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Independence of events

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51010910
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 \(6\%\).
51364810
Lucas flips a fair coin five times, and every flip lands heads. He says, “The probability of tails on the sixth flip must be greater than \(50\%\), because the results have to balance out over time.” Evaluate Lucas’s claim and justify your answer using appropriate probability terms.

Hints

- Does a coin remember earlier results? - Recall what it means for events to be independent. - Distinguish a long-run relative frequency from the probability of the next single trial.

Solution

1. The coin flips are independent events, so the result of one flip does not affect the next flip. 2. The probability of tails on every individual flip remains \(P(\text{tails}) = 0.5 = 50\%\). 3. The law of large numbers says that the relative frequency of tails tends to approach \(0.5\) over a very large number of flips. It does not mean that the coin “remembers” a short streak and corrects it on the next flip. 4. Therefore, Lucas is incorrect. The sixth flip is no more likely to be tails than any other flip.

Answer

Lucas is incorrect. Because coin flips are independent, the probability of tails on the sixth flip is still exactly \(50\%\).
51535710
A streaming service analyzes \(500\) customers. Customers are classified as under age \(25\) \((U)\) or age \(25\) and older \((A)\), and by whether they have a premium subscription \((P)\). <table> <tr><td></td><td>\(U\)</td><td>\(A\)</td><td>Total</td></tr> <tr><td>\(P\)</td><td>\(120\)</td><td>\(80\)</td><td>\(200\)</td></tr> <tr><td>\(\overline{P}\)</td><td>\(180\)</td><td>\(120\)</td><td>\(300\)</td></tr> <tr><td>Total</td><td>\(300\)</td><td>\(200\)</td><td>\(500\)</td></tr> </table> Determine whether premium-subscription status is independent of age group by comparing \(P(P\mid U)\) with \(P(P)\).

Hints

- Find the proportion of premium customers among all customers. - Then find the proportion of premium customers within the under-\(25\) group. - Equality of the conditional and marginal probabilities indicates independence.

Solution

1. The overall premium probability is \(P(P)=\frac{200}{500}=0.40\). 2. Among customers under age \(25\), \(P(P\mid U)=\frac{120}{300}=0.40\). 3. Since \(P(P\mid U)=P(P)\), the event \(U\) does not change the probability of \(P\). Therefore, premium status and age group are independent in this data set.

Answer

\(P(P)=0.40\) and \(P(P\mid U)=0.40\). Because these probabilities are equal, premium-subscription status and age group are independent in this data set.
52209410
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 \(P(\overline{U})=1-0.60=0.40\). 2. If \(\overline{U}\) and \(S\) are independent, then \(P(\overline{U}\cap S)=P(\overline{U})P(S)\). 3. Compute the product: \(P(\overline{U})P(S)=0.40\cdot 0.30=0.12\). 4. This equals the given value \(P(\overline{U}\cap S)=0.12\), so \(\overline{U}\) 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 \(P(\overline{U}\cap S)=0.12=P(\overline{U})P(S)\).
52213210
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 \(P(L\cap T)=\frac{1}{8}\neq\frac{1}{6}=P(L)P(T)\).
52214110
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.
53207510
A bag contains the \(12\) balls shown. Each label uses B for blue or R for red, followed by the ball's number. 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 \(P(A\cap E)=\frac{1}{12}=P(A)P(E)\).
53748810
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 \(P(B\mid\overline{A})=0.25\) 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: \(P(B\mid A)=P(B\mid\overline{A})=0.25\). Therefore, the events are independent.
53748910
A market survey records purchase interest in a new product (event \(B\)) by age group. Event \(U\) means under age \(30\), and \(\overline{U}\) means age \(30\) or older. The table and probability tree show the same relative frequencies. <table> <tr><td></td><td>\(B\)</td><td>\(\overline{B}\)</td><td>Total</td></tr> <tr><td>\(U\)</td><td>\(0.06\)</td><td>\(0.14\)</td><td>\(0.20\)</td></tr> <tr><td>\(\overline{U}\)</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 either representation that shows quickly that \(U\) and \(B\) are independent.
Figure for problem 537489

Hints

- Compare an inner table cell with the product of its row and column totals. - Alternatively, compare the conditional proportions within the two rows. - In the tree, look for matching second-stage probabilities.

