AP Statistics Quiz: Conditional Probability

What this quiz covers

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

Question 1

1. It increases the likelihood of $U$ if $P(U\mid C) > P(U)$. (correct answer)
2. It increases the likelihood of $U$ if $P(C\mid U) > P(C)$.
3. It has no effect because $P(U\mid C)$ must equal $P(C\mid U)$.
4. It has no effect because events like $C$ and $U$ are always independent.
5. It decreases the likelihood of $U$ because listening to $C$ uses time that could be spent upgrading.

Explanation: This question examines how knowing about curated playlist usage affects upgrade likelihood. When we condition on C (user listened to curated playlist), we're asking about P(U|C) - the probability of upgrading given playlist listening occurred. The service suspects playlists encourage upgrades, meaning P(U|C) > P(U). Option A correctly states that knowing C occurred increases the likelihood of U when P(U|C) > P(U). Option B confuses the direction of conditioning, while C and D make false claims about probability relationships. The key insight is that P(U|C) and P(C|U) are different quantities related by Bayes' theorem.

Question 2

1. It makes $P$ more likely exactly when $P(P\mid R) > P(P)$. (correct answer)
2. It makes $P$ more likely exactly when $P(R\mid P) > P(P)$.
3. It makes no difference because $P(P\mid R)$ must equal $P(P)$ whenever $P(R\mid P)$ is large.
4. It makes $P$ less likely exactly when $P(P\mid R) > P(P)$.
5. It makes no difference because $P(P\mid R)=P(R\mid P)$.

Explanation: This question involves how clicking product reviews affects purchase probability. When we know R occurred (customer clicked reviews), we're evaluating P(P|R). The retailer believes review readers are more serious shoppers, suggesting P(P|R) > P(P). Option A correctly states that knowing R occurred makes P more likely exactly when P(P|R) > P(P). Option B confuses the conditioning direction by referencing P(R|P), while C, D, and E make incorrect claims about probability relationships. Understanding that we're conditioning on R to find the probability of P is key to selecting the correct answer.

Question 3

1. It makes $C$ more likely if $P(C\mid D) > P(C)$. (correct answer)
2. It makes $C$ more likely if $P(D\mid C) > P(D)$, regardless of $P(C\mid D)$.
3. It makes no difference because $P(C\mid D)=P(D\mid C)$.
4. It makes no difference because using a coupon cannot affect what is purchased.
5. It makes $C$ less likely if $P(C\mid D) > P(C)$.

Explanation: This question tests understanding of how a digital coupon for store-brand cereal affects purchase probability. When we know D occurred (shopper used the coupon), we're evaluating P(C|D) - the probability of buying store-brand cereal given coupon use. Since the coupon is specifically for this cereal, we expect P(C|D) > P(C). Option A correctly states that knowing D occurred makes C more likely if P(C|D) > P(C). Option B confuses conditioning direction, C incorrectly equates different conditional probabilities, and D makes an illogical claim. The key is recognizing that conditioning restricts our sample space to coupon users.

Question 4

1. It increases the likelihood of $S$ if $P(S\mid E) > P(S)$. (correct answer)
2. It decreases the likelihood of $S$ if $P(S\mid E) > P(S)$.
3. It has no effect because $P(S\mid E)$ always equals $P(S)$.
4. It has no effect because $P(S\mid E)=P(E\mid S)$.
5. It increases the likelihood of $S$ only if $P(E\mid S) < P(E)$.

Explanation: This question tests how choosing a vegetarian entree affects salad selection probability. When we know E occurred (student chose vegetarian entree), we're evaluating P(S|E). The menu layout suggests association between these choices, so P(S|E) > P(S). Option A correctly states that knowing E occurred increases the likelihood of S if P(S|E) > P(S). Option B reverses the effect, C and D make false claims about probability relationships, and E introduces an irrelevant condition. The key is recognizing that physical proximity in the menu layout can create positive association between food choices.

