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Conditional probability

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54734912
Suppose \(P(B)=0\). Is \(P(A\mid B)\) defined by the usual ratio formula? Explain.

Hints

- Inspect the denominator in the conditional-probability definition. - A conditioning event must have positive probability for the elementary ratio. - Do not assign a value merely from the intersection.

Solution

1. The ratio would require division by \(P(B)=0\). 2. Division by zero is undefined. 3. Therefore the elementary conditional probability \(P(A\mid B)\) is not defined when the conditioning event has probability zero.

Answer

No. The usual conditional probability is undefined because its denominator is \(0\).
55619812
A fair six-sided die is rolled. Given that the result is greater than \(3\), what is the probability that the result is even?

Hints

- First restrict the sample space using the information after “given that.” - Count the outcomes in the restricted sample space. - Then count how many of those restricted outcomes satisfy the second event.

Solution

1. The condition restricts the possible outcomes to \(\{4,5,6\}\). 2. Within that restricted set, the even outcomes are \(4\) and \(6\). 3. Therefore the conditional probability is \(\frac23\).

Answer

\(\frac23\)
55619912
Events \(A\) and \(B\) satisfy \(P(B)=0.50\) and \(P(A\cap B)=0.20\). Find \(P(A\mid B)\).

Hints

- Identify which event is the condition. - The numerator is the probability of both events occurring. - Divide the intersection probability by the probability of the conditioning event.

Solution

1. Conditional probability uses the restricted probability of \(B\) as the denominator. 2. \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.20}{0.50}=0.40\).

Answer

\(0.40\)
54731212
Two tokens are drawn without replacement from the urn shown. Given that the first token is red, find the probability that the second token is red.
Figure for problem 547312

Hints

- Conditioning means the sample space has been narrowed to a specified event. - For sequential settings, update what remains after the stated condition occurs. - Explain what knowing the condition changes about the probability question.

Solution

1. After a red token is drawn, \(4\) red tokens remain out of \(11\) total tokens. 2. Thus \(P(\text{second red}\mid\text{first red})=\frac{4}{11}\).

Answer

\(\frac{4}{11}\).
54731712
A randomly chosen order is from the weekend with probability \(0.40\). Among weekend orders, \(35\%\) include express shipping. Find the joint probability that an order is from the weekend and includes express shipping.

Hints

- One probability gives the size of the conditioning group. - The other gives the share inside that group. - Convert the subgroup share to a share of all orders.

Solution

1. Use the conditional multiplication relationship. 2. \(P(\text{weekend and express})=0.40\cdot0.35=0.14\).

Answer

The joint probability is \(0.14\).
54732312
Use the probability tree shown to find \(P(A\cap B\cap C)\).
Figure for problem 547323

Hints

- Multiply probabilities along one fully specified path. - Each later probability is conditional on the preceding path. - Keep the event order aligned with the given conditions.

Solution

1. Follow the specified path through the stages. 2. \(P(A\cap B\cap C)=0.60\cdot0.50\cdot0.80=0.24\).

Answer

The probability is \(0.24\).
54734012
Give a numerical example showing that conditioning can increase an event’s probability. Use a sample of \(100\) observations and state both \(P(A)\) and \(P(A\mid B)\).

Hints

- Choose a subgroup with a higher concentration of \(A\) than the full sample. - Keep the intersection no larger than either event. - Compare the subgroup rate with the marginal rate.

Solution

1. Let \(A\) occur for \(30\) of \(100\) observations, so \(P(A)=0.30\). 2. Let \(B\) contain \(20\) observations, \(15\) of which are in \(A\). 3. Then \(P(A\mid B)=\frac{15}{20}=0.75>0.30\).

Answer

For example, \(P(A)=0.30\) and \(P(A\mid B)=0.75\) when \(15\) of \(20\) observations in \(B\) also satisfy \(A\).
55620012
A shipping company finds that \(0.75\) of its packages leave the warehouse on the scheduled day. Among packages that leave on the scheduled day, \(0.92\) arrive by the promised date. What is the probability that a randomly selected package both leaves on the scheduled day and arrives by the promised date?

Hints

- One probability describes the first event, and the other describes the second event within the first event. - The question asks for both events to occur. - Connect a conditional probability to a joint probability using the multiplication relationship.

Solution

1. Let \(S\) be leaving on the scheduled day and \(A\) be arriving by the promised date. 2. The given values are \(P(S)=0.75\) and \(P(A\mid S)=0.92\). 3. Use the multiplication rule: \(P(S\cap A)=P(S)P(A\mid S)=0.75\cdot0.92=0.69\).

Answer

\(0.69\)
55620112
The displayed probability tree describes two machine types, R and S, and whether a machine fails inspection. Given that a randomly selected machine failed inspection, what is the probability that it was type S?
Figure for problem 556201

Hints

- Identify all terminal paths that are consistent with the condition “failed inspection.” - Find the probability represented by each relevant path before reversing the conditioning. - Within the total failure probability, determine the share contributed by type S.

Solution

1. From the tree, \(P(R)=0.60\), \(P(S)=0.40\), \(P(F\mid R)=0.05\), and \(P(F\mid S)=0.10\). 2. The failed type-R path has probability \(0.60\cdot0.05=0.03\), and the failed type-S path has probability \(0.40\cdot0.10=0.04\). 3. Thus \(P(F)=0.03+0.04=0.07\). 4. Therefore \(P(S\mid F)=\frac{0.04}{0.07}=\frac47\).