Solution

1. Using the table, multiply the marginal probabilities: \(P(U)P(B)=0.20\cdot 0.30=0.06\). 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 \(\frac{0.24}{0.80}=0.30\). The probability tree displays this as matching second-stage branch probabilities.

Answer

One valid feature is \(P(U\cap B)=0.06=0.20\cdot 0.30=P(U)P(B)\). Another is that both age groups have the same conditional purchase-interest rate, \(0.30\).
53756610
Suppose \(P(A)=0.4\), \(P(B)=0.5\), and \(P(A\cap B)=0.2\). Determine whether \(A\) and \(B\) are independent.
Figure for problem 537566

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.
53756810
In the probability tree, \(P(A)=0.40\), \(P(B\mid A)=x\), and \(P(B\mid\overline{A})=0.30\). Find \(x\) so that \(A\) and \(B\) are independent.
Figure for problem 537568

Hints

- For independent events, the second-stage probability of \(B\) does not depend on the first-stage outcome. - You can verify the value by finding the overall probability of \(B\) from both paths.

Solution

1. If \(A\) and \(B\) are independent, the probability of \(B\) must be the same after either first-stage outcome. 2. Therefore, \(P(B\mid A)=P(B\mid\overline{A})\), so \(x=0.30\). 3. Verification: \(P(B)=0.40\cdot 0.30+0.60\cdot 0.30=0.30=P(B\mid A)\).

Answer

\(x=0.30\)
53757710
Among patients who have a particular disease, a laboratory uses two tests, A and B. The tree shows \(P(B^+\mid A^+)=0.95\) and \(P(B^+\mid A^-)=0.60\). 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 probability is \(0.95\), but after \(A^-\), it is \(0.60\). 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 \(P(B^+\mid A^+)=0.95\neq 0.60=P(B^+\mid A^-)\). The two test results are conditionally dependent in this group.
51364910
A fair six-sided number cube is rolled \(600\) times. The result \(6\) occurs \(85\) times. a) Find the relative frequency of rolling a \(6\). b) A student says, “Because the theoretical probability of rolling a \(6\) is \(\frac{1}{6}\), the next \(600\) rolls must include more than \(100\) sixes so the results balance out.” Explain why this statement shows a misunderstanding of the law of large numbers.

Hints

- Divide the number of sixes by the total number of rolls. - Ask whether an earlier roll changes the probability on a later roll. - Distinguish long-run stabilization from guaranteed short-term balancing.

Solution

1. a) The relative frequency is \(\frac{85}{600} = \frac{17}{120} \approx 0.1417\), or about \(14.17\%\). 2. The theoretical probability is \(P(6) = \frac{1}{6} \approx 0.1667\), so the observed relative frequency is below the theoretical probability. 3. b) The law of large numbers says that relative frequency tends to stabilize near theoretical probability over many independent trials. It does not require a short-term “deficit” to be corrected in a later group of rolls. 4. Each future roll is independent of the first \(600\) rolls, and the probability of a \(6\) on each roll remains \(\frac{1}{6}\).

