AP Statistics Quiz: Independent Events And Unions Of Events

What this quiz covers

This quiz focuses on Independent Events And Unions Of Events, 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. Events A and B are mutually exclusive, so $P(A\cup B)=P(A)+P(B)$.
2. Events A and B are independent because $P(A\cap B)$ is 10%.
3. Events A and B are not independent because $P(A\cap B)$ is not equal to $P(A)P(B)$. (correct answer)
4. Events A and B are independent because "prefers X" and "prefers Y" are different options.
5. Events A and B are independent because $P(A\cup B)=P(A)+P(B)+P(A\cap B)$.

Explanation: This question tests independence with overlapping preferences. Given P(A) = 0.48, P(B) = 0.42, and P(A∩B) = 0.10, we check if P(A∩B) = P(A)·P(B). We calculate P(A)·P(B) = 0.48 × 0.42 = 0.2016. Since 0.10 ≠ 0.2016, events A and B are not independent. The lower intersection probability (0.10 < 0.2016) indicates negative association - people tend to prefer one brand or the other, not both. This shows that events representing choices or preferences often exhibit dependence, as people's preferences tend to be exclusive rather than independent.

Question 2

1. Events A and B are not independent because the same bag is used twice.
2. Events A and B are independent because replacing the first chip keeps the second-draw probabilities the same. (correct answer)
3. Events A and B are mutually exclusive because only one chip can be white.
4. Events A and B are not independent because A and B can both occur.
5. Events A and B are independent only if the first chip is black.

Explanation: This question contrasts sampling with and without replacement. When drawing with replacement, each draw is from the same distribution: P(white) = 5/10 = 1/2 for both draws. Events A and B are independent because P(B|A) = P(B|A') = P(B) = 1/2 - the first draw's outcome doesn't affect the second draw's probabilities since the chip is replaced. We can verify: P(A∩B) = P(A)·P(B) = (1/2)·(1/2) = 1/4. The replacement ensures that each draw is independent, maintaining the same probability distribution throughout the experiment.

Question 3

1. Events A and B are independent because there are wrapped candies in both categories.
2. Events A and B are not independent because $P(A\cap B)\ne P(A)P(B)$. (correct answer)
3. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
4. Events A and B are independent because $P(A\cap B)=P(A)+P(B)$.
5. Events A and B are not independent because chocolate candies and wrapped candies cannot overlap.

Explanation: This problem involves calculating independence with overlapping categories. From the given information: P(A) = 15/25, P(B) = 8/25, and P(A∩B) = 3/25 (wrapped chocolate candies). To check independence: P(A)·P(B) = (15/25)·(8/25) = 120/625 = 24/125, while P(A∩B) = 3/25 = 15/125. Since 24/125 ≠ 15/125, events A and B are not independent. Choice E incorrectly claims chocolate and wrapped candies cannot overlap, but the problem explicitly states 3 chocolate candies are wrapped. The calculation shows these events are dependent—knowing a candy is chocolate changes the probability it's wrapped.

Question 4

1. Events A and B are independent because a ticket cannot be both blue and a prize ticket.
2. Events A and B are independent because $P(A\cap B)=P(A)P(B)$.
3. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
4. Events A and B are not independent because $P(A\cap B)\ne P(A)P(B)$. (correct answer)
5. Events A and B are not independent because Events A and B overlap.

Explanation: This question tests understanding of independent events versus mutually exclusive events. To check if events A and B are independent, we need to verify if P(A∩B) = P(A)·P(B). From the given information: P(A) = 18/40, P(B) = 14/40, and P(A∩B) = 6/40 (blue prize tickets). Calculating: P(A)·P(B) = (18/40)·(14/40) = 252/1600 = 63/400, while P(A∩B) = 6/40 = 60/400. Since 63/400 ≠ 60/400, the events are not independent. Choice A incorrectly confuses independence with mutual exclusivity—the fact that some tickets are both blue AND prize tickets shows they're not mutually exclusive.

