinner shown has \(10\) equal sections. It is spun twice. a) Find the probability that the same color appears on both spins. b) Find the probability that red appears at least once. c) An urn contains \(10\) balls in the same colors and quantities as the spinner sections. Two balls are drawn without replacement. Explain why the probability of drawing the same color twice is less than the probability in part a.
Figure for problem 535936

Hints

- Read the one-spin color probabilities from the spinner. - For part a, add the probabilities of the three same-color paths. - For part b, use the complement of getting no red. - For part c, compare what changes after the first ball is removed.

Solution

1. From the spinner, \(P(\text{green})=\frac{4}{10}\), \(P(\text{blue})=\frac{3}{10}\), and \(P(\text{red})=\frac{3}{10}\). The two spinner results are independent. 2. The same color appears twice through green-green, blue-blue, or red-red. Therefore, \(P(\text{same color})=\left(\frac{4}{10}\right)^2+\left(\frac{3}{10}\right)^2+\left(\frac{3}{10}\right)^2=\frac{17}{50}=34\%\). 3. For part b, the complement of at least one red is no red. Thus, \(P(\text{at least one red})=1-\left(\frac{7}{10}\right)^2=\frac{51}{100}=51\%\). 4. Without replacement, the second draw depends on the first because one ball of the first color has been removed. The probability of matching colors is \(\frac{4}{10}\cdot\frac{3}{9}+\frac{3}{10}\cdot\frac{2}{9}+\frac{3}{10}\cdot\frac{2}{9}=\frac{4}{15}\approx26.7\%\), which is less than \(34\%\).

Answer

a) \(\frac{17}{50}=34\%\) b) \(\frac{51}{100}=51\%\) c) Without replacement, the second draw is dependent on the first and has one fewer ball of the first color available. The same-color probability is \(\frac{4}{15}\approx26.7\%\), which is less than \(34\%\).
Independence of events 51
A bag contains \(2\) red balls and \(2\) blue balls. Two balls are drawn in sequence without replacement. Let \(A\) be the event that the first ball is red, and let \(B\) be the event that the second ball is red. Determine whether \(A\) and \(B\) are independent.

Hints

- Find the probability of red on each draw. - Find the probability that both draws are red. - Compare the joint probability with the product of the marginal probabilities.

Solution

1. Since \(2\) of the \(4\) balls are red, \(P(A)=\frac{2}{4}=\frac{1}{2}\). 2. By symmetry, \(P(B)=\frac{1}{2}\). Equivalently, \(P(B)=\frac{2}{4}\cdot\frac{1}{3}+\frac{2}{4}\cdot\frac{2}{3}=\frac{1}{2}\). 3. The event \(A\cap B\) means that both selected balls are red, so \(P(A\cap B)=\frac{2}{4}\cdot\frac{1}{3}=\frac{1}{6}\). 4. Since \(P(A)P(B)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\neq\frac{1}{6}\), the events are dependent.

Answer

The events are dependent because \(P(A\cap B)=\frac{1}{6}\neq\frac{1}{4}=P(A)P(B)\).
Independence of events 52
A bag contains \(2\) red balls and \(3\) blue balls. Two balls are drawn in sequence with replacement. Let \(A\) be the event that the first ball is red, and let \(B\) be the event that exactly one red ball is drawn. Determine whether \(A\) and \(B\) are independent.

Hints

- List the two ordered color sequences that contain exactly one red ball. - Find \(P(A)\), \(P(B)\), and \(P(A\cap B)\). - With replacement, the color probabilities remain the same on the second draw. - Compare the joint probability with the product of the marginal probabilities.

Solution

1. On either draw, \(P(\text{red})=\frac{2}{5}=0.40\) and \(P(\text{blue})=\frac{3}{5}=0.60\). 2. Therefore, \(P(A)=0.40\). 3. Exactly one red ball occurs in the sequences red-blue or blue-red, so \(P(B)=0.40\cdot0.60+0.60\cdot0.40=0.48\). 4. The event \(A\cap B\) requires red on the first draw and blue on the second, so \(P(A\cap B)=0.40\cdot0.60=0.24\). 5. Since \(P(A)P(B)=0.40\cdot0.48=0.192\neq0.24\), the events are dependent.