Answer

a) The relative frequency is \(\frac{85}{600} \approx 0.1417\), or about \(14.17\%\). b) The claim is incorrect. The rolls are independent, so earlier results do not make a \(6\) more likely on future rolls. The law of large numbers describes long-run stabilization, not forced compensation.
51365010
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.
51474810
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 \(P(D\cap\overline{B})=0.05-0.01=0.04\), \(P(\overline{D}\cap B)=0.03-0.01=0.02\), and \(P(\overline{D}\cap\overline{B})=1-(0.01+0.04+0.02)=0.93\). 2. \(P(D\mid B)=\frac{P(D\cap B)}{P(B)}=\frac{0.01}{0.03}=\frac{1}{3}\). 3. Since \(P(D\mid B)=\frac{1}{3}\neq 0.05=P(D)\), 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>\(\overline{B}\)</td><td>Total</td></tr> <tr><td>\(D\)</td><td>\(0.01\)</td><td>\(0.04\)</td><td>\(0.05\)</td></tr> <tr><td>\(\overline{D}\)</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) \(P(D\mid B)=\frac{1}{3}\approx 0.3333\), or about \(33.33\%\). c) Because \(P(D\mid B)\neq P(D)\), display and battery defects are dependent in these production data.
51476610
A survey of \(200\) households records dog ownership \((D)\) and yard ownership \((Y)\). <table> <tr><td></td><td>\(Y\)</td><td>\(\overline{Y}\)</td><td>Total</td></tr> <tr><td>\(D\)</td><td>\(48\)</td><td>\(12\)</td><td>\(60\)</td></tr> <tr><td>\(\overline{D}\)</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 \(P(Y\mid\overline{D})\). 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, \(P(Y\mid\overline{D})=\frac{56}{140}=0.40\). 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) \(P(Y\mid\overline{D})=0.40\), or \(40\%\). Yard ownership is twice as common among dog-owning households in this sample, indicating dependence.
52209310
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>\(\overline{M}\) (no training)</td><td>Total</td></tr> <tr><td>\(E\) (passed)</td><td>\(150\)</td><td>\(200\)</td><td>\(350\)</td></tr> <tr><td>\(\overline{E}\) (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 \(P(M)P(E)=0.40\cdot 0.70=0.28\). 3. Since \(0.30\neq 0.28\), 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 \(P(M\cap E)=0.30\neq 0.28=P(M)P(E)\). In this sample, training participation is positively associated with passing the exam, but this observational comparison does not by itself prove causation.
52210410
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 \(P(E\cap F)=0.12=P(E)P(F)\).
52210510
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 \(P(A\cap B)=0.36=P(A)P(B)\).
52210610
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 \(P(G)P(S)=0.24\).
52210710
The following two-way probability table for independent events \(E\) and \(F\) is incomplete. <table> <tr><td></td><td>\(F\)</td><td>\(\overline{F}\)</td><td>Total</td></tr> <tr><td>\(E\)</td><td></td><td>\(0.12\)</td><td></td></tr> <tr><td>\(\overline{E}\)</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 \(\overline{F}\) 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 \(\overline{F}\) column: \(P(\overline{F})=0.12+0.18=0.30\), so \(P(F)=1-0.30=0.70\). 2. Since \(E\) and \(F\) are independent, \(E\) and \(\overline{F}\) are also independent. Thus, \(P(E\cap\overline{F})=P(E)P(\overline{F})\), so \(0.12=P(E)(0.30)\) and \(P(E)=0.40\). 3. Therefore, \(P(\overline{E})=0.60\). 4. Use independence to find the remaining joint probabilities: \(P(E\cap F)=0.40\cdot 0.70=0.28\) and \(P(\overline{E}\cap F)=0.60\cdot 0.70=0.42\).

Answer

<table> <tr><td></td><td>\(F\)</td><td>\(\overline{F}\)</td><td>Total</td></tr> <tr><td>\(E\)</td><td>\(0.28\)</td><td>\(0.12\)</td><td>\(0.40\)</td></tr> <tr><td>\(\overline{E}\)</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>
52211310
A medical study compares a treatment group \((T)\), which received a new medication, with a placebo group \((\overline{T})\). Recovery after two weeks is represented by \(R\). The table shows the numbers of participants. <table> <tr><td></td><td>\(R\)</td><td>\(\overline{R}\)</td></tr> <tr><td>\(T\)</td><td>\(120\)</td><td>\(80\)</td></tr> <tr><td>\(\overline{T}\)</td><td>\(45\)</td><td>\(y\)</td></tr> </table> 1. Find the value of \(y\) for which recovery is independent of group assignment. 2. Interpret the result in the context of the study.

Hints

- For independence in a \(2\times 2\) table, compare the products of the diagonally opposite cell counts. - Set up an equation involving \(y\). - Compare the recovery rates in the two groups, then distinguish association from causation.

Solution

1. For a \(2\times 2\) table with cell counts \(a,b,c,d\), independence is equivalent to \(ad=bc\). 2. Substitute the values: \(120y=80\cdot 45=3600\). 3. Solve to get \(y=30\). 4. With \(y=30\), the recovery rate in the treatment group is \(\frac{120}{200}=0.60\), and the recovery rate in the placebo group is \(\frac{45}{75}=0.60\). 5. The equal recovery rates show no association between group assignment and recovery in this table. This calculation alone does not provide a significance test or establish a causal effect.