Answer

\(\frac47\)
55620212
In a group of \(200\) job applicants, \(50\) have a professional certificate. Of those \(50\), \(30\) are hired. Altogether, \(80\) applicants are hired. a) Find the probability that an applicant is hired given that the applicant has the certificate. b) Find the probability that an applicant has the certificate given that the applicant is hired. c) Explain why the two conditional probabilities are different.

Hints

- Read the event after “given that” first in each part. - The same overlap count can be divided by different totals when the conditioning direction changes. - Compare the two restricted groups rather than assuming reversed conditionals are equal.

Solution

1. For \(P(\text{hired}\mid\text{certificate})\), restrict to the \(50\) certificate holders. The probability is \(\frac{30}{50}=0.60\). 2. For \(P(\text{certificate}\mid\text{hired})\), restrict to the \(80\) hired applicants. The probability is \(\frac{30}{80}=0.375\). 3. The numerators count the same \(30\) applicants, but the conditioning groups—and therefore the denominators—are different.

Answer

a) \(0.60\) b) \(0.375\) c) The conditioning events are different, so the denominators are \(50\) and \(80\), respectively.
53083912
A simplified survival table shows how many men from an initial group of \(100{,}000\) live births reached selected ages in two different years. <table> <tr><th>Age in years</th><th>\(1960\)</th><th>\(2020\)</th></tr> <tr><td>\(0\)</td><td>\(100{,}000\)</td><td>\(100{,}000\)</td></tr> <tr><td>\(40\)</td><td>\(92{,}400\)</td><td>\(98{,}800\)</td></tr> <tr><td>\(70\)</td><td>\(52{,}100\)</td><td>\(84{,}500\)</td></tr> <tr><td>\(85\)</td><td>\(12{,}300\)</td><td>\(42{,}600\)</td></tr> </table> a) For each year, estimate the probability that a man who has reached age \(40\) will reach age \(70\). b) For each year, estimate the probability that a man who has reached age \(40\) will reach at least age \(85\). c) Explain why these are empirical probabilities rather than probabilities based on equally likely outcomes.

Hints

- For each year, use the number reaching the later age as the numerator. - Because the person has already reached age \(40\), use the age-\(40\) count as the denominator. - Distinguish probabilities estimated from data from probabilities derived from equally likely theoretical outcomes. - Identify the population represented by each column.

Solution

1. For 1960, the conditional proportion reaching age \(70\) among those reaching age \(40\) is \(\frac{52100}{92400}\approx 0.5639\). For 2020, it is \(\frac{84500}{98800}\approx 0.8553\). 2. For 1960, the conditional proportion reaching age \(85\) among those reaching age \(40\) is \(\frac{12300}{92400}\approx 0.1331\). For 2020, it is \(\frac{42600}{98800}\approx 0.4312\). 3. The values are estimated from observed relative frequencies in population data. They are not derived from a sample space in which all elementary outcomes are assumed to be equally likely.

Answer

a) 1960: \(0.5639\), or about \(56.39\%\). 2020: \(0.8553\), or about \(85.53\%\). b) 1960: \(0.1331\), or about \(13.31\%\). 2020: \(0.4312\), or about \(43.12\%\). c) They are empirical probabilities because they are based on observed relative frequencies.
53084012
An electronics company tracks the reliability of \(50{,}000\) smartphones. The table shows how many phones are still fully functional after selected numbers of months. <table> <tr><th>Months in use</th><th>Functional phones</th></tr> <tr><td>\(0\)</td><td>\(50{,}000\)</td></tr> <tr><td>\(12\)</td><td>\(48{,}500\)</td></tr> <tr><td>\(24\)</td><td>\(42{,}000\)</td></tr> <tr><td>\(36\)</td><td>\(25{,}000\)</td></tr> <tr><td>\(48\)</td><td>\(8000\)</td></tr> </table> a) Estimate the probability that a phone still working after \(12\) months is also working after \(36\) months. b) Estimate the probability that a phone working after \(24\) months fails during the third year, between months \(24\) and \(36\). c) Explain why failure time cannot be modeled here by assuming all possible months of failure are equally likely.

Hints

- In a conditional probability, use only the phones that satisfy the stated starting condition. - Find the number of failures during a time interval by subtracting the later survivor count from the earlier survivor count. - Consider whether wear makes failure equally likely at every age.

Solution

1. For part a, restrict the group to the \(48{,}500\) phones working after \(12\) months. Of those, \(25{,}000\) are working after \(36\) months, so the estimate is \(\frac{25000}{48500}\approx 0.5155\). 2. Between months \(24\) and \(36\), \(42{,}000-25{,}000=17{,}000\) phones fail. Among phones working at month \(24\), the estimated conditional probability is \(\frac{17000}{42000}\approx 0.4048\). 3. Failure rates can change with age and wear. The data do not support treating failure in each month as an equally likely outcome.