Question 5

1. It increases the likelihood of $N$ exactly when $P(N\mid L) > P(N)$. (correct answer)
2. It increases the likelihood of $N$ exactly when $P(L\mid N) > P(N)$.
3. It has no effect because $P(N\mid L)=P(L\mid N)$.
4. It has no effect because renewals and interlibrary loans are always independent.
5. It decreases the likelihood of $N$ exactly when $P(N\mid L) > P(N)$.

Explanation: This question examines how interlibrary loan usage affects book renewal probability. When we know L occurred (patron used interlibrary loan), we're evaluating P(N|L). The context suggests that patrons who make the effort for interlibrary loans value the books more and renew more often, so P(N|L) > P(N). Option A correctly states that knowing L occurred increases the likelihood of N exactly when P(N|L) > P(N). Option B confuses the conditioning direction, C incorrectly equates different conditional probabilities, and E reverses the effect. Understanding that extra effort correlates with higher renewal rates is key.

Question 6

1. It makes $A$ less likely because $P(A\mid B)=P(A\cap B)$.
2. It does not change the likelihood because $P(A\mid B)=P(A)$.
3. It makes $A$ more likely because $P(A\mid B)>P(A)$. (correct answer)
4. It makes $A$ less likely because $P(B\mid A)>P(B)$.
5. It cannot be determined without computing because $P(A\mid B)=P(B\mid A)$.

Explanation: This AP Statistics question on conditional probability looks at resolution speed (event A) given priority (event B). High-priority tickets resolve faster, so P(A|B) > P(A), increasing likelihood. Conditioning on B limits the sample space to high-priority tickets, handled more efficiently. A distractor is choice A, incorrectly using joint for conditional, but P(A|B) divides by P(B). Mini-lesson: Conditioning can boost P(A) when B correlates with favorable outcomes, as P(A|B) focuses on B's enhanced subset. Here, priority status raises the quick-resolution probability. This applies to efficiency in service metrics.

Question 7

1. It makes $A$ more likely because $P(A\mid B)>P(A)$. (correct answer)
2. It makes $A$ less likely because $P(A\mid B)<P(A)$.
3. It does not change the likelihood because $P(A\mid B)=P(A)$.
4. It makes $A$ less likely because $P(A\mid B)=P(B\mid A)$.
5. It makes $A$ more likely because $P(A\mid B)=P(A\cap B)$.

Explanation: This question in AP Statistics examines conditional probability of transit use (A) during rush hour (B). Higher transit fraction in rush hour means P(A|B) > P(A), increasing likelihood. Conditioning on B restricts to rush-hour trips, where transit is more common. Choice E is a distractor, wrongly setting conditional equal to joint, but normalization by P(B) is key. Mini-lesson: Conditioning elevates P(A) if B selects times with higher A rates, via P(A|B) = P(A ∩ B)/P(B). Knowing it's rush hour thus boosts the transit probability. This reflects temporal patterns in commuting data.

Question 8

1. It makes $A$ less likely because $P(A\mid B)<P(A)$.
2. It makes $A$ more likely because $P(A\mid B)>P(A)$. (correct answer)
3. It does not change the likelihood because $P(A\mid B)=P(A)$.
4. It makes $A$ more likely because $P(A\mid B)=P(A\cap B)$.
5. It cannot be determined without knowing $P(B\mid A)$, since $P(A\mid B)=P(B\mid A)$.

Explanation: In AP Statistics, this question tests conditional probability by comparing defect rates (event A) given the machine source (event B). Machine 2's higher defect rate means P(A|B) > P(A), increasing the defect likelihood when B is known. Conditioning on B limits the sample space to items from Machine 2, elevating the defect proportion. Choice D is a distractor, wrongly using joint probability instead of the conditional formula P(A|B) = P(A ∩ B)/P(B). Mini-lesson: Conditioning updates probabilities by restricting to B's occurrences, which can raise P(A) if B is linked to higher A rates. Thus, knowing the item is from Machine 2 makes defects more probable than the factory average. This illustrates how conditional probability helps in quality control analysis.

