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Statistics Quiz

Statistics Quiz: Conditional Probability And Independence

Practice Conditional Probability And Independence in Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

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 $P(A)$ .

What this quiz covers

This quiz focuses on Conditional Probability And Independence, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

No; $P(A\mid B)=0.75$ , which is not equal to $P(A)$ . (correct answer)
No; $P(A\mid B)=0.40$ , which is not equal to $P(A)$ .
Yes; $P(A\mid B)=0.30$ , so $P(A\mid B)=P(A\cap B)$ .
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.

Question 2

In a study of 300 households, 120 have a pet (event $B$ ). Of those, 45 also have a garden (event $A\cap B$ ), where event $A$ is “the household has a garden.” Based on the information given, what is $P(A\mid B)$ ? Give your answer as a percent.

$25\%$
$15\%$
$37.5\%$ (correct answer)
$40\%$

Explanation: This question requires conditional probability. P(A|B) is the fraction of pet households (B) that also have a garden (A). With 120 pet households and 45 with both, P(A|B) = 45/120 = 0.375 or 37.5%. This means 37.5% of pet-owning households have a garden, identifying the correct answer. Misconceptions include using total households (300) as denominator, giving 45/300 = 15%, but we condition on B. Generally, underline 'given' (have a pet, 120), use as denominator group, and compare those with gardens. This focuses on the specified subset.

Question 3

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; $P(A\mid B)=0.24$ , which is not equal to $P(A)$ . (correct answer)
Yes; $P(A\mid B)=0.30$ , so $P(A\mid B)=P(A)$ .
Yes; $P(A\mid B)=0.50$ , so $P(A\mid B)=P(B)$ .
No; $P(A\mid B)=0.12$ , 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.

Question 4

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.

$\frac{8}{50}$
$\frac{8}{30}$
$\frac{8}{20}$ (correct answer)
$\frac{20}{50}$

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.

Question 5

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 $P(A)$ .

Yes; $P(A\mid B)=0.25$ , so $P(A\mid B)=P(B)$ .
No; $P(A\mid B)=0.10$ , which is not equal to $P(A)$ .
Yes; $P(A\mid B)=0.40$ , so $P(A\mid B)=P(A)$ . (correct answer)
No; $P(A\mid B)=0.50$ , which is not equal to $P(A)$ .

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.

Question 6

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.

$\frac{4}{10}$ (correct answer)
$\frac{14}{40}$
$\frac{10}{40}$
$\frac{4}{40}$

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.

Question 7

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.

$\frac{12}{48}$
$\frac{12}{80}$
$\frac{32}{80}$
$\frac{12}{32}$ (correct answer)

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.

Question 8

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.

15%
65%
30% (correct answer)
7.5%

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.

Question 9

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.

$0.18$
$0.60$ (correct answer)
$0.12$
$0.54$

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.

Question 10

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 and plays a sport is $\frac{0.06}{0.20}$ .
Among those who play a sport, the probability a student is in band is $\frac{0.06}{0.20}$ . (correct answer)
The probability a student is in band or 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.

Question 11

In a group of 200 employees, 80 work remotely (event $B$ ), and 28 both work remotely and are in management (event $A \cap B$ ), where event $A$ is “the employee is in management.” Based on the information given, what is $P(A \mid B)$ ? Give your answer as a fraction in simplest form.

$\frac{14}{25}$
$\frac{2}{5}$
$\frac{7}{20}$ (correct answer)
$\frac{7}{50}$

Explanation: This question involves conditional probability. $P(A|B)$ represents the probability of A given B, or among the B outcomes, the fraction that are also A. With 80 employees working remotely (B) and 28 of them in management ( $A \cap B$ ), $P(A|B) = 28/80 = 7/20$ . Thus, among remote workers, the probability an employee is in management is 7/20, making that the correct answer. A distractor might arise from using the total employees (200) as the denominator instead of just the remote ones, leading to errors like 28/200 = 7/50. For transfer, underline the 'given' (remotely, so 80), use that as the denominator group, and compare the subset that also fits A (management). This ensures focus on the conditioned subgroup.

Question 12

A factory reports $P(A\cap B)=0.08$ and $P(B)=0.16$ , where event $A$ is “an item is defective” and event $B$ is “an item came from Machine 2.” Based on the information given, what is $P(A\mid B)$ ? Give your answer as a decimal.