Answer

1. \(y=30\) 2. The recovery rate is \(60\%\) in both groups, so recovery and group assignment are independent in this table. This result alone does not establish statistical significance or a causal effect of the medication.
52211810
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.
52211910
Container A holds \(5\) red chips and \(15\) blue chips. Container B holds \(10\) red chips and \(30\) blue chips. One container is selected at random, with each container equally likely, and then one chip is drawn. Let \(A\) be the event that Container A is selected, and let \(R\) be the event that a red chip is drawn. a) Show that \(A\) and \(R\) are independent. b) A total of \(15\) additional red chips will be divided between the containers. How many should be placed in each container so that \(A\) and \(R\) remain independent? c) Use the change to Container A from part b, but leave Container B with its original red chips and add \(10\) blue chips to Container B instead. Determine whether \(A\) and \(R\) are independent.

Hints

- Compare the proportion of red chips in the two containers. - For part b, let a variable represent the number of new red chips placed in Container A. - Set the two conditional probabilities of drawing red equal to each other. - Check the final red-chip proportions in both containers.

Solution

1. Initially, \(P(R\mid A)=\frac{5}{20}=0.25\) and \(P(R\mid\overline{A})=\frac{10}{40}=0.25\). Because the conditional probabilities are equal, the events are independent. 2. Let \(x\) be the number of additional red chips placed in Container A. Then \(15-x\) are placed in Container B. Independence requires \(\frac{5+x}{20+x}=\frac{25-x}{55-x}\). 3. Cross-multiply and solve: \((5+x)(55-x)=(25-x)(20+x)\), \(275+50x-x^2=500+5x-x^2\), \(45x=225\), so \(x=5\). Therefore, Container A receives \(5\) red chips and Container B receives \(10\). 4. For part c, Container A has \(10\) red and \(15\) blue chips, so \(P(R\mid A)=\frac{10}{25}=0.40\). Container B has \(10\) red and \(40\) blue chips, so \(P(R\mid\overline{A})=\frac{10}{50}=0.20\). 5. Since the conditional probabilities are not equal, the events are dependent.

Answer

a) \(P(R\mid A)=P(R\mid\overline{A})=0.25\), so the events are independent. b) Place \(5\) red chips in Container A and \(10\) red chips in Container B. c) The events are dependent because \(P(R\mid A)=0.40\neq 0.20=P(R\mid\overline{A})\).
52212010
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 \(\overline{T}\) and \(\overline{B}\) 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 \(120\cdot 0.30=36\) teenagers and \(80\cdot 0.30=24\) 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 \(P(B\mid\overline{T})=\frac{20}{80}=0.25\). 4. Since the conditional probabilities differ, \(T\) and \(B\) are dependent. 5. Dependence is preserved when both events are replaced by their complements. Therefore, \(\overline{T}\) and \(\overline{B}\) are also dependent.

Answer

a) \(36\) teenagers and \(24\) adults. b) \(P(B\mid T)=\frac{1}{3}\neq 0.25=P(B\mid\overline{T})\), so the events are dependent. c) \(\overline{T}\) and \(\overline{B}\) are dependent because replacing both events with their complements preserves independence or dependence.
52212110
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 \(P(A\cap C)=\frac{1}{12}\neq\frac{5}{72}=P(A)P(C)\).
52212310
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>\(\overline{L}\)</td><td>Total</td></tr> <tr><td>\(S\)</td><td>\(25\%\)</td><td></td><td>\(55\%\)</td></tr> <tr><td>\(\overline{S}\)</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 \(P(S\mid\overline{L})\). 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: \(P(L\cap\overline{S})=0.40-0.25=0.15\), \(P(\overline{L}\cap S)=0.55-0.25=0.30\), and \(P(\overline{L}\cap\overline{S})=1-(0.25+0.15+0.30)=0.30\). The remaining margins are \(P(\overline{L})=0.60\) and \(P(\overline{S})=0.45\). 2. For independence, \(P(L\cap S)\) must equal \(P(L)P(S)\). Here, \(P(L)P(S)=0.40\cdot 0.55=0.22\), 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 \(P(S\mid\overline{L})=\frac{0.30}{0.60}=0.50\). 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>\(\overline{L}\)</td><td>Total</td></tr> <tr><td>\(S\)</td><td>\(25\%\)</td><td>\(30\%\)</td><td>\(55\%\)</td></tr> <tr><td>\(\overline{S}\)</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 \(P(L\cap S)=0.25\neq 0.22=P(L)P(S)\). c) \(P(S\mid L)=0.625\), or \(62.5\%\), is the satisfaction rate among respondents who own the company's laptop. \(P(S\mid\overline{L})=0.50\), or \(50\%\), is the satisfaction rate among those who do not. d) Laptop ownership and service satisfaction are positively associated in this sample.
52212410
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 \(P(\overline{D})=0.95\). 2. Complete the inner cells: \(P(A\cap\overline{D})=0.60-0.03=0.57\), \(P(B\cap D)=0.05-0.03=0.02\), and \(P(B\cap\overline{D})=0.40-0.02=0.38\). 3. Since \(P(A)P(D)=0.60\cdot 0.05=0.03=P(A\cap D)\), 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>\(\overline{D}\)</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.
52213110
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.
52213410
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 \(P(C\mid\overline{S})\). 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 \(P(C\mid\overline{S})\). - Distinguish a statistical association from evidence of causation.