Answer

a) \(\frac{25000}{48500}\approx 0.5155\), or about \(51.55\%\). b) \(\frac{17000}{42000}\approx 0.4048\), or about \(40.48\%\). c) The probability of failure changes over time, so the possible failure months are not equally likely.
53095112
A survival table gives the number \(l_x\) of people from an initial population who reach age \(x\). <table> <tr><th>Age \(x\)</th><th>Number \(l_x\) reaching age \(x\)</th></tr> <tr><td>\(25\)</td><td>\(98{,}412\)</td></tr> <tr><td>\(45\)</td><td>\(95{,}120\)</td></tr> <tr><td>\(65\)</td><td>\(80{,}455\)</td></tr> </table> Estimate each probability. 1. A person who is currently \(25\) reaches age \(65\). 2. Two people who are currently \(45\) both reach age \(65\). Assume their survival events are independent. 3. A person who is currently \(25\) dies after reaching age \(45\) but before reaching age \(65\).

Hints

- Estimate a conditional probability by dividing the later-age count by the count for the current age. - For two independent events that must both occur, multiply their probabilities. - To count deaths between two ages, subtract the later survivor count from the earlier survivor count. - Use the population that has already reached the stated current age as the denominator.

Solution

1. Among people reaching age \(25\), the estimated proportion reaching age \(65\) is \(\frac{80455}{98412}\approx 0.8175\). 2. For one person currently age \(45\), the estimated probability of reaching age \(65\) is \(\frac{80455}{95120}\approx 0.8458\). Under the stated independence assumption, the probability that both people reach age \(65\) is \(\left(\frac{80455}{95120}\right)^2\approx 0.7154\). 3. The number reaching age \(45\) but not age \(65\) is \(95{,}120-80{,}455=14{,}665\). Relative to those reaching age \(25\), the estimated probability is \(\frac{14665}{98412}\approx 0.1490\).

Answer

1. \(0.8175\), or about \(81.75\%\). 2. \(0.7154\), or about \(71.54\%\). 3. \(0.1490\), or about \(14.90\%\).
54726412
In a randomized-response survey, a participant flips a fair coin. On heads, the participant answers a sensitive yes/no question truthfully. On tails, the participant must answer “yes.” In a large survey, \(62\%\) of responses are “yes.” Estimate the true proportion \(p\) whose truthful answer is yes. Explain how the randomization protects individual responses while allowing population estimation.

Hints

- Split a reported “yes” into the two disjoint ways it can occur. - One branch depends on the unknown prevalence; the other does not. - Use the known randomization probability to solve for the population rate.

Solution

1. A “yes” response occurs either from a heads flip followed by a truthful yes or from any tails flip. 2. Therefore \(0.62=0.50p+0.50\). 3. Solving gives \(0.50p=0.12\), so \(p=0.24\). 4. A particular yes response may be truthful or forced by the coin, so it does not reveal the individual’s status. Across many responses, the known coin mechanism can be removed algebraically.

Answer

The estimated truthful-yes proportion is \(p=0.24\), or \(24\%\). An individual “yes” may be truthful or forced by the coin, protecting the person’s status, while the known randomization rate allows the population proportion to be recovered.
54731012
Let \(A\subseteq C\) and let \(P(B)>0\). Prove that \(P(A\mid B)\le P(C\mid B)\). Explain why ordinary event inclusion remains monotone after conditioning on the same event.

Hints

- Intersect both nested events with the conditioning event. - Inclusion is preserved under intersection with a fixed set. - The conditional probabilities share the same positive denominator.

Solution

1. From \(A\subseteq C\), intersecting both sets with \(B\) gives \(A\cap B\subseteq C\cap B\). 2. Probability monotonicity gives \(P(A\cap B)\le P(C\cap B)\). 3. Divide both sides by the same positive number \(P(B)\) to obtain \(P(A\mid B)\le P(C\mid B)\).

Answer

\(P(A\mid B)\le P(C\mid B)\). Intersecting both events with the same conditioning event preserves inclusion, and dividing by the same positive probability preserves the inequality.
54731112
Suppose \(P(A\mid B)=0.40\). Is \(P(A\cap B)\) determined? Give two probability models with different positive values of \(P(B)\) that both satisfy the conditional probability, and find the corresponding joint probabilities.

Hints

- Rearrange the conditional-probability relationship without assuming the size of the conditioning event. - Choose two valid positive probabilities for the conditioning event. - Check the ratio in each constructed model.

Solution

1. The relationship is \(P(A\cap B)=0.40P(B)\). 2. If \(P(B)=0.50\), then \(P(A\cap B)=0.20\). 3. If \(P(B)=0.20\), then \(P(A\cap B)=0.08\). 4. Both models have \(P(A\mid B)=0.40\), so the conditional probability alone does not determine the joint probability.

Answer

No. For example: - If \(P(B)=0.50\), then \(P(A\cap B)=0.20\). - If \(P(B)=0.20\), then \(P(A\cap B)=0.08\). Both produce \(P(A\mid B)=0.40\).
54731312
Assume \(P(A)>0\) and \(P(B)>0\). Starting from the definition of conditional probability, prove the two forms of the multiplication rule: \(P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A)\). Explain why the two products describe the same joint event even though they condition in opposite directions.

Hints

- Write each conditional probability as a ratio. - Clear the denominator in each equation. - Compare the joint event appearing in both definitions.

Solution

1. From \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\), multiply by \(P(B)\) to obtain \(P(A\cap B)=P(A\mid B)P(B)\). 2. From \(P(B\mid A)=\frac{P(A\cap B)}{P(A)}\), multiply by \(P(A)\) to obtain \(P(A\cap B)=P(B\mid A)P(A)\). 3. Both expressions calculate the probability of the same intersection; they only factor it using different starting events.