Question 5

1. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
2. Events A and B are independent because $P(A\cap B)=P(A)P(B)$.
3. Events A and B are not independent because the events overlap.
4. Events A and B are not independent because $P(A\cap B)\ne P(A)P(B)$. (correct answer)
5. Events A and B are independent because playing a sport implies having a job.

Explanation: This problem tests independence with survey data. Given: P(A) = 90/200, P(B) = 120/200, and P(A∩B) = 50/200. For independence, we need P(A∩B) = P(A)·P(B). Calculating: P(A)·P(B) = (90/200)·(120/200) = 10800/40000 = 54/200. However, P(A∩B) = 50/200. Since 54/200 ≠ 50/200, events A and B are not independent. Choice C incorrectly states that overlapping events cannot be independent—events can overlap and still be independent if they satisfy the multiplication rule. The key is checking the probability relationship, not just whether events can occur together.

Question 6

1. Events A and B are independent because $P(A\cap B)$ is less than both $P(A)$ and $P(B)$.
2. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
3. Events A and B are not independent because $P(A\cap B)$ is not equal to $P(A)P(B)$. (correct answer)
4. Events A and B are mutually exclusive because some students do both.
5. Events A and B are independent because band and sports are different activities.

Explanation: This question tests the independence formula P(A∩B) = P(A)·P(B) with given probabilities. We have P(A) = 0.40, P(B) = 0.30, and P(A∩B) = 0.15. For independence, we need P(A∩B) = P(A)·P(B) = 0.40 × 0.30 = 0.12. Since 0.15 ≠ 0.12, events A and B are not independent. The fact that P(A∩B) > P(A)·P(B) indicates positive association - students who play sports are more likely than average to be in band. This demonstrates that independence requires the exact equality P(A∩B) = P(A)·P(B), not just that events can occur together.

Question 7

1. Events A and B are independent because 12 employees are in both categories.
2. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
3. Events A and B are not independent because $P(A\cap B)\ne P(A)P(B)$. (correct answer)
4. Events A and B are independent because remote work and sales are unrelated job features.
5. Events A and B are not independent because Events A and B are mutually exclusive.

Explanation: This question involves checking independence using given frequencies. From the data: P(A) = 30/80, P(B) = 28/80, and P(A∩B) = 12/80. To verify independence, we check if P(A∩B) = P(A)·P(B). Calculating: P(A)·P(B) = (30/80)·(28/80) = 840/6400 = 21/160, while P(A∩B) = 12/80 = 24/160. Since 21/160 ≠ 24/160, the events are not independent. The fact that 12 employees are in both categories (choice A) doesn't determine independence—we must check if the probability relationship holds. Events that overlap can still be independent if they satisfy the multiplication rule.

Question 8

1. Events A and B are independent because there are wrapped candies in both categories.
2. Events A and B are not independent because $P(A\cap B)\ne P(A)P(B)$. (correct answer)
3. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
4. Events A and B are independent because $P(A\cap B)=P(A)+P(B)$.
5. Events A and B are not independent because chocolate candies and wrapped candies cannot overlap.

Explanation: This problem involves calculating independence with overlapping categories. From the given information: P(A) = 15/25, P(B) = 8/25, and P(A∩B) = 3/25 (wrapped chocolate candies). To check independence: P(A)·P(B) = (15/25)·(8/25) = 120/625 = 24/125, while P(A∩B) = 3/25 = 15/125. Since 24/125 ≠ 15/125, events A and B are not independent. Choice E incorrectly claims chocolate and wrapped candies cannot overlap, but the problem explicitly states 3 chocolate candies are wrapped. The calculation shows these events are dependent—knowing a candy is chocolate changes the probability it's wrapped.