$0.50$ (correct answer)
$0.08$
$0.24$
$2.00$

Explanation: This question focuses on conditional probability. P(A|B) means among items from Machine 2, the fraction defective. We find P(A|B) = P(A∩B)/P(B) = 0.08/0.16 = 0.50. So, given from Machine 2, the defect probability is 0.50, which is the correct decimal. Errors might include not dividing, picking 0.08, or reversing to divide wrongly like 0.16/0.08=2.00, but conditioning is on B. Broadly, underline 'given' (Machine 2, P(B)=0.16), use as denominator, and find defective fraction. This isolates the conditioned probability.

Question 13

A website tracks user behavior. Event $A$ is “a user clicks the newsletter link,” and event $B$ is “a user is on a mobile device.” The data show $P(A\cap B)=0.14$ and $P(B)=0.35$ . Based on the information given, what is $P(A\mid B)$ ? Give your answer as a fraction in simplest form.

$\frac{7}{25}$
$\frac{14}{35}$
$\frac{1}{5}$
$\frac{2}{5}$ (correct answer)

Explanation: This question involves conditional probability. P(A|B) is the probability of clicking the newsletter given mobile device, or among mobile users, the fraction clicking. We calculate P(A|B) = P(A∩B)/P(B) = 0.14/0.35 = 0.40 = 2/5. Thus, among mobile users, 2/5 click the link, confirming the correct simplified fraction. A wrong denominator, like not dividing properly, might yield 14/35 (unsimplified) or other errors, but we simplify after conditioning on B. To apply elsewhere, underline 'given' (mobile, P(B)=0.35), set as denominator, and compute the intersection fraction. This ensures accurate conditioning.

Question 14

In a class of 40 students, 16 are taking calculus (event $B$ ). Of those, 10 also play a varsity sport (event $A\cap B$ ), where event $A$ is “the student plays a varsity sport.” Based on the information given, what is $P(A\mid B)$ ? Give your answer as a fraction in simplest form.

$\frac{3}{4}$
$\frac{5}{8}$ (correct answer)
$\frac{8}{5}$
$\frac{10}{40}$

Explanation: This question uses conditional probability. P(A|B) is among calculus students, the fraction playing varsity sports. With 16 in calculus and 10 both, P(A|B) = 10/16 = 5/8. This indicates 5/8 of calculus students play varsity, confirming the correct fraction. A distractor is using total students (40) as denominator, giving 10/40=1/4, but we condition on B. To generalize, underline 'given' (calculus, 16), set as denominator group, and compare sports players among them. This approach emphasizes the restricted sample.

Question 15

A jar contains 50 marbles. Event $B$ is “a randomly selected marble is blue,” and event $A$ is “a randomly selected marble is striped.” There are 20 blue marbles, and 6 marbles are both blue and striped. Based on the information given, what is $P(A\mid B)$ ? Give your answer as a fraction in simplest form.

$\frac{3}{10}$ (correct answer)
$\frac{6}{50}$
$\frac{7}{25}$
$\frac{6}{20}$

Explanation: This question deals with conditional probability. P(A|B) is the probability of A given B, meaning among blue marbles, the fraction that are striped. With 20 blue marbles and 6 both blue and striped, P(A|B) = 6/20 = 3/10. So, given a marble is blue, the probability it is striped is 3/10, which is the correct choice. A common error is using the total marbles (50) as the denominator, yielding 6/50, but we condition only on blue ones. For broader application, underline 'given' (blue, so 20), set as denominator, and count the striped among them. This method restricts to the relevant group.

Question 16

A school surveyed 200 students about whether they play a sport and whether they are in the band. Let event $A$ be “the student plays a sport” and event $B$ be “the student is in the band.” Based on the information given that $P(A\cap B)=0.18$ and $P(B)=0.30$ , what is $P(A\mid B)$ ? Give your answer as a simplified fraction.

$\frac{3}{5}$ (correct answer)
$\frac{5}{6}$
$\frac{3}{10}$
$\frac{9}{50}$

Explanation: This problem asks for the conditional probability P(A|B), where A is "plays a sport" and B is "is in the band." The conditional probability P(A|B) means "among students who are in the band, what fraction also play a sport?" Using the formula P(A|B) = P(A∩B)/P(B), we substitute the given values: P(A|B) = 0.18/0.30 = 18/30 = 3/5. This means that 3/5 (or 60%) of band students also play a sport. A common mistake is to reverse the condition and calculate P(B|A) instead, or to use P(A) as the denominator. The key strategy is to identify what's given ("in the band") and make that the denominator group, then find what fraction of that group also satisfies the other condition.