Answer

\(P(A\cap B)=P(A\mid B)P(B)=P(B\mid A)P(A)\). The factors differ, but both products equal the same intersection probability.
54731912
Use the probability tree shown. Given that a randomly selected item is defective, find the probability that it came from Machine 3.
Figure for problem 547319

Hints

- Find each machine’s contribution to the defective items. - Add those contributions to form the conditioning group. - Compare Machine 3’s defective contribution with the total defective probability.

Solution

1. The joint defective shares are \(0.50\cdot0.01=0.005\), \(0.30\cdot0.02=0.006\), and \(0.20\cdot0.05=0.010\). 2. The total defective probability is \(0.005+0.006+0.010=0.021\). 3. \(P(\text{Machine 3}\mid\text{defective})=\frac{0.010}{0.021}=\frac{10}{21}\approx 0.476\).

Answer

\(\frac{10}{21}\approx 0.476\).
54732412
A point is selected uniformly from the square shown. Given that the point lies in the shaded region, find the probability that \(x<\frac12\).
Figure for problem 547324

Hints

- Treat the condition as a restricted geometric region. - Find the area satisfying both conditions. - Divide intersection area by conditioning-region area.

Solution

1. The conditioning region is the triangle above \(y=x\), with area \(\frac12\). 2. In the half-strip \(x<\frac12\), the region above the line has area \(\frac12-\frac12\cdot\frac12\cdot\frac12=\frac38\). 3. The conditional probability is \(\frac{\frac38}{\frac12}=\frac34\).

Answer

The probability is \(\frac34\).
54732512
A fair six-sided die is rolled repeatedly until the first \(6\). Given that the first \(6\) occurs within the first three rolls, find the probability that it occurs on the second roll.

Hints

- Describe the exact sequence required for the first success to occur on the specified roll. - Find the probability of the entire stopping-time condition. - The favorable event is contained inside the conditioning event.

Solution

1. The probability that the first \(6\) occurs on roll \(2\) is \(\frac56\cdot\frac16=\frac{5}{36}\). 2. The probability that a \(6\) occurs within three rolls is \(1-(\frac56)^3=\frac{91}{216}\). 3. The conditional probability is \(\frac{\frac{5}{36}}{\frac{91}{216}}=\frac{30}{91}\).

Answer

The probability is \(\frac{30}{91}\approx 0.330\).
54732712
Two fair dice are rolled. Given that the product is even, find the probability that the sum is even.

Hints

- Characterize parity patterns for product and sum. - Remove patterns excluded by the condition. - Count ordered die outcomes for the surviving pattern.

Solution

1. Product even excludes only odd-odd outcomes, so the conditioning set has \(36-9=27\) outcomes. 2. An even sum occurs for even-even or odd-odd. Within the conditioning set, only even-even remains, with \(9\) outcomes. 3. The conditional probability is \(\frac{9}{27}=\frac13\).

Answer

The probability is \(\frac13\).
54732812
Within event \(B\), the odds in favor of event \(A\) are \(3\) to \(7\). Also, \(P(B)=0.40\). Find \(P(A\cap B)\) and \(P(A^c\cap B)\).

Hints

- Convert the two parts of the odds into shares of the conditioning event. - The two requested intersections partition event \(B\). - Check that the two joint probabilities add to the given probability of \(B\).

Solution

1. Odds of \(3\) to \(7\) within \(B\) mean \(P(A\mid B)=\frac{3}{10}\) and \(P(A^c\mid B)=\frac{7}{10}\). 2. \(P(A\cap B)=P(B)P(A\mid B)=0.40\cdot0.30=0.12\). 3. \(P(A^c\cap B)=0.40\cdot0.70=0.28\).

Answer

\(P(A\cap B)=0.12\) and \(P(A^c\cap B)=0.28\).
54732912
Assume \(P(A)>0\) and \(P(B)>0\). Can both statements hold? \(P(A\mid B)=1\) and \(P(B\mid A)=0\). Prove your conclusion using intersection probabilities.

Hints

- Express both conditionals through the same intersection. - Use the positivity of each conditioning event. - Compare the two consequences for the intersection probability.

Solution

1. \(P(A\mid B)=1\) implies \(P(A\cap B)=P(B)>0\). 2. \(P(B\mid A)=0\) implies \(P(A\cap B)=0\). 3. The same intersection cannot have both positive probability and probability \(0\). Therefore the two statements are incompatible under the positive-margin assumptions.

Answer

No. The first statement forces \(P(A\cap B)=P(B)>0\), while the second forces \(P(A\cap B)=0\).
54733012
Suppose \(P(A)=0.55\), \(P(B)=0.40\), and \(P(A\cap B)=0.25\). Find \(P(A\mid B^c)\).

Hints

- Split event \(A\) into its parts inside and outside \(B\). - Use the complement of the conditioning event as the denominator. - Verify that the intersection part does not exceed the complement.

Solution

1. \(P(A\cap B^c)=P(A)-P(A\cap B)=0.55-0.25=0.30\). 2. \(P(B^c)=1-0.40=0.60\). 3. \(P(A\mid B^c)=\frac{0.30}{0.60}=0.50\).