Question 9

1. Events A and B are independent because suit and rank are unrelated categories.
2. Events A and B are mutually exclusive because a heart cannot be a face card.
3. Events A and B are independent because $P(A\cap B)=P(A)P(B)$. (correct answer)
4. Events A and B are not independent because $P(A\cup B)=P(A)+P(B)+P(A\cap B)$.
5. Events A and B are not independent because $P(A\cap B)=P(A)+P(B)$.

Explanation: This question tests identifying independent events in a card draw by verifying P(A ∩ B) = P(A)P(B). Event A (heart) has P(A) = 13/52 = 0.25, Event B (face card) has P(B) = 12/52 ≈ 0.231, and P(A ∩ B) = 3/52 ≈ 0.0577, equaling 0.25 * 0.231 ≈ 0.0577, confirming independence. The relationship reflects that suit and rank are unrelated in a standard deck. Distractor choice B confuses independence with exclusivity, as hearts can be face cards. Choice D uses the union formula incorrectly as a test for independence. Mini-lesson: Independence is confirmed when the intersection probability is the product of individuals, unlike mutual exclusivity where the intersection is zero; these are distinct, as independent events can overlap, while exclusive ones cannot.

Question 10

1. Events A and B are independent because $P(A\cap B)=P(A)P(B)$. (correct answer)
2. Events A and B are not independent because they overlap.
3. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
4. Events A and B are not independent because $P(A\cap B)=0$.
5. Events A and B are independent because "even" and "greater than 4" describe different properties.

Explanation: This problem tests calculating independence for events with a fair die. Event A (even) includes {2, 4, 6}, so P(A) = 3/6 = 1/2. Event B (greater than 4) includes {5, 6}, so P(B) = 2/6 = 1/3. The intersection A∩B = {6}, so P(A∩B) = 1/6. Checking independence: P(A)·P(B) = (1/2)·(1/3) = 1/6 = P(A∩B). Since the equation holds, events A and B are independent. Choice C incorrectly applies the addition rule for mutually exclusive events, but these events overlap at outcome 6. Independence means knowing one event occurred doesn't change the probability of the other, which is true here.

Question 11

1. Events A and B are independent because prime and even describe different properties.
2. Events A and B are mutually exclusive, so $P(A\cup B)=P(A)+P(B)$.
3. Events A and B are not independent because knowing a number is even changes the chance it is prime. (correct answer)
4. Events A and B are independent because $P(A\cap B)=P(A)+P(B)$.
5. Events A and B are independent because the selection is random.

Explanation: This question examines independence between number properties. From {1,2,3,4,5,6}, the primes are {2,3,5} and the evens are {2,4,6}. Thus P(A) = 3/6 = 1/2, P(B) = 3/6 = 1/2, and P(A∩B) = 1/6 (only 2 is both prime and even). For independence, we need P(A∩B) = P(A)·P(B) = (1/2)·(1/2) = 1/4. Since 1/6 ≠ 1/4, events A and B are not independent. Intuitively, knowing a number is even reduces the chance it's prime (only 2 among evens vs. half overall), confirming dependence between these mathematical properties.

Question 12

1. Events A and B are independent because $P(A\cap B)=P(A)P(B)$ for this die roll. (correct answer)
2. Events A and B are independent because they are not mutually exclusive.
3. Events A and B are not independent because A implies B.
4. Events A and B are mutually exclusive, so they are independent.
5. Events A and B are not independent because $P(A\cup B)=P(A)P(B)$.

Explanation: This question involves checking independence for specific die roll events. Event A (greater than 4) = {5,6} and Event B (even) = {2,4,6}, so P(A) = 2/6 = 1/3, P(B) = 3/6 = 1/2, and P(A∩B) = 1/6 (only 6 satisfies both). For independence, we need P(A∩B) = P(A)·P(B) = (1/3)·(1/2) = 1/6. Since this equality holds, events A and B are independent. This is a special case where specific numeric properties happen to be independent - knowing the roll is even doesn't change the probability it's greater than 4 (still 1/3 among evens: only 6 out of {2,4,6}).