Answer

The events are dependent because \(P(A\cap B)=0.24\neq0.192=P(A)P(B)\).
Independence of events 53
The probability tree represents two independent events \(A\) and \(B\). Find \(P(B)\).
Figure for problem 537482

Hints

- Use the first-stage probability to find \(P(A^c)\). - Read the joint probability at the end of the \(A^c\)-then-\(B\) path. - Independence lets you write the path probability as a product. - Solve the resulting equation for \(P(B)\).

Solution

1. Since \(P(A)=0.80\), the complement has probability \(P(A^c)=1-0.80=0.20\). 2. The tree gives \(P(A^c\cap B)=0.06\). 3. Because \(A\) and \(B\) are independent, \(A^c\) and \(B\) are also independent. Thus, \(P(A^c\cap B)=P(A^c)P(B)\). 4. Solve \(0.06=0.20P(B)\) to get \(P(B)=0.30\).

Answer

\(P(B)=0.30\)
Independence of events 54
A city library surveys patrons about e-book use. The tree diagram groups the respondents as younger \((Y)\) or older \((O)\), then as e-book users \((E)\) or nonusers \((E^c)\). a) Assume \(1000\) people were surveyed. Create a complete two-way table of counts. b) Find \(P(Y\mid E)\) and interpret it in context. c) Determine whether age group and e-book use are independent. Support your conclusion with probabilities.
Figure for problem 537487

Hints

- Multiply along each path to find the joint probabilities. - Multiply each joint probability by \(1000\) to obtain a count. - For the conditional probability, identify the conditioned group first. - Independence requires \(P(Y\cap E)=P(Y)P(E)\).

Solution

1. The age-group counts are \(n(Y)=0.40\cdot1000=400\) and \(n(O)=0.60\cdot1000=600\). 2. The joint counts are \(n(Y\cap E)=400\cdot0.75=300\), \(n(Y\cap E^c)=400\cdot0.25=100\), \(n(O\cap E)=600\cdot0.30=180\), and \(n(O\cap E^c)=600\cdot0.70=420\). 3. Thus, \(n(E)=300+180=480\) and \(n(E^c)=100+420=520\). 4. \(P(Y\mid E)=\frac{300}{480}=0.625\). Therefore, \(62.5\%\) of the e-book users are in the younger group. 5. For independence, compare \(P(Y\cap E)=0.300\) with \(P(Y)P(E)=0.40\cdot0.48=0.192\). Since they are not equal, the events are dependent.

Answer

a) <table> <tr><th></th><th>\(E\)</th><th>\(E^c\)</th><th>Total</th></tr> <tr><th>\(Y\)</th><td>\(300\)</td><td>\(100\)</td><td>\(400\)</td></tr> <tr><th>\(O\)</th><td>\(180\)</td><td>\(420\)</td><td>\(600\)</td></tr> <tr><th>Total</th><td>\(480\)</td><td>\(520\)</td><td>\(1000\)</td></tr> </table> b) \(P(Y\mid E)=0.625\). This means that \(62.5\%\) of the e-book users are in the younger group. c) The events are dependent because \(P(Y\cap E)=0.300\neq0.192=P(Y)P(E)\).
Independence of events 55
A market survey records purchase interest in a new product (event \(B\)) by age group. Event \(U\) means under age \(30\), and \(U^c\) means age \(30\) or older. The table shows the relative frequencies. <table> <tr><td></td><td>\(B\)</td><td>\(B^c\)</td><td>Total</td></tr> <tr><td>\(U\)</td><td>\(0.06\)</td><td>\(0.14\)</td><td>\(0.20\)</td></tr> <tr><td>\(U^c\)</td><td>\(0.24\)</td><td>\(0.56\)</td><td>\(0.80\)</td></tr> <tr><td>Total</td><td>\(0.30\)</td><td>\(0.70\)</td><td>\(1.00\)</td></tr> </table> Name one feature of the table that shows quickly that \(U\) and \(B\) are independent.

Hints

- Compare an inner table cell with the product of its row and column totals. - Alternatively, compare the conditional proportions within the two rows.

Solution

1. Multiply the marginal probabilities: \(P(U)P(B)=0.20\cdot0.30=0.06\). 2. This equals the joint probability \(P(U\cap B)=0.06\), so the events are independent. 3. Equivalently, the conditional purchase-interest rates are equal in both age groups: \(\frac{0.06}{0.20}=0.30\) and \(\frac{0.24}{0.80}=0.30\).