Solution

1. The number of servers with the system is \(0.85\cdot 5000=4250\), so \(750\) servers do not have it. 2. Of the \(350\) compromised servers, \(350-150=200\) did not have the system. 3. Therefore, \(P(C\mid S)=\frac{150}{4250}=\frac{3}{85}\approx 0.0353\), and \(P(C\mid\overline{S})=\frac{200}{750}=\frac{4}{15}\approx 0.2667\). 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) \(P(C\mid S)=\frac{3}{85}\approx 3.53\%\), and \(P(C\mid\overline{S})=\frac{4}{15}\approx 26.67\%\). b) The events are dependent because \(P(C\mid S)\approx 0.0353\neq 0.07=P(C)\). 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.
52214210
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.
53085510
Two fair eight-sided dice labeled \(1\) through \(8\) are rolled. Find the probability that at least one die: (1) shows \(8\); (2) shows a prime number; (3) shows a number less than \(3\).

Hints

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

Solution

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

Answer

(1) \(\frac{15}{64}\) (2) \(\frac{3}{4}\) (3) \(\frac{7}{16}\)
53616110
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.
Figure for problem 536161

Hints

- Find the probability of red on each draw. - A tree diagram can organize the outcomes without replacement. - 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 \(P(A\cap B)=\frac{1}{6}\neq\frac{1}{4}=P(A)P(B)\).
53617710
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.
Figure for problem 536177

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 \(P(B)=0.40\cdot 0.60+0.60\cdot 0.40=0.48\). 4. The event \(A\cap B\) requires red on the first draw and blue on the second, so \(P(A\cap B)=0.40\cdot 0.60=0.24\). 5. Since \(P(A)P(B)=0.40\cdot 0.48=0.192\neq 0.24\), the events are dependent.

Answer

The events are dependent because \(P(A\cap B)=0.24\neq 0.192=P(A)P(B)\).
53748210
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 \(P(\overline{A})\). - Read the joint probability at the end of the \(\overline{A}\)-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 \(P(\overline{A})=1-0.80=0.20\). 2. The tree gives \(P(\overline{A}\cap B)=0.06\). 3. Because \(A\) and \(B\) are independent, \(\overline{A}\) and \(B\) are also independent. Thus, \(P(\overline{A}\cap B)=P(\overline{A})P(B)\). 4. Solve \(0.06=0.20P(B)\) to get \(P(B)=0.30\).

Answer

\(P(B)=0.30\)
53748710
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 \((\overline{E})\). 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 \(n(Y)=0.40\cdot 1000=400\) and \(n(O)=0.60\cdot 1000=600\). 2. The joint counts are \(n(Y\cap E)=400\cdot 0.75=300\), \(n(Y\cap\overline{E})=400\cdot 0.25=100\), \(n(O\cap E)=600\cdot 0.30=180\), and \(n(O\cap\overline{E})=600\cdot 0.70=420\). 3. Thus, \(n(E)=300+180=480\) and \(n(\overline{E})=100+420=520\). 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 \(P(Y)P(E)=0.40\cdot 0.48=0.192\). Since they are not equal, the events are dependent.