Answer

\(P(A\mid B^c)=0.50\).
54733212
Assume \(P(A)>0\), \(P(B)>0\), and \(P(A\cap B)>0\). Prove that \(P(A\mid B)=P(B\mid A)\) if and only if \(P(A)=P(B)\).

Hints

- Write both reverse conditionals with their common numerator. - The positive intersection allows cancellation. - Identify what must be equal after the common factor is removed.

Solution

1. The two conditionals are \(\frac{P(A\cap B)}{P(B)}\) and \(\frac{P(A\cap B)}{P(A)}\). 2. Because the common numerator is positive, equality of the ratios is equivalent to equality of the denominators. 3. Thus the reverse conditionals are equal exactly when \(P(A)=P(B)\).

Answer

\(P(A\mid B)=P(B\mid A)\) exactly when \(P(A)=P(B)\), under the stated positivity assumptions.
54733612
A student claims that \(P(A^c\mid B^c)=1-P(A\mid B)\) for all events with positive conditioning probabilities. Use the uniform sample space \(S=\{1,2,3,4\}\), with \(A=\{1\}\) and \(B=\{1,2\}\), to refute the claim. State the correct conditional-complement identity.

Hints

- Compute each conditional probability from its own restricted sample space. - Changing the event and changing the condition are different operations. - A conditional complement must use the same conditioning event.

Solution

1. \(P(A\mid B)=\frac12\), so \(1-P(A\mid B)=\frac12\). 2. \(B^c=\{3,4\}\) and \(A^c=\{2,3,4\}\). Thus \(P(A^c\mid B^c)=\frac22=1\). 3. Since \(1\ne\frac12\), the claim is false. 4. The correct identity keeps the conditioning event fixed: \(P(A^c\mid B)=1-P(A\mid B)\).

Answer

The proposed identity is false: \(P(A^c\mid B^c)=1\), while \(1-P(A\mid B)=\frac12\). The correct identity is \(P(A^c\mid B)=1-P(A\mid B)\).
54733812
In the uniform sample space \(S=\{1,2,3,4\}\), let \(B=\{1,2\}\), \(A=\{1,3\}\), and \(C=\{2,3\}\). a) Are \(A\) and \(C\) mutually exclusive in the original sample space? b) After restricting attention to outcomes in \(B\), can \(A\) and \(C\) occur together? Compute \(P(A\cap C\mid B)\) to justify your answer. c) Explain how the answers to a) and b) can differ.

Hints

- Find the intersection \(A\cap C\) in the original sample space first. - Then check whether that shared outcome is still available after restricting to \(B\). - Express the restricted overlap with the conditional-probability ratio.

Solution

1. In the original sample space, \(A\cap C=\{3\}\), so \(A\) and \(C\) are not mutually exclusive. 2. The conditioning event \(B\) excludes outcome \(3\). Therefore \(A\cap C\cap B=\varnothing\). 3. Since \(P(B)>0\), \(P(A\cap C\mid B)=\frac{P(A\cap C\cap B)}{P(B)}=0\). 4. Conditioning changes the reference set: an overlap that lies entirely outside \(B\) contributes no probability within the restricted sample space.

Answer

a) No; \(A\cap C=\{3\}\). b) No. Within \(B\), they cannot occur together, and \(P(A\cap C\mid B)=0\). c) Their only shared outcome, \(3\), is excluded when attention is restricted to \(B\).
54733912
Events \(A\) and \(B\) satisfy \(P(A)=0.50\), \(P(B)=0.40\), and \(P(A\triangle B)=0.30\), where \(A\triangle B\) means exactly one of the events occurs. Find \(P(A\cap B)\) and \(P(A\mid B)\).

Hints

- “Exactly one” counts the two one-sided differences but excludes the intersection. - Express the symmetric difference using the two margins and the overlap. - Use the recovered intersection as the conditional numerator.

Solution

1. The symmetric-difference identity is \(P(A\triangle B)=P(A)+P(B)-2P(A\cap B)\). 2. Thus \(0.30=0.50+0.40-2x\), so \(2x=0.60\) and \(x=0.30\). 3. \(P(A\mid B)=\frac{0.30}{0.40}=0.75\).

Answer

\(P(A\cap B)=0.30\) and \(P(A\mid B)=0.75\).
54734112
Among \(200\) students, \(80\) are in event \(A\), \(70\) are in event \(B\), and \(50\) are in event \(C\). Also, \(|A\cap B|=30\), \(|A\cap C|=25\), \(|B\cap C|=20\), and \(|A\cap B\cap C|=10\). Find \(P(A\mid B\cup C)\).

Hints

- First determine the size of the conditioning union. - Rewrite the favorable event by distributing the intersection over the union. - Watch for observations counted in both favorable pieces.

Solution

1. \(|B\cup C|=70+50-20=100\). 2. \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\). 3. Its size is \(30+25-10=45\), subtracting the triple intersection once. 4. Therefore \(P(A\mid B\cup C)=\frac{45}{100}=0.45\).

Answer

\(P(A\mid B\cup C)=0.45\).
54734412
In a summer survey of \(100\) people, \(40\) bought ice cream, \(30\) got sunburned, and \(25\) did both. a) Find \(P(\text{sunburned}\mid\text{bought ice cream})\). b) A headline says, “Buying ice cream causes sunburn.” Explain why the conditional probability does not justify that causal conclusion, and name one plausible common cause.

