Conditional Probability and Independence
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Statistics › Conditional Probability and Independence
A university reports $P(A)=0.30$, $P(B)=0.50$, and $P(A\cap B)=0.12$, where event $A$ is “a student studies abroad” and event $B$ is “a student is in the honors program.” Based on the information given, are $A$ and $B$ independent? Use conditional probability by comparing $P(A\mid B)$ and $P(A)$.
No; $P(A\mid B)=0.12$, which is not equal to $P(A)$.
Yes; $P(A\mid B)=0.50$, so $P(A\mid B)=P(B)$.
Yes; $P(A\mid B)=0.30$, so $P(A\mid B)=P(A)$.
No; $P(A\mid B)=0.24$, which is not equal to $P(A)$.
Explanation
This question checks independence with conditional probability. P(A|B) is the fraction of honors students (B) who study abroad (A). Compute P(A|B) = 0.12 / 0.50 = 0.24, not equal to P(A) = 0.30, so not independent. Honors decreases abroad probability from 30% to 24%, supporting 'no' with 0.24 ≠ 0.30. Misconception: comparing to P(B) instead of P(A), as in D. Underline 'given' (honors), find rate in that group, compare to overall P(A). Mismatch indicates dependence.
In a group of 50 employees, 20 work remotely (event $B$) and 8 both work remotely and are on the morning shift (event $A\cap B$), where event $A$ is “an employee is on the morning shift.” Based on the information given, what is $P(A\mid B)$? Give your answer as a simplified fraction.
$\frac{8}{20}$
$\frac{8}{50}$
$\frac{20}{50}$
$\frac{8}{30}$
Explanation
This question involves conditional probability. P(A|B) represents the probability of A given B, or among those in group B (remote workers), the fraction that are also in A (morning shift). We calculate it as the number in A∩B divided by the number in B: 8 / 20 = 8/20 (simplified to 2/5 if needed, but 8/20 matches). Thus, among remote employees, 8 out of 20 are on the morning shift, making 8/20 the correct answer. A frequent distractor is using the total employees as the denominator, like 8/50, which ignores the conditioning on B. For transfer, underline the 'given' (remote workers), use that group as the denominator, and count how many in that group satisfy the condition. This ensures you focus on the restricted sample rather than the whole.
A website’s analytics show $P(A)=0.40$, $P(B)=0.25$, and $P(A\cap B)=0.10$, where event $A$ is “a visitor clicks an ad” and event $B$ is “a visitor is on a mobile device.” Based on the information given, are $A$ and $B$ independent? Use conditional probability by comparing $P(A\mid B)$ and $P(A)$.
No; $P(A\mid B)=0.10$, which is not equal to $P(A)$.
Yes; $P(A\mid B)=0.25$, so $P(A\mid B)=P(B)$.
No; $P(A\mid B)=0.50$, which is not equal to $P(A)$.
Yes; $P(A\mid B)=0.40$, so $P(A\mid B)=P(A)$.
Explanation
This question tests independence using conditional probability. P(A|B) is the fraction of B outcomes (mobile visitors) that are also A (ad clicks). Compute P(A|B) = P(A∩B) / P(B) = 0.10 / 0.25 = 0.40, which equals P(A) = 0.40, so A and B are independent. This means mobile device use doesn't affect ad-click probability, supporting the 'yes' answer with P(A|B) = 0.40 matching P(A). A misconception is comparing P(A|B) to P(B) instead of P(A), as in choice D. To generalize, underline the 'given' (B), compute the conditional rate in that group, and compare to the overall P(A). If they match, independence holds.
A company finds $P(A)=0.50$, $P(B)=0.40$, and $P(A\cap B)=0.30$, where event $A$ is “an employee completed training” and event $B$ is “an employee received a promotion.” Based on the information given, are $A$ and $B$ independent? Use conditional probability by comparing $P(A\mid B)$ and $P(A)$.
No; $P(A\mid B)=0.40$, which is not equal to $P(A)$.
Yes; $P(A\mid B)=0.30$, so $P(A\mid B)=P(A\cap B)$.
No; $P(A\mid B)=0.75$, which is not equal to $P(A)$.
Yes; $P(A\mid B)=0.50$, so $P(A\mid B)=P(A)$.
Explanation
This question assesses independence via conditional probability. P(A|B) is the fraction of promoted employees (B) who completed training (A). Compute P(A|B) = 0.30 / 0.40 = 0.75, which does not equal P(A) = 0.50, so not independent. Promotion increases training likelihood from 50% to 75%, confirming 'no' with 0.75 ≠ 0.50. Distractor: confusing with P(A|B) = P(A∩B), as in A, but independence requires P(A|B) = P(A). Underline 'given' (promotion), compute rate in that group, compare to overall P(A). If unequal, dependence exists.
A jar contains 40 marbles. Event $B$ is “the marble is blue,” and event $A$ is “the marble is striped.” There are 10 blue marbles, and 4 of the blue marbles are striped (so $A\cap B$ occurs for 4 marbles). Based on the information given, what is $P(A\mid B)$? Give your answer as a simplified fraction.
