This question tests conditional probability using the definition P(A|B) = P(A and B)/P(B). The sample space is the 6 equally likely outcomes from a die roll: 1 through 6. Event B (greater than 4) has outcomes 5 and 6, so P(B) = 2/6 = 1/3; A and B (even and >4) is just 6, so P(A and B) = 1/6, yielding P(A|B) = (1/6)/(1/3) = 1/2. Counting within B, 1 of 2 outcomes is even. This is justified as conditional probability restricts the space to B. A tempting incorrect option is 1/3, perhaps from P(A) overall instead of conditional. This fails because it ignores the conditioning on B, which changes the relevant sample space.

This question tests the probability of a union of two events when rolling a die. The sample space consists of the six equally likely outcomes: {1, 2, 3, 4, 5, 6}. Event A (multiples of 3) = {3, 6} and event B (even numbers) = {2, 4, 6}. To find P(A ∪ B), we need the outcomes in either A or B: {2, 3, 4, 6}, which contains 4 elements. Therefore, P(A ∪ B) = 4/6 = 2/3. A common mistake would be to add P(A) + P(B) = 2/6 + 3/6 = 5/6, which double-counts the outcome 6 that appears in both events.

This question tests probability in sampling with replacement, where draws are independent. The sample space is all possible outcomes of two draws from 8 balls (3 red, 5 yellow), with 88 = 64 equally likely results. The probability of red on first draw is 3/8, and again on second is 3/8, so (3/8)(3/8) = 9/64. This matches the proportion of red-red outcomes. The result is justified because replacement ensures independence, allowing multiplication of probabilities. A tempting incorrect option is 3/28, perhaps from using without replacement formula incorrectly. This fails as it assumes dependence, which does not apply here with replacement.

This question tests probability with independent events (with replacement). The sample space contains 12 balls: 5 red and 7 blue. Since we replace the ball after each draw, the probability remains constant: P(red) = 5/12 for each draw. The probability of drawing two red balls is (5/12) × (5/12) = 25/144. A common mistake would be to treat this as without replacement, calculating (5/12) × (4/11) = 20/132 = 5/33, which incorrectly assumes the first ball isn't replaced.

This question tests the complement rule in probability. The sample space consists of 2³ = 8 equally likely sequences: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. To find P(at least one head), it's easier to use the complement: P(at least one H) = 1 - P(no heads) = 1 - P(TTT). Since P(TTT) = 1/8, we get P(at least one H) = 1 - 1/8 = 7/8. A common error would be to think P(at least one H) = 3 × (1/2) = 3/2, which incorrectly adds probabilities and exceeds 1.

This question tests probability with dependent events (without replacement). The sample space initially contains 10 balls: 4 black and 6 white. For the first draw, P(black) = 4/10. After removing one black ball, 9 balls remain with 3 black, so P(second black | first black) = 3/9. The probability of both events is (4/10) × (3/9) = 12/90 = 2/15. A tempting error would be to use (4/10) × (4/10) = 16/100, which incorrectly assumes replacement and independence between draws.

This question tests conditional probability in sequential selection without replacement. The sample space initially has 10 employees: 4 managers and 6 non-managers. Given that the first selected is a manager, we now have 9 employees left: 3 managers and 6 non-managers. The probability that the second is a manager, given the first is a manager, is 3/9 = 1/3. A common mistake would be to use the original probability 4/10 = 2/5, which fails to account for the changed composition after the first selection.

This question tests probability of complementary events in categorized sampling. The sample space is all combinations of 2 chips from 10 (4A, 3B, 3C), totaling C(10,2) = 45 equally likely pairs. Same labels: C(4,2) + C(3,2) + C(3,2) = 6 + 3 + 3 = 12, so different is 45 - 12 = 33, probability 33/45 = 11/15. This is justified as the complement simplifies counting diverse pairs. A tempting incorrect option is 4/9, perhaps from averaging probabilities incorrectly. This fails because it might assume equal group sizes, ignoring the actual distributions.

This question tests basic probability in a binary outcome scenario. The sample space consists of all items produced, where each is either acceptable or defective, with defects occurring at 8%. The probability of selecting a defective item is 0.08, so the probability of acceptable is 1 - 0.08 = 0.92. This complementary approach is efficient here. The result is justified as the events are exhaustive and mutually exclusive, based on historical data assuming stability. A tempting incorrect option is 0.08, confusing acceptable with defective. This fails because it directly misreads the question, emphasizing the need to distinguish between the event and its complement.

This question tests probability of joint events from independent experiments. The sample space consists of 12 equally likely pairs: (H or T) with (1 through 6). Favorable outcomes are tails with 5 or 6, giving 2 outcomes. Thus, the probability is 2/12 = 1/6. This is justified as the coin and die are independent, so P(tails and >4) = P(tails) * P(>4) = (1/2)*(2/6) = 1/6. A tempting incorrect option is 1/12, perhaps from miscounting only one die outcome >4. This fails because >4 includes both 5 and 6, underlining the need to accurately define the event.