Question 9

1. It makes $A$ less likely because $P(A\mid B)=P(B\mid A)$.
2. It makes $A$ more likely because $P(A\mid B)>P(A)$. (correct answer)
3. It does not change the likelihood because $P(A\mid B)=P(A)$.
4. It makes $A$ more likely because $P(A\mid B)=P(A\cap B)$.
5. It makes $A$ less likely because $P(A\mid B)<P(A)$.

Explanation: This question assesses the concept of conditional probability in AP Statistics, specifically how knowing one event affects another's likelihood. Here, event A is drawing a prize ticket, and event B is purchasing a snack bundle, with the problem stating that bundle buyers are more likely to get prizes, implying P(A|B) > P(A). Conditioning on B restricts the sample space to only students who bought snacks, where extra prize tickets increase the chance of A. A common distractor is choice D, which incorrectly equates conditional probability with joint probability, but P(A|B) = P(A ∩ B)/P(B), not P(A ∩ B). In a mini-lesson on conditioning: conditional probability measures how the occurrence of B updates the probability of A by focusing only on B's subset of the sample space. Since P(A|B) > P(A), knowing B makes A more likely, as organizers favored bundle buyers. Thus, the correct answer explains this positive dependence accurately.

Question 10

1. It makes $A$ more likely because $P(A\mid B)=P(B\mid A)$.
2. It makes $A$ less likely because $P(A\mid B)<P(A)$. (correct answer)
3. It does not change the likelihood because $P(A\mid B)=P(A)$.
4. It makes $A$ more likely because $P(A\mid B)>P(A)$.
5. It makes $A$ less likely because $P(A\mid B)=P(A\cap B)$.

Explanation: Addressing conditional probability in AP Statistics, this question involves marble color (A: red) given size (B: large). The lower red proportion among larges means P(A|B) < P(A), reducing likelihood. Conditioning on B confines the sample space to large marbles, with fewer reds relatively. Choice E distracts by mixing conditional with joint, but P(A|B) ≠ P(A ∩ B) generally. Mini-lesson: Conditioning adjusts P(A) downward if B selects a subgroup with lower A prevalence, using P(A|B) = P(A ∩ B)/P(B). Knowing it's large thus decreases the red probability below overall. This demonstrates negative association in probability.

Question 11

1. It becomes $P(A\mid B)$, the admission probability among early-deadline applicants. (correct answer)
2. It becomes $P(A\cap B)$ because knowing $B$ occurred means we count only applicants who are both early and admitted.
3. It becomes $P(B\mid A)$, the probability of applying early among admitted applicants.
4. It stays $P(A)$ because admission decisions are independent of application timing.
5. It becomes $P(A)+P(B)$ because early application and admission are related events.

Explanation: This question examines conditional probability in college admissions. When we know B occurred (applicant applied early), the probability of A (being admitted) becomes P(A|B), which is the admission rate among early-deadline applicants. The problem notes that early applicants differ in preparation and selectivity, suggesting the admission rate for early applicants may differ from the overall rate. Option C reverses the conditioning direction, while option B confuses conditioning with intersection probability. Option D incorrectly assumes independence between application timing and admission decisions, contradicting the problem setup. The conditioning process restricts our sample space to only early applicants and asks what fraction of them are admitted, which may be higher or lower than the overall admission rate depending on the applicant pool characteristics.

Question 12

1. It becomes $P(A\mid B)$, which could be higher or lower than $P(A)$ depending on the association between cash payments and snack purchases. (correct answer)
2. It becomes $P(B\mid A)$, since we are told the customer bought a snack.
3. It stays $P(A)$ because learning $B$ never changes the probability of $A$.
4. It becomes $P(A\cap B)$ because both events are now known to have happened.
5. It becomes $P(A)/P(B)$ because we divide by the probability of the given event.

Explanation: This question tests understanding of conditional probability notation and interpretation. When we learn that event B occurred (customer paid with cash), we update our probability of event A (customer bought a snack) from P(A) to P(A|B), which is the probability of buying a snack given that we know the customer paid with cash. The problem states that cash customers often buy small items while card users buy larger bundles, suggesting an association between payment method and purchase type. Option C incorrectly claims that conditioning never changes probability, which is only true when events are independent. Option D misunderstands conditioning as intersection, while option E incorrectly applies a formula fragment. The key concept is that P(A|B) restricts our sample space to only those customers who paid with cash, then asks what fraction of those bought snacks.