Question 13

1. Events A and B are independent because both events involve even numbers.
2. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
3. Events A and B are not independent because $P(B\mid A)\ne P(B)$. (correct answer)
4. Events A and B are independent because $P(A\cap B)=0$.
5. Events A and B are not independent because they are mutually exclusive.

Explanation: This question examines independence when one event is a subset of another. Event A (multiples of 2) = {2, 4, 6, 8}, so P(A) = 4/8 = 1/2. Event B (multiples of 4) = {4, 8}, so P(B) = 2/8 = 1/4. Since every multiple of 4 is also a multiple of 2, B⊆A, meaning A∩B = B = {4, 8}, so P(A∩B) = 2/8 = 1/4. The conditional probability P(B|A) = P(A∩B)/P(A) = (1/4)/(1/2) = 1/2, while P(B) = 1/4. Since P(B|A) ≠ P(B), the events are not independent. When one event is a subset of another, they cannot be independent unless one has probability 0 or 1.

Question 14

1. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
2. Events A and B are independent because $P(A\cap B)=P(A)P(B)$.
3. Events A and B are not independent because the events overlap.
4. Events A and B are not independent because $P(A\cap B)\ne P(A)P(B)$. (correct answer)
5. Events A and B are independent because playing a sport implies having a job.

Explanation: This problem tests independence with survey data. Given: P(A) = 90/200, P(B) = 120/200, and P(A∩B) = 50/200. For independence, we need P(A∩B) = P(A)·P(B). Calculating: P(A)·P(B) = (90/200)·(120/200) = 10800/40000 = 54/200. However, P(A∩B) = 50/200. Since 54/200 ≠ 50/200, events A and B are not independent. Choice C incorrectly states that overlapping events cannot be independent—events can overlap and still be independent if they satisfy the multiplication rule. The key is checking the probability relationship, not just whether events can occur together.

Question 15

1. The counselor is correct only if Events $A$ and $B$ are independent.
2. The counselor is correct only if Events $A$ and $B$ are disjoint (mutually exclusive). (correct answer)
3. The counselor is correct because "$A$ or $B$" means "$A$ and $B$."
4. The counselor is correct because $P(A\cup B)=P(A)P(B)$ for any events.
5. The counselor is correct only if $P(A\cap B)=P(A)+P(B)$.

Explanation: This question tests understanding of when the addition rule P(A∪B) = P(A) + P(B) applies. This formula is only valid when events A and B are disjoint (mutually exclusive), meaning they cannot occur together. For general events, we need P(A∪B) = P(A) + P(B) - P(A∩B). Choice B correctly identifies that the counselor's statement is only true if the events are disjoint. Since an AP course could meet in the morning, the events can overlap, so they may not be disjoint. Choice A incorrectly suggests independence is the requirement. Choice C confuses union with intersection. The distinction between disjoint events and independent events is crucial: disjoint events satisfy the simple addition rule, while independent events satisfy the multiplication rule.

Question 16

1. Events $A$ and $B$ are independent because $P(B\mid A)=P(B)$. (correct answer)
2. Events $A$ and $B$ must be disjoint because $P(B\mid A)=P(B)$.
3. Events $A$ and $B$ are independent only if $P(A\cup B)=P(A)P(B)$.
4. Events $A$ and $B$ are independent only if $P(A\cap B)=P(A)+P(B)$.
5. Events $A$ and $B$ are not independent because $A$ happens before $B$.

Explanation: This question tests understanding of conditional probability and independence. The given condition P(B|A) = P(B) is exactly the definition of independence - it means that knowing a visitor is on mobile doesn't change the probability they make a purchase. This is equivalent to saying P(A∩B) = P(A)P(B), which is the multiplication rule for independent events. Choice A correctly identifies this relationship. Choices C and D give incorrect formulas for independence. Choice B incorrectly suggests the events must be disjoint, but independent events can (and often do) occur together. Choice E incorrectly assumes temporal order affects independence.

