This problem asks for the probability of getting exactly two heads in three coin flips. The possible ways to get exactly two heads are: HHT, HTH, and THH. Each specific sequence has probability (1/2)³ = 1/8, and there are 3 such sequences. Therefore, P(exactly 2 heads) = 3 × (1/8) = 3/8. This is a binomial probability problem with n=3, k=2, and p=1/2. The key is counting all possible arrangements of 2 heads and 1 tail.

This problem asks for the probability of landing on a multiple of 3 OR a number greater than 6 on a spinner numbered 1-8. Multiples of 3 are {3, 6} and numbers greater than 6 are {7, 8}, with no overlap between these sets. Therefore, P(multiple of 3 or >6) = P(multiple of 3) + P(>6) = 2/8 + 2/8 = 4/8 = 1/2. Since there's no overlap, we don't need to subtract anything. The key is carefully identifying all favorable outcomes: 3, 6, 7, and 8.

This problem involves drawing tiles with replacement, asking for the probability of getting exactly one white tile in two draws from a bag with 8 black and 2 white tiles. There are two ways this can happen: white then black, or black then white. P(white then black) = (2/10) × (8/10) = 16/100, and P(black then white) = (8/10) × (2/10) = 16/100. The total probability is 16/100 + 16/100 = 32/100 = 8/25. The key insight is that with replacement, each draw has the same probabilities, and we must consider both orderings.

This problem asks for the probability of rolling an even number OR a number greater than 4 on a standard six-sided die. The even numbers are {2, 4, 6} and numbers greater than 4 are {5, 6}, with 6 appearing in both sets. Using P(A or B) = P(A) + P(B) - P(A and B), we get P(even) = 3/6, P(>4) = 2/6, and P(even and >4) = 1/6 (only the 6). Therefore, P(even or >4) = 3/6 + 2/6 - 1/6 = 4/6 = 2/3. The key is recognizing that 6 satisfies both conditions and must not be double-counted.

This problem asks for the probability of drawing two blue marbles without replacement from a jar containing 6 red, 5 blue, and 4 green marbles (15 total). For the first draw, P(blue) = 5/15 = 1/3. After removing one blue marble, there are 4 blue marbles left out of 14 total marbles, so P(second blue | first blue) = 4/14 = 2/7. The probability of both events is (5/15) × (4/14) = (1/3) × (2/7) = 2/21. A common mistake is treating this as replacement, which would give (5/15)² = 25/225. When drawing without replacement, always adjust both the numerator and denominator for the second draw.

This problem asks for the probability of answering all 4 multiple-choice questions incorrectly when guessing randomly. Each question has 5 choices with 1 correct answer, so P(incorrect) = 4/5 for each question. Since the questions are independent, P(all 4 incorrect) = (4/5)⁴ = 256/625. This matches answer C. A common error would be calculating (1/5)⁴ = 1/625, which is the probability of getting all 4 correct. When dealing with independent events, multiply the individual probabilities.

This problem asks for the probability of drawing two balls of different colors without replacement from 4 red and 6 yellow balls. There are two ways: red then yellow, or yellow then red. P(red then yellow) = (4/10) × (6/9) = 24/90, and P(yellow then red) = (6/10) × (4/9) = 24/90. Total probability = 24/90 + 24/90 = 48/90 = 8/15. The key insight is considering both orderings and adjusting the denominator for the second draw without replacement.

This question involves drawing two candies without replacement, asking for the probability that both are cherry. The jar has 5 lemon + 7 cherry = 12 total candies. For the first draw, P(cherry) = 7/12. After removing one cherry candy, there are 6 cherry candies left out of 11 total, so P(second cherry|first cherry) = 6/11. Therefore, P(both cherry) = 7/12 × 6/11. The key distinction is "without replacement," which means the second probability changes based on the first outcome. A common error is using 7/12 × 7/12, which would only be correct for drawing with replacement.

This question asks for the probability of drawing a card that is both a heart AND a face card from a standard 52-card deck. A standard deck has 13 hearts and 12 face cards total (3 per suit: J, Q, K), but we need cards that satisfy both conditions. The heart suit contains exactly 3 face cards: jack of hearts, queen of hearts, and king of hearts. Therefore, P(heart AND face card) = 3/52. A common error is multiplying P(heart) × P(face card) = 13/52 × 12/52, which assumes independence and gives the wrong answer. For AND probabilities, count only the outcomes that satisfy both conditions simultaneously.

This question asks for the probability of NOT drawing a blue marble from a bag. The bag contains 6 red + 4 blue + 5 green = 15 total marbles, and we want any color except blue. The favorable outcomes are red or green marbles: 6 + 5 = 11 marbles. Therefore, P(not blue) = 11/15. A common mistake is calculating P(blue) = 4/15 but forgetting to subtract from 1, or miscounting the total marbles. When dealing with "not" probabilities, you can either count favorable outcomes directly or use the complement rule: P(not blue) = 1 - P(blue) = 1 - 4/15 = 11/15.