Answer

One valid feature is \(P(U\cap B)=0.06=0.20\cdot0.30=P(U)P(B)\). Another is that both age groups have the same conditional purchase-interest rate, \(0.30\).
Independence of events 56
A bag contains \(4\) green balls and \(2\) yellow balls. Compare the probability of drawing the same color twice when drawing with replacement and without replacement. Tree diagram a) represents drawing with replacement, and tree diagram b) represents drawing without replacement.
Figure for problem 537504

Hints

- Calculate the green-green and yellow-yellow paths in each model. - Determine whether the color proportions change after the first draw.

Solution

1. With replacement, the probability of the same color twice is \(\left(\frac{4}{6}\right)^2 + \left(\frac{2}{6}\right)^2 = \frac{5}{9}\). 2. Without replacement, the probability is \(\frac{4}{6} \cdot \frac{3}{5} + \frac{2}{6} \cdot \frac{1}{5} = \frac{7}{15}\). 3. Since \(\frac{5}{9} > \frac{7}{15}\), drawing the same color twice is more likely with replacement. With replacement, the two draws are independent; without replacement, the second draw depends on the first.

Answer

With replacement: \(\frac{5}{9}\); without replacement: \(\frac{7}{15}\). The event is more likely with replacement.
Independence of events 57
For two events, \(P(A)=0.6\), \(P(B)=0.4\), and \(P(A\cap B)=0.3\). Determine whether the events are independent, and find \(P(B\mid A)\).

Hints

- Compare the observed intersection with the product of the two individual probabilities. - For the conditional probability, restrict the sample space to event \(A\).

Solution

1. If the events were independent, the intersection probability would be \(P(A)P(B)=0.6\cdot0.4=0.24\). 2. Since \(0.3\neq0.24\), the events are not independent. 3. The conditional probability is \(P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{0.3}{0.6}=0.5\).

Answer

The events are not independent, and \(P(B\mid A)=0.5\).
Independence of events 58
Explain how this probability tree shows that events \(A\) and \(B\) are independent even though \(P(A)\) is not known.
Figure for problem 537569

Hints

- Look at the two branches that lead to \(B\). - Add the two path probabilities to express the overall chance of \(B\). - Does that sum depend on the unknown first-stage probability?

Solution

1. Let \(P(A)=p\). The other first-stage probability is \(1-p\). 2. Add the two paths leading to \(B\): \(P(B)=0.70p+0.70(1-p)=0.70\). 3. Because the tree gives \(P(B\mid A)=0.70=P(B)\), the events are independent.

Answer

Both branches leading to \(B\) have conditional probability \(0.70\), so the overall probability of \(B\) is also \(0.70\), regardless of \(P(A)\). The events are independent.
Independence of events 59
A bag contains \(2\) red balls and \(3\) blue balls. Compare two draws with replacement in tree a) and two draws without replacement in tree b). Let \(R_1\) be the event that the first ball is red, and let \(R_2\) be the event that the second ball is red. In which model are \(R_1\) and \(R_2\) independent? Explain.
Figure for problem 537579

Hints

- Compare \(P(R_2\mid R_1)\) with the overall probability \(P(R_2)\) in each model. - Decide in which model the first draw changes the composition of the bag.

Solution

1. With replacement, \(P(R_2)=\frac{2}{5}\), and \(P(R_2\mid R_1)=\frac{2}{5}\). Therefore, the events are independent. 2. Without replacement, the marginal probability is still \(P(R_2)=\frac{2}{5}\), but after a red first draw, \(P(R_2\mid R_1)=\frac{1}{4}\). 3. Since \(\frac{1}{4}\neq\frac{2}{5}\), the events are dependent without replacement. The first draw changes the contents of the bag.

Answer

The events are independent only when drawing with replacement. Without replacement, the first draw changes the probability of red on the second draw.
Independence of events 60
Two fair coins are tossed. Let \(A\) mean the first coin shows heads. a) Let \(B\) mean the two coins match. Are \(A\) and \(B\) independent? Show your probabilities. b) Now let \(C\) mean both coins show heads. Are \(A\) and \(C\) independent? Explain what changed.

Hints

- Write out the four ordered outcomes. - For each event pair, first list the outcomes that satisfy both conditions. - Can one of the new events happen without the other?