Answer

a) <table> <tr><th></th><th>\(E\)</th><th>\(\overline{E}\)</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 \(P(Y\cap E)=0.300\neq 0.192=P(Y)P(E)\).
53750410
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.
53756710
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 \(P(B\mid A)\).
Figure for problem 537567

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 \(0.3\ne0.24\), 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 \(P(B\mid A)=0.5\).
53756910
The probability tree shows \(P(B\mid A)=P(B\mid\overline{A})=0.70\). Without knowing \(P(A)\), explain why \(A\) and \(B\) are independent.
Figure for problem 537569

Hints

- Represent the unknown probability \(P(A)\) with a variable. - Find \(P(B)\) by adding the probabilities of the two paths that end in \(B\). - Compare the result with \(P(B\mid A)\).

Solution

1. Let \(P(A)=p\), so \(P(\overline{A})=1-p\). 2. Use the law of total probability: \(P(B)=p\cdot 0.70+(1-p)\cdot 0.70=0.70\). 3. Thus, \(P(B\mid A)=0.70=P(B)\). Therefore, \(A\) and \(B\) are independent.

Answer

\(A\) and \(B\) are independent because the conditional probability of \(B\) is \(0.70\) after either outcome of the first stage, so the overall probability is also \(0.70\).
53757910
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.
53758810
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.
Figure for problem 537588

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.
52210810
Events \(A\) and \(B\) are independent. You know that \(P(A\cap B)=0.12\) and \(P(A\cup B)=0.68\). Create a complete two-way probability table for the events. There are two possible assignments of the marginal probabilities; give either one.

Hints

- Use the addition rule for \(P(A\cup B)\). - Independence gives an equation involving the product \(P(A)P(B)\). - Use the sum and product of the two unknown marginal probabilities to determine them. - After finding the margins, subtract to complete the inner cells.

Solution

1. Apply the addition rule: \(0.68=P(A)+P(B)-0.12\), so \(P(A)+P(B)=0.80\). 2. Independence gives \(P(A)P(B)=P(A\cap B)=0.12\). 3. Let \(x=P(A)\). Then \(P(B)=0.80-x\), so \(x(0.80-x)=0.12\). This gives \(x^2-0.80x+0.12=0\), whose solutions are \(x=0.60\) and \(x=0.20\). 4. Choose \(P(A)=0.60\) and \(P(B)=0.20\). Then \(P(A\cap\overline{B})=0.60-0.12=0.48\), \(P(\overline{A}\cap B)=0.20-0.12=0.08\), and \(P(\overline{A}\cap\overline{B})=1-(0.12+0.48+0.08)=0.32\).

Answer

One possible table is: <table> <tr><td></td><td>\(B\)</td><td>\(\overline{B}\)</td><td>Total</td></tr> <tr><td>\(A\)</td><td>\(0.12\)</td><td>\(0.48\)</td><td>\(0.60\)</td></tr> <tr><td>\(\overline{A}\)</td><td>\(0.08\)</td><td>\(0.32\)</td><td>\(0.40\)</td></tr> <tr><td>Total</td><td>\(0.20\)</td><td>\(0.80\)</td><td>\(1\)</td></tr> </table> The other valid table results from interchanging the marginal probabilities of \(A\) and \(B\).
52211410
A \(2\times 2\) table for two categorical variables has cell counts \(a,b,c,d\), arranged as shown. <table> <tr><td>\(a\)</td><td>\(b\)</td></tr> <tr><td>\(c\)</td><td>\(d\)</td></tr> </table> Consider this claim: “If every count in the first row is multiplied by the same positive integer \(k\), independence is preserved whenever the original variables were independent.” Determine whether the claim is true in general by using the criterion \(ad=bc\).

Hints

- Write the cross-product condition for the original table. - Replace \(a\) and \(b\) with their new values after multiplying the first row by \(k\). - Simplify the new condition and compare it with the original one.

Solution

1. For the original table, independence is equivalent to \(ad=bc\). 2. After multiplying the first row by \(k\), the new counts are \(ka,kb,c,d\). 3. Apply the independence criterion to the new table: \((ka)d=(kb)c\), which is equivalent to \(kad=kbc\). 4. Because \(k>0\), dividing both sides by \(k\) gives \(ad=bc\), the original condition. Therefore, if the original table represents independent variables, the modified table does also.