Question 13

1. It becomes $P(A\mid B)$, the missed-payment probability among those with automatic payments enabled. (correct answer)
2. It becomes $P(B\mid A)$, the automatic-payment probability among those who missed a payment.
3. It stays $P(A)$ because enabling automatic payments provides no information.
4. It becomes $P(A\cap B)$ because we restrict to accounts where both are true.
5. It becomes $P(A)/P(B)$, since conditional probability is always a ratio of marginals.

Explanation: This question tests conditional probability in banking. When we know B occurred (automatic payments enabled), the probability of A (missing a payment) becomes P(A|B), representing the missed-payment rate among accounts with automatic payments. The bank expects these events to be associated, likely with automatic payments reducing missed payments. Option B reverses the conditioning, asking for the automatic payment rate among those who missed payments. Option C incorrectly claims automatic payments provide no information, contradicting the problem's statement about association. Option E mischaracterizes conditional probability as always being a ratio of marginal probabilities, missing the intersection term. The key concept is that enabling automatic payments likely reduces the probability of missing a payment, so P(A|B) < P(A).

Question 14

1. It becomes $P(A\mid B)$, which equals $P(A)$ only if $A$ and $B$ are independent. (correct answer)
2. It becomes $P(B\mid A)$ because commuting choice determines distance to the station.
3. It becomes $P(A\cap B)$ because both events are being considered together.
4. It stays $P(A)$ because knowing $B$ occurred cannot change the probability of $A$.
5. It becomes $P(A)/P(B)$ regardless of any relationship between $A$ and $B$.

Explanation: This question tests understanding of conditional probability and independence. When we know B occurred (resident lives within 1 mile of a subway station), the probability of A (commuting by public transit) becomes P(A|B). The problem hints that distance influences commuting choices, suggesting the events are not independent. Importantly, P(A|B) equals P(A) only when A and B are independent events; otherwise, knowing B changes our assessment of A's likelihood. Option B reverses the conditioning direction, while option C confuses conditioning with intersection. Option D incorrectly claims conditioning never changes probability, and option E misapplies a formula. The key insight is that living near a subway station likely increases the probability of using public transit, so P(A|B) ≠ P(A).

Question 15

1. It makes $A$ more likely because $P(A\mid B)=P(A\cap B)$.
2. It makes $A$ less likely because $P(A\mid B)<P(A)$.
3. It does not change the likelihood because $P(A\mid B)=P(A)$. (correct answer)
4. It cannot be determined because independence implies $P(B\mid A)=P(A)$.
5. It makes $A$ more likely because $P(A\mid B)=P(B\mid A)$.

Explanation: In AP Statistics, this question probes conditional probability for quiz scores (A: ≥90%) given studying (B), stated as independent. Independence means P(A|B) = P(A), so no effect on likelihood. Conditioning on B narrows to studiers, but independence keeps the score probability the same. Choice D distracts by misstating independence as P(B|A) = P(A), which is incorrect. Mini-lesson: For independent events, conditioning yields P(A|B) = P(A), as there's no informational update from B. Thus, studying doesn't change the high-score chance in this scenario. This underscores how independence simplifies probability calculations.

Question 16

1. It makes $A$ more likely because $P(A\mid B)>P(A)$.
2. It makes $A$ less likely because $P(A\mid B)<P(A)$.
3. It does not change the likelihood because $P(A\mid B)=P(A)$. (correct answer)
4. It cannot be determined without computing because $P(A\mid B)=P(B\mid A)$.
5. It makes $A$ more likely because $P(A\mid B)=P(A\cap B)$.