Solution

1. The equally likely outcomes are \(HH,HT,TH,TT\). Event \(A\) has two outcomes, so \(P(A)=\frac{1}{2}\). 2. Event \(B\) contains \(HH,TT\), giving \(P(B)=\frac{1}{2}\). Only \(HH\) satisfies both \(A\) and \(B\), so \(P(A\cap B)=\frac{1}{4}=P(A)P(B)\); they are independent. 3. Event \(C\) consists only of \(HH\), so \(P(C)=\frac{1}{4}\). Its intersection with \(A\) is still \(HH\), with probability \(\frac{1}{4}\). 4. Since \(P(A)P(C)=\frac{1}{8}\ne\frac{1}{4}=P(A\cap C)\), these events are dependent. Learning \(C\) guarantees \(A\).

Answer

a) Independent: \(P(A\cap B)=\frac{1}{4}=\frac{1}{2}\cdot\frac{1}{2}\). b) Dependent: \(P(A\cap C)=\frac{1}{4}\ne\frac{1}{8}=P(A)P(C)\). The change is that knowing both coins show heads guarantees the first coin shows heads.
Independence of events 61
In a survey, \(60\%\) of respondents regularly drink coffee (event \(K\)), \(30\%\) are frequently late (event \(Z\)), and \(18\%\) satisfy both conditions. Determine whether the events are independent, and explain whether the result establishes a cause-and-effect relationship.

Hints

- Compare the observed intersection with the product of the two individual probabilities. - Keep a statistical relationship separate from a claim about cause and effect.

Solution

1. The product of the marginal probabilities is \(P(K)P(Z)=0.6\cdot0.3=0.18\). 2. Since this equals \(P(K\cap Z)=0.18\), the events are independent in this data model. 3. Independence is a statistical relationship. It does not establish that either behavior causes or prevents the other.

Answer

The events are independent in the survey data, but this result does not establish causation.
Independence of events 62
A school surveyed students about wearing a sun hat and spending time outdoors. <table><tr><th></th><th>Outdoors</th><th>Indoors</th><th>Total</th></tr><tr><th>Sun hat</th><td>10</td><td>2</td><td>12</td></tr><tr><th>No sun hat</th><td>2</td><td>6</td><td>8</td></tr><tr><th>Total</th><td>12</td><td>8</td><td>20</td></tr></table> Are wearing a sun hat and being outdoors independent in these data? Does this table prove that wearing a hat makes a student go outdoors?

Hints

- Compare the outdoor fraction among hat wearers with the overall outdoor fraction. - Does the survey tell you whether students put on hats before or after deciding to go outdoors?

Solution

1. Among hat wearers, the outdoor proportion is \(\frac{10}{12}=\frac{5}{6}\). 2. Across all students, the outdoor proportion is \(\frac{12}{20}=\frac{3}{5}\). 3. The proportions differ, so the events are associated, not independent in this table. 4. The table does not prove causation; going outdoors may itself make students more likely to wear hats.

Answer

Not independent: \(\frac{5}{6}\ne\frac{3}{5}\). No, the table does not establish that hats cause outdoor activity.
Independence of events 63
A music school records whether students are in Grade \(9\) (event \(A\)) or Grade \(10\) (event \(A^c\)), and whether they play piano (event \(B\)). <table><tr><th></th><th>Piano</th><th>Not piano</th></tr><tr><th>Grade 9</th><td>\(6\)</td><td>\(4\)</td></tr><tr><th>Grade 10</th><td>\(9\)</td><td>\(6\)</td></tr></table> 1. Show that piano playing and grade group are independent. 2. Suppose both Grade \(9\) counts are multiplied by the same positive whole number \(k\), while Grade \(10\) counts stay the same. Will independence still hold for every \(k\)? Explain using the percentages within each grade.

Hints

- For each grade, what fraction of students play piano? - When a whole row is multiplied by the same factor, what happens to its numerator and denominator? - Compare the two within-grade rates after the change.

Solution

1. Among Grade \(9\) students, the piano rate is \(\frac{6}{6+4}=\frac{3}{5}=0.60\). Among Grade \(10\) students, it is \(\frac{9}{9+6}=\frac{3}{5}=0.60\). Equal rates show independence in this table. 2. After scaling, the Grade \(9\) piano rate is \(\frac{6k}{6k+4k}=\frac{6}{10}=0.60\) for every positive \(k\). The Grade \(10\) rate remains \(\frac{9}{15}=0.60\). 3. The within-grade rates are still equal, so independence is preserved.

Answer

1. Independent: the piano rate is \(60\%\) in both grades. 2. Yes. Scaling the entire Grade \(9\) row changes its counts but not its within-row piano rate, which remains \(60\%\).

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