Question 17

A factory inspected 100 light bulbs. Event $A$ is “the bulb is defective,” and event $B$ is “the bulb is from Machine 1.” Of the 40 bulbs from Machine 1, 6 were defective. Based on the information given, what is $P(A\mid B)$ ? Give your answer as a simplified fraction.

$\frac{3}{20}$ (correct answer)
$\frac{2}{15}$
$\frac{20}{3}$
$\frac{6}{100}$

Explanation: This problem asks for P(A|B), the probability a bulb is defective given it's from Machine 1. The conditional probability P(A|B) means "among bulbs from Machine 1, what fraction are defective?" We're told there are 40 bulbs from Machine 1 (our denominator group) and 6 of these are defective (our numerator). Therefore, P(A|B) = 6/40 = 3/20. This tells us that 3/20 (or 15%) of Machine 1's bulbs are defective. A common mistake is using the total number of bulbs (100) in the calculation instead of focusing only on Machine 1's output. The strategy is to restrict attention to the given condition (Machine 1 bulbs) and find what fraction of those meet the other criterion.

Question 18

A jar contains 50 marbles. Event $A$ is “a randomly selected marble is red,” and event $B$ is “a randomly selected marble is large.” There are 12 marbles that are both red and large, and there are 20 large marbles total. Based on the information given, what is $P(A\mid B)$ ? Give your answer as a simplified fraction.

$\frac{12}{50}$
$\frac{3}{5}$ (correct answer)
$\frac{2}{3}$
$\frac{5}{12}$

Explanation: We need to find P(A|B), the probability a marble is red given that it's large. The conditional probability P(A|B) means "among the large marbles, what fraction are red?" We're told there are 20 large marbles total (this is our denominator group) and 12 marbles that are both red and large (this is our numerator). Therefore, P(A|B) = 12/20 = 3/5. This tells us that 3/5 of the large marbles are red. A common error is using the total number of marbles (50) as the denominator instead of focusing only on the large marbles. Remember: when finding P(A|B), we restrict our attention to only the B outcomes and ask what fraction of those also satisfy A.

Question 19

A streaming service tracks users. Event $A$ is “a user watches a comedy,” and event $B$ is “a user has a premium subscription.” Based on the information given that $P(A\cap B)=0.12$ and $P(B)=0.20$ , what is $P(A\mid B)$ ? Give your answer as a percent.

12%
32%
60% (correct answer)
8%

Explanation: We need to find P(A|B), the probability a user watches comedy given they have a premium subscription. The conditional probability P(A|B) means "among premium subscribers, what fraction watch comedy?" Using the formula P(A|B) = P(A∩B)/P(B), we get P(A|B) = 0.12/0.20 = 12/20 = 0.60 = 60%. This means 60% of premium subscribers watch comedy. A common error is confusing the intersection probability (12%) with the conditional probability, or reversing the condition to find P(B|A) instead. Remember to identify what's given (premium subscription) and make that your denominator group, then determine what fraction of that group satisfies the other condition.

Question 20

A clinic records whether patients received a flu shot and whether they later got the flu. Event $A$ is “the patient got the flu,” and event $B$ is “the patient received a flu shot.” Based on the information given that $P(A\cap B)=0.06$ and $P(B)=0.40$ , what is $P(A\mid B)$ ? Give your answer as a decimal.

0.15 (correct answer)
0.24
0.67
0.06

Explanation: This problem asks for P(A|B), the probability a patient got the flu given they received a flu shot. The conditional probability P(A|B) means "among patients who got the flu shot, what fraction still got the flu?" Using P(A|B) = P(A∩B)/P(B), we calculate P(A|B) = 0.06/0.40 = 6/40 = 0.15. This means 15% of vaccinated patients still got the flu. A common error is using P(A∩B) = 0.06 as the answer, forgetting to divide by P(B). The key insight is that we're looking at only the vaccinated group (40% of all patients) and finding what fraction of that specific group got the flu.