Hints

- Use the group named after “given” as the denominator. - Distinguish observing a relationship from assigning a treatment. - Look for a third variable that could influence both events.

Solution

1. The conditional probability is \(\frac{25}{40}=0.625\). 2. The data are observational and establish association, not the effect of an intervention. 3. Hot, sunny weather can increase both ice-cream purchases and sun exposure, creating the observed association without ice cream causing sunburn.

Answer

a) \(P(\text{sunburned}\mid\text{bought ice cream})=0.625\). b) The survey shows association only. Hot, sunny weather is a plausible common cause.
54734612
A health survey sampled \(1000\) people, but only \(600\) responded to the smoking-status question. Among respondents, \(180\) reported smoking. Find \(P(\text{smoker}\mid\text{responded})\). Can the marginal probability \(P(\text{smoker})\) for all sampled people be determined from these numbers? State an assumption that would justify using the respondent rate for the full sample.

Hints

- Use only respondents in the conditional denominator. - Identify which group’s smoking outcomes are unobserved. - State what relationship between response and smoking would permit generalization.

Solution

1. The respondent group is the conditioning event, so \(P(\text{smoker}\mid\text{responded})=\frac{180}{600}=0.30\). 2. Smoking status is unknown for the \(400\) nonrespondents, so the full-sample marginal rate is not determined. 3. Using \(0.30\) for the full sample would require an assumption such as response status being independent of smoking status within the sampled population.

Answer

\(P(\text{smoker}\mid\text{responded})=0.30\). The full-sample smoking probability is not identifiable without an assumption about nonrespondents; independence of response and smoking would justify using \(0.30\).
54734812
Before an alarm, the odds that a product came from Line 2 rather than Line 1 are \(1\) to \(4\). An alarm is \(6\) times as likely for a Line 2 product as for a Line 1 product. Use odds form to find the posterior odds and posterior probability that an alarmed product came from Line 2.

Hints

- Treat the given comparison as odds, not a probability. - Update the odds by how much more likely the evidence is under one source. - Convert the resulting odds to a probability.

Solution

1. Prior odds for Line 2 versus Line 1 are \(1\) to \(4\). 2. Multiply the odds by the likelihood ratio \(6\), giving posterior odds of \(6\) to \(4\), or \(3\) to \(2\). 3. Odds of \(3\) to \(2\) correspond to posterior probability \(\frac{3}{3+2}=\frac35=0.60\).

Answer

Posterior odds are \(3\) to \(2\), so the posterior probability of Line 2 is \(\frac35=0.60\).
54729812
Events \(A\) and \(B\) are mutually exclusive, and \(C\) is an event with \(P(C)>0\). Prove that \(P(A\cup B\mid C)=P(A\mid C)+P(B\mid C)\). State why no independence assumption is needed.

Hints

- Intersect both original events with the conditioning event. - Verify that disjoint events remain disjoint after the same intersection. - Divide the ordinary addition rule by the conditioning probability.

Solution

1. Since \(A\cap B=\varnothing\), the events \(A\cap C\) and \(B\cap C\) are also disjoint. 2. \((A\cup B)\cap C=(A\cap C)\cup(B\cap C)\). 3. Additivity gives \(P((A\cup B)\cap C)=P(A\cap C)+P(B\cap C)\). 4. Divide by \(P(C)>0\) to obtain the conditional identity. 5. Independence is irrelevant; the proof uses only disjointness and the definition of conditional probability.

Answer

\(P(A\cup B\mid C)=P(A\mid C)+P(B\mid C)\). Disjointness is sufficient; independence is not required.
54731412
A transit company operates \(10\) small buses carrying \(10\) riders each and \(5\) large buses carrying \(40\) riders each. All buses are full. Two distinct riders are selected uniformly from all riders, conditional on the two riders being on the same bus. Find the probability that they are on a large bus.

Hints

- Represent the random selection in a way that keeps the original elementary outcomes equally likely. - Apply the condition “the two riders are on the same bus” before comparing bus sizes. - Check whether every bus contributes the same number of eligible selections after conditioning.

Solution

1. A small bus contributes \(\binom{10}{2}=45\) same-bus rider pairs, so all small buses contribute \(10\cdot45=450\). 2. A large bus contributes \(\binom{40}{2}=780\) pairs, so all large buses contribute \(5\cdot780=3900\). 3. Under the condition that the pair shares a bus, the eligible pair count is \(450+3900=4350\). 4. The probability of a large bus is \(\frac{3900}{4350}=\frac{26}{29}\).

Answer

The conditional probability is \(\frac{26}{29}\approx0.897\).
54731512
Suppose \(P(A)=0.60\) and \(P(B)=0.50\). No other information is known. Find the smallest and largest possible values of \(P(A\mid B)\), and show that both bounds are attainable.

Hints

- First bound the intersection using only the two marginal probabilities. - Conditional probability rescales the intersection by \(P(B)\). - Describe event arrangements that attain the smallest and largest intersections.

Solution

1. The intersection satisfies \(\max(0,P(A)+P(B)-1)\le P(A\cap B)\le\min(P(A),P(B))\). 2. Thus \(0.10\le P(A\cap B)\le0.50\). 3. Divide by \(P(B)=0.50\): \(0.20\le P(A\mid B)\le1\). 4. The lower endpoint is attained when the union has probability \(1\), and the upper endpoint is attained when \(B\subseteq A\).