Question 17

1. Events $A$ and $B$ are independent because a cup cannot be both berry and low sugar.
2. Events $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)P(B)$. (correct answer)
3. Events $A$ and $B$ are independent if and only if $P(A\cup B)=P(A)+P(B)$.
4. Events $A$ and $B$ are independent because some berry cups are low sugar.
5. Events $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)+P(B)$.

Explanation: This question tests understanding of the definition of independent events. The correct answer is B because independence is defined by the multiplication rule: P(A∩B) = P(A)P(B). The problem states that some berry cups are low sugar, meaning events A and B can occur together, so they are not mutually exclusive. Choice A incorrectly confuses mutual exclusivity with independence - events that cannot occur together are mutually exclusive, not independent. Choice C gives the formula for the union of mutually exclusive events, not independent events. Independence means that knowing one event occurred doesn't change the probability of the other event, which is captured by the multiplication rule.

Question 18

1. Events A and B are independent because a ticket cannot be both blue and a prize ticket.
2. Events A and B are independent because $P(A\cap B)=P(A)P(B)$.
3. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
4. Events A and B are not independent because $P(A\cap B)\ne P(A)P(B)$. (correct answer)
5. Events A and B are not independent because Events A and B overlap.

Explanation: This question tests understanding of independent events versus mutually exclusive events. To check if events A and B are independent, we need to verify if P(A∩B) = P(A)·P(B). From the given information: P(A) = 18/40, P(B) = 14/40, and P(A∩B) = 6/40 (blue prize tickets). Calculating: P(A)·P(B) = (18/40)·(14/40) = 252/1600 = 63/400, while P(A∩B) = 6/40 = 60/400. Since 63/400 ≠ 60/400, the events are not independent. Choice A incorrectly confuses independence with mutual exclusivity—the fact that some tickets are both blue AND prize tickets shows they're not mutually exclusive.

Question 19

1. Events A and B are independent because both events involve even numbers.
2. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
3. Events A and B are not independent because $P(B\mid A)\ne P(B)$. (correct answer)
4. Events A and B are independent because $P(A\cap B)=0$.
5. Events A and B are not independent because they are mutually exclusive.

Explanation: This question examines independence when one event is a subset of another. Event A (multiples of 2) = {2, 4, 6, 8}, so P(A) = 4/8 = 1/2. Event B (multiples of 4) = {4, 8}, so P(B) = 2/8 = 1/4. Since every multiple of 4 is also a multiple of 2, B⊆A, meaning A∩B = B = {4, 8}, so P(A∩B) = 2/8 = 1/4. The conditional probability P(B|A) = P(A∩B)/P(A) = (1/4)/(1/2) = 1/2, while P(B) = 1/4. Since P(B|A) ≠ P(B), the events are not independent. When one event is a subset of another, they cannot be independent unless one has probability 0 or 1.

Question 20

1. Events A and B are independent because $P(A\cap B)=P(A)P(B)$. (correct answer)
2. Events A and B are not independent because they overlap.
3. Events A and B are independent because $P(A\cup B)=P(A)+P(B)$.
4. Events A and B are not independent because $P(A\cap B)=0$.
5. Events A and B are independent because "even" and "greater than 4" describe different properties.

Explanation: This problem tests calculating independence for events with a fair die. Event A (even) includes {2, 4, 6}, so P(A) = 3/6 = 1/2. Event B (greater than 4) includes {5, 6}, so P(B) = 2/6 = 1/3. The intersection A∩B = {6}, so P(A∩B) = 1/6. Checking independence: P(A)·P(B) = (1/2)·(1/3) = 1/6 = P(A∩B). Since the equation holds, events A and B are independent. Choice C incorrectly applies the addition rule for mutually exclusive events, but these events overlap at outcome 6. Independence means knowing one event occurred doesn't change the probability of the other, which is true here.