This problem involves selecting 2 students without replacement from 30 total, where 18 submitted homework on time. For the first selection, P(on time) = 18/30 = 3/5, and for the second selection (after removing one on-time student), P(on time) = 17/29. The probability both are on time is (18/30) × (17/29) = 306/870 = 51/145. The key is recognizing that without replacement changes both the numerator and denominator for the second draw. A common error is using 18/30 for both draws, which would apply only with replacement.

This question asks for the probability of rolling an even number OR a number greater than 4 on a fair 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(both) = 1/6 (only 6). Therefore, P(even or >4) = 3/6 + 2/6 - 1/6 = 4/6 = 2/3. A common mistake is double-counting the 6 by not subtracting the overlap. When using "or" in probability, always check for overlapping outcomes.

This question asks for the probability of NOT landing on a multiple of 3 when spinning a spinner with 8 equal sections (1-8). First, identify the multiples of 3 in this range: {3, 6}, which gives 2 favorable outcomes for multiples of 3. The probability of landing on a multiple of 3 is 2/8 = 1/4, so the probability of NOT landing on a multiple of 3 is 1 - 1/4 = 3/4. A common error