Explanation: This AP Statistics question tests conditional probability with a card draw, where A is heart and B is face card. In a standard deck, P(A|B) = 3/12 = 1/4, matching P(A) = 13/52 = 1/4, so no change. Conditioning on B reduces the sample space to 12 face cards, with hearts proportionally the same. Choice E distracts by suggesting uncertainty without P(B|A), but independence means P(A|B) = P(A) directly. Mini-lesson: When events are independent, conditioning doesn't alter probabilities, as P(A|B) = P(A ∩ B)/P(B) = P(A)P(B)/P(B) = P(A). Here, suit and face status are independent, leaving the heart probability unchanged. This exemplifies independence in uniform random selections.

Question 17

1. It makes $C$ more likely if $P(C\mid R) < P(C)$.
2. It makes $C$ less likely if $P(C\mid R) < P(C)$. (correct answer)
3. It makes no difference because $P(C\mid R)=P(R\mid C)$.
4. It makes no difference because knowing $R$ occurred cannot change probabilities.
5. It makes $C$ less likely only if $P(R\mid C) > P(R)$.

Explanation: This problem involves how remote work affects attendance at in-person coffee chats. When we know R occurred (employee worked remotely), we evaluate P(C|R). Since coffee chats are in-person only, remote workers cannot attend, making P(C|R) < P(C). Option B correctly identifies that knowing R occurred makes C less likely if P(C|R) < P(C). Option A reverses the logic, C incorrectly equates different conditional probabilities, and D makes a false claim. The physical constraint that remote workers cannot attend in-person events creates the negative association between R and C.

Question 18

1. It makes $I$ more likely if $P(I\mid V) < P(I)$.
2. It makes $I$ less likely if $P(I\mid V) < P(I)$. (correct answer)
3. It makes no difference because $P(I\mid V)=P(V\mid I)$.
4. It makes no difference because vaccination and influenza are always independent in probability models.
5. It makes $I$ less likely only if $P(V\mid I) > P(V)$.

Explanation: This problem examines how flu vaccination affects the probability of testing positive for influenza. When we know V occurred (patient was vaccinated), we evaluate P(I|V). Since vaccination reduces influenza risk, we expect P(I|V) < P(I). Option B correctly identifies that knowing V occurred makes I less likely if P(I|V) < P(I). Option A reverses the logic, C incorrectly equates different conditional probabilities, and D makes a false claim about independence. The key insight is that effective vaccination creates negative association between V and I, reducing conditional probability.

Question 19

1. It makes $T$ less likely because $P(T\mid R)$ must be smaller than $P(T)$.
2. It makes $T$ more likely if $P(T\mid R) > P(T)$. (correct answer)
3. It does not change the likelihood because $P(T\mid R)=P(R\mid T)$.
4. It does not change the likelihood because $P(T\mid R)=P(T)$ for any events.
5. It makes $T$ more likely only when $P(R\mid T) > P(R)$.

Explanation: This question tests understanding of conditional probability and how additional information affects likelihood. When we know event R occurred (student attended Resume workshop), we restrict our sample space to only those students who attended R. The problem states that many Resume attendees also choose Time-Management because sessions are back-to-back, suggesting P(T|R) > P(T). Option B correctly identifies that knowing R occurred makes T more likely when the conditional probability P(T|R) exceeds the marginal probability P(T). Options C and D incorrectly claim conditional probabilities must equal other values, while A incorrectly states conditioning always decreases probability.

Question 20

1. It makes $L$ more likely if $P(L\mid A) < P(L)$.
2. It makes $L$ less likely if $P(L\mid A) < P(L)$. (correct answer)
3. It makes no difference because $P(L\mid A)=P(A\mid L)$.
4. It makes no difference because conditional probability never differs from marginal probability.
5. It makes $L$ less likely only if $P(A\mid L) > P(A)$.

Explanation: This problem examines how checking real-time transit updates affects lateness probability. When we know A occurred (rider checked updates), we evaluate P(L|A). The context suggests checking updates helps avoid delays, so P(L|A) < P(L). Option B correctly identifies that knowing A occurred makes L less likely if P(L|A) < P(L). Option A reverses the logic, C incorrectly equates different conditional probabilities, and D makes a false general claim. The insight is that when P(L|A) < P(L), the conditional probability is lower than the marginal, making lateness less likely given update checking.