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.

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.

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.

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.

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.

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.

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.

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.

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.

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.