$\frac{14}{40}$
$\frac{10}{40}$
$\frac{4}{10}$
$\frac{4}{40}$
Explanation
This question is about conditional probability. P(A|B) means among the blue marbles (B), the fraction that are striped (A). We compute it as the number in A∩B over the number in B: 4 / 10 = 4/10 (simplifies to 2/5). So, among blue marbles, 4 out of 10 are striped, confirming 4/10 as correct. A common error is using the total marbles as denominator, like 4/40, which is P(A∩B) not conditional on B. For broader application, underline the 'given' event (blue), treat that as the new total, and find the proportion that are also striped. This refocuses on the subset.
In a sample of 80 households, 32 have a pet (event $B$). Of those, 12 also subscribe to a streaming service (event $A \cap B$), where event $A$ is “a household subscribes to a streaming service.” Based on the information given, what is $P(A \mid B)$? Give your answer as a simplified fraction.
$\frac{12}{80}$
$\frac{12}{32}$
$\frac{32}{80}$
$\frac{12}{48}$
Explanation
This question deals with conditional probability. $P(A \mid B)$ is the probability of subscribing to streaming (A) given pet ownership (B), or the fraction of pet households that subscribe. Compute as $12 / 32 = \frac{12}{32}$ (simplifies to $\frac{3}{8}$). Among pet households, 12 out of 32 subscribe, so $\frac{12}{32}$ is correct. Misconception: using total households as denominator, like $\frac{12}{80}$, which is joint not conditional. Underline the 'given' (pet owners), use that count as denominator, then find how many in that group subscribe. This method isolates the conditioned subset.
A clinic reports $P(A\cap B)=0.06$ and $P(B)=0.20$, where event $A$ is “a patient has elevated blood pressure” and event $B$ is “a patient is over age 60.” Based on the information given, which statement correctly interprets $P(A\mid B)$?
30% of all patients are both over age 60 and have elevated blood pressure.
Among those who have elevated blood pressure, 30% are over age 60.
Among those who are over age 60, 30% have elevated blood pressure.
Among those who are not over age 60, 30% have elevated blood pressure.
Explanation
This question interprets conditional probability. P(A|B) is the fraction of patients over 60 (B) who have elevated blood pressure (A). Calculate P(A|B) = P(A∩B) / P(B) = 0.06 / 0.20 = 0.30, or 30%. This interprets as 'among those over age 60, 30% have elevated blood pressure,' matching choice A. A distractor misconception is reversing to 'among those with elevated BP, 30% are over 60,' which would require P(B|A). To transfer, underline the 'given' (over 60), make that the base group, and compute the rate of the other event in it. This clarifies the direction of conditioning.
A bookstore’s records show $P(A\cap B)=0.15$ and $P(B)=0.50$, where event $A$ is “a customer buys fiction” and event $B$ is “a customer is a member of the rewards program.” Based on the information given, what is $P(A\mid B)$? Give your answer as a percent.
7.5%
15%
30%
65%
Explanation
This question is on conditional probability. P(A|B) means among rewards members (B), the fraction buying fiction (A). Compute P(A|B) = 0.15 / 0.50 = 0.30, or 30%. So, 30% of members buy fiction, making 30% the answer. Common error: thinking 15% is the conditional, but that's joint; or reversing to P(B|A). Underline the 'given' (members), use that as denominator, find fraction buying fiction. This focuses on the subgroup probability.
A campus survey found that $P(A\cap B)=0.18$ and $P(B)=0.30$, where event $A$ is “a randomly selected student bikes to campus” and event $B$ is “a randomly selected student lives on campus.” Based on the information given, what is $P(A\mid B)$? Give your answer as a decimal.
$0.54$
$0.12$
$0.18$
$0.60$
Explanation
This question focuses on conditional probability. The conditional probability P(A|B) is defined as the probability that event A occurs given that event B has occurred, which can be interpreted as the fraction of B outcomes that are also A. Here, we compute P(A|B) = P(A∩B) / P(B) = 0.18 / 0.30 = 0.60. This means that among students who live on campus, 60% bike to campus, so the correct answer is 0.60. A common misconception is reversing the conditioning to compute P(B|A) instead, which might lead to incorrect choices like 0.18, but we must condition on B as the given event. To apply this generally, underline the 'given' event (B, living on campus), make it the denominator group, and then count or compare the fraction that also satisfy A (biking to campus). This strategy helps avoid using the wrong denominator.
A school reports that $P(A\cap B)=0.06$ and $P(B)=0.20$, where event $A$ is “a randomly selected student is in band” and event $B$ is “a randomly selected student plays a sport.” Based on the information given, which statement correctly interprets $P(A\mid B)$?
Among those who are in band, the probability a student plays a sport is $\frac{0.06}{0.20}$.
The probability a student is in band or plays a sport is $\frac{0.06}{0.20}$.
Among those who play a sport, the probability a student is in band is $\frac{0.06}{0.20}$.
The probability a student is in band and plays a sport is $\frac{0.06}{0.20}$.
Explanation
This question interprets conditional probability. P(A|B) means among those in B (sports), the fraction also in A (band). Given P(A∩B) = 0.06 and P(B) = 0.20, P(A|B) = 0.06/0.20, which describes the probability a student is in band given they play a sport. This matches the statement about among sports players, the probability of being in band, so that's correct. A distractor is reversing to P(B|A), like saying among band students, the probability of sports, but the conditioning is on B. To transfer, underline the 'given' (plays a sport), make it the denominator, and find the fraction also in band. This clarifies the direction of conditioning.