Answer

The claim is true. The new condition \(kad=kbc\) is equivalent to the original condition \(ad=bc\) because \(k>0\).
52212510
Bag A contains \(6\) red balls and \(4\) blue balls. Bag B contains \(k\) red balls and \(2k\) blue balls, where \(k\) is a positive integer. First, one ball is selected at random from Bag A and placed in Bag B. Then one ball is selected at random from Bag B and placed in Bag A. a) Find, in terms of \(k\), the probability that Bag A again contains exactly \(6\) red balls after both transfers. b) For one value of \(k\), the probability from part a is \(\frac{19}{35}\). Find \(k\). c) Let \(R_1\) be the event that the first ball selected from Bag A is red. Let \(G\) be the event that Bag A returns to its original composition after the second transfer. Determine whether \(R_1\) and \(G\) are independent for any positive integer \(k\).

Hints

- Identify the two color sequences that restore Bag A's original composition. - After the first transfer, Bag B contains \(3k+1\) balls. - For part b, set the expression from part a equal to the given fraction. - For part c, compare \(P(G\mid R_1)\) with \(P(G)\).

Solution

1. Bag A returns to \(6\) red balls if a red ball is transferred in each direction or a blue ball is transferred in each direction. 2. If the first ball is red, Bag B then has \(k+1\) red balls out of \(3k+1\). If the first ball is blue, Bag B then has \(2k+1\) blue balls out of \(3k+1\). 3. Therefore, \(P(G)=\frac{6}{10}\cdot\frac{k+1}{3k+1}+\frac{4}{10}\cdot\frac{2k+1}{3k+1}=\frac{7k+5}{15k+5}\). 4. For part b, solve \(\frac{7k+5}{15k+5}=\frac{19}{35}\): \(35(7k+5)=19(15k+5)\), \(245k+175=285k+95\), so \(k=2\). 5. Given \(R_1\), restoring the original composition requires drawing a red ball from Bag B, so \(P(G\mid R_1)=\frac{k+1}{3k+1}\). 6. Independence would require \(P(G\mid R_1)=P(G)\). Substitution gives \(\frac{k+1}{3k+1}=\frac{7k+5}{15k+5}\), which reduces to \(k=0\) after excluding values that make a denominator zero. Because \(k\) must be positive, the events are dependent for every allowed value of \(k\).

Answer

a) \(P(G)=\frac{7k+5}{15k+5}\) b) \(k=2\) c) The events are dependent for every positive integer \(k\). Independence would require \(k=0\), which is not allowed.
53593610
The spinner shown has \(10\) equal sections: \(4\) green, \(3\) blue, and \(3\) red. 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

- Find the probability of each color on one spin. - A tree diagram can organize the possible two-stage outcomes. - 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. On one spin, \(P(\text{green}) = \frac{4}{10}\), \(P(\text{blue}) = \frac{3}{10}\), 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 \(\frac{4}{10} \cdot \frac{3}{9} + \frac{3}{10} \cdot \frac{2}{9} + \frac{3}{10} \cdot \frac{2}{9} = \frac{4}{15} \approx 26.7\%\), 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 \(\frac{4}{15} \approx 26.7\%\), which is less than \(34\%\).
53758010
Two fair coins are tossed. Let \(A\) be the event that the first coin shows heads, \(B\) the event that the second coin shows heads, and \(C\) the event that both coins show the same side. Show that \(A\), \(B\), and \(C\) are pairwise independent but not mutually independent.
Figure for problem 537580

Hints

- Match each event to the four ordered outcomes \(HH,HT,TH,TT\). - Check each two-event intersection first. - Then compare the three-event intersection with the product of all three marginal probabilities.

Solution

1. Each event has probability \(\frac{1}{2}\): \(P(A)=P(B)=P(C)=\frac{1}{2}\). 2. For each pair, the intersection contains one of the four equally likely outcomes: \(P(A\cap B)=P(A\cap C)=P(B\cap C)=\frac{1}{4}\). 3. Since \(\frac{1}{4}=\frac{1}{2}\cdot\frac{1}{2}\), every pair is independent. 4. The three events occur together only for \(HH\), so \(P(A\cap B\cap C)=\frac{1}{4}\). 5. However, \(P(A)P(B)P(C)=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{8}\). Because \(\frac{1}{4}\neq\frac{1}{8}\), the three events are not mutually independent.

Answer

The events are pairwise independent because every pair has joint probability \(\frac{1}{4}\), equal to the product of its marginal probabilities. They are not mutually independent because \(P(A\cap B\cap C)=\frac{1}{4}\neq\frac{1}{8}=P(A)P(B)P(C)\).

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