Answer

\(P(A\mid B)\in[0.20, 1]\), and both endpoints are possible.
54731612
Let \(P(B)>0\). Prove the conditional intersection bound \(P(A\cap C\mid B)\ge P(A\mid B)+P(C\mid B)-1\). Then use it when \(P(A\mid B)=0.75\) and \(P(C\mid B)=0.65\).

Hints

- Work entirely within the outcomes for which \(B\) occurs. What quantities in the statement become ordinary event probabilities in that restricted setting? - Look for a probability relationship involving both \(A\cup C\) and \(A\cap C\). - What universal upper bound on an event probability can turn that relationship into a lower bound for the intersection?

Solution

1. Within the conditional probability measure given \(B\), the union bound gives \(P(A\cup C\mid B)\le1\). 2. Conditional inclusion-exclusion gives \(P(A\cup C\mid B)=P(A\mid B)+P(C\mid B)-P(A\cap C\mid B)\). 3. Rearranging yields the stated lower bound. 4. Numerically, \(P(A\cap C\mid B)\ge0.75+0.65-1=0.40\).

Answer

The general lower bound is \(P(A\mid B)+P(C\mid B)-1\). For the given values, \(P(A\cap C\mid B)\ge0.40\).
54731812
Suppose \(P(B)>0\) and \(P(A\mid B)=1\). What does this imply about \(P(B\setminus A)\)? Does it necessarily imply the strict set inclusion \(B\subseteq A\) when probability-zero outcomes are possible? Explain.

Hints

- Decompose the conditioning event into the part inside and outside \(A\). - Use the fact that the conditional probability equals \(1\). - Distinguish an empty set from a nonempty probability-zero set.

Solution

1. \(P(A\mid B)=1\) gives \(P(A\cap B)=P(B)\). 2. Since \(B=(A\cap B)\cup(B\setminus A)\) is a disjoint union, \(P(B\setminus A)=P(B)-P(A\cap B)=0\). 3. Thus \(B\) is contained in \(A\) up to a probability-zero part. 4. Strict set inclusion need not follow if \(B\setminus A\) is nonempty but has probability \(0\).

Answer

\(P(B\setminus A)=0\). The condition implies \(B\subseteq A\) only up to probability-zero outcomes, not necessarily as a literal set inclusion.
54732012
There is no general monotonicity rule for the conditioning event. In the uniform sample space \(S=\{1,2,3,4\}\), let \(A=\{1,2\}\). a) Find events \(B\subset C\) such that \(P(A\mid B)>P(A\mid C)\). b) Find events \(D\subset E\) such that \(P(A\mid D)<P(A\mid E)\). c) Explain what the two examples show.

Hints

- In this uniform sample space, conditional probability can be computed from the fraction of the conditioning set that lies in \(A\). - You need two nested-set examples in which that fraction moves in opposite directions. - Focus on how adding outcomes changes both the intersection with \(A\) and the size of the conditioning event.

Solution

1. Take \(B=\{1\}\) and \(C=\{1,3\}\). Then \(P(A\mid B)=1\) and \(P(A\mid C)=\frac12\). 2. Take \(D=\{3\}\) and \(E=\{1,3\}\). Then \(P(A\mid D)=0\) and \(P(A\mid E)=\frac12\). 3. Enlarging the conditioning event can either decrease or increase a conditional probability, depending on which new outcomes are added.

Answer

a) One example is \(B=\{1\}\subset C=\{1,3\}\), giving \(1>\frac12\). b) One example is \(D=\{3\}\subset E=\{1,3\}\), giving \(0<\frac12\). c) Conditional probability is not monotone in the conditioning event.
54732112
Suppose events \(B\) and \(C\) satisfy \(P(B\triangle C)=0\) and have positive probability. Prove that for every event \(A\), \(P(A\mid B)=P(A\mid C)\).

Hints

- Break the symmetric difference into the two one-sided differences. - Compare the probabilities of the conditioning events and of their intersections with \(A\). - Probability-zero discrepancies do not change either conditional ratio.

Solution

1. Since \(B\triangle C\) has probability \(0\), the differences \(B\setminus C\) and \(C\setminus B\) each have probability \(0\). 2. Therefore \(P(B)=P(B\cap C)=P(C)\). 3. The sets \(A\cap B\) and \(A\cap C\) can differ only inside \(B\triangle C\), so their probabilities are equal. 4. Equal numerators and equal positive denominators give \(P(A\mid B)=P(A\mid C)\).

Answer

For every event \(A\), \(P(A\mid B)=P(A\mid C)\). Conditioning is unchanged when the conditioning events differ only on a null set.
54732612
Let \(B\) and \(C\) be mutually exclusive events with \(P(B)>0\) and \(P(C)>0\). Prove that \(P(A\mid B\cup C)=P(A\mid B)P(B\mid B\cup C)+P(A\mid C)P(C\mid B\cup C)\).

Hints

- Split the conditioned intersection across the two disjoint parts. - Insert each subgroup probability as a factor and its reciprocal. - Interpret the result as a weighted average within the union.

Solution

1. Since \(B\) and \(C\) are disjoint, \(A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\) is a disjoint union. 2. Divide by \(P(B\cup C)\): \(P(A\mid B\cup C)=\frac{P(A\cap B)}{P(B\cup C)}+\frac{P(A\cap C)}{P(B\cup C)}\). 3. Factor each term as \(\frac{P(A\cap B)}{P(B)}\cdot\frac{P(B)}{P(B\cup C)}\) and similarly for \(C\). 4. These factors are the conditional probabilities in the stated identity.

Answer

\(P(A\mid B\cup C)=P(A\mid B)P(B\mid B\cup C)+P(A\mid C)P(C\mid B\cup C)\).
54734712
Let \(C\subseteq B\), with \(P(C)>0\). Prove the nested-conditioning identity \(P(A\mid C)=\frac{P(A\cap C\mid B)}{P(C\mid B)}\). Explain why the assumption \(C\subseteq B\) is essential to this form of the identity.

Hints

- Rewrite both conditional probabilities in the ratio using the same denominator. - Use the subset relationship to simplify intersections with \(B\). - Consider what the numerator and denominator would represent if part of \(C\) lay outside \(B\).

Solution

1. Since \(C\subseteq B\), \((A\cap C)\cap B=A\cap C\) and \(C\cap B=C\). 2. Therefore \(P(A\cap C\mid B)=\frac{P(A\cap C)}{P(B)}\) and \(P(C\mid B)=\frac{P(C)}{P(B)}\). 3. Dividing cancels \(P(B)\): \(\frac{P(A\cap C\mid B)}{P(C\mid B)}=\frac{P(A\cap C)}{P(C)}=P(A\mid C)\). 4. Without \(C\subseteq B\), conditioning on \(B\) replaces \(C\) by \(B\cap C\), so the ratio generally describes conditioning on \(B\cap C\), not on \(C\).

Answer

\(P(A\mid C)=\frac{P(A\cap C\mid B)}{P(C\mid B)}\). The subset condition ensures that conditioning first on \(B\) does not remove any outcomes from \(C\).
54735012
Events satisfy \(P(A\mid B)=0.60\), \(P(B\mid A)=0.40\), and \(P(A\cup B)=0.70\). Find \(P(A)\), \(P(B)\), and \(P(A\cap B)\).

Hints

- Express both marginal probabilities in terms of the shared intersection. - Substitute those expressions into the union formula. - Solve for the intersection before recovering the marginals.

Solution

1. Let \(I=P(A\cap B)\). Then \(P(B)=\frac{I}{0.60}\) and \(P(A)=\frac{I}{0.40}\). 2. The union equation is \(\frac{I}{0.40}+\frac{I}{0.60}-I=0.70\). 3. This gives \(\frac{19}{6}I=0.70\), so \(I=\frac{21}{95}\), \(P(A)=\frac{21}{38}\), and \(P(B)=\frac{7}{19}\).

Answer

\(P(A\cap B)=\frac{21}{95}\), \(P(A)=\frac{21}{38}\), and \(P(B)=\frac{7}{19}\).
54735112
Assume events \(A,B,C\) have positive probabilities and positive pairwise intersections. A report claims \(P(A\mid B)=0.60\), \(P(B\mid A)=0.30\), \(P(B\mid C)=0.50\), \(P(C\mid B)=0.50\), \(P(C\mid A)=0.40\), and \(P(A\mid C)=0.40\). Can all six claims hold simultaneously?

Hints

- For one pair of events, write both directions of the conditional probability definition and notice what quantity appears in both numerators. - What necessary relationship between the two marginal probabilities can be recovered from one pair of reverse conditionals? - Try to combine the analogous necessary relationships for all three pairs and ask what must cancel in any internally consistent model.

Solution

1. For positive intersections, \(\frac{P(A\mid B)}{P(B\mid A)}=\frac{P(A)}{P(B)}\), and similarly around the cycle. 2. Multiplying the three ratios must give \(\frac{P(A)}{P(B)}\cdot\frac{P(B)}{P(C)}\cdot\frac{P(C)}{P(A)}=1\). 3. The reported ratios give \(\frac{0.60}{0.30}\cdot\frac{0.50}{0.50}\cdot\frac{0.40}{0.40}=2\), not \(1\). 4. Therefore the claims are inconsistent.

Answer

No. The cycle of reverse-conditional ratios must multiply to \(1\), but the reported values give \(2\).
54885512
A normal variable has mean \(70\) and standard deviation \(10\). Given that an observation is above \(70\), find the conditional probability that it is above \(80\). Use a standard normal table or calculator.
Figure for problem 548855

Hints

- Identify the new reference group created by the given condition. - Express the conditional probability as a ratio of nested tail areas. - Use symmetry to find the area above the mean.

Solution

1. The event \(X>80\) is contained in the event \(X>70\), so \(P(X>80\mid X>70)=\frac{P(X>80)}{P(X>70)}\). 2. The standard score for \(80\) is \(z=\frac{80-70}{10}=1\), so \(P(X>80)=P(Z>1)\approx0.1587\). 3. By symmetry, \(P(X>70)=0.5\). 4. Using the unrounded tail probability, the conditional probability is approximately \(\frac{0.158655}{0.5}\approx0.3173\).

Answer

\(P(X>80\mid X>70)\approx0.3173\), or about \(31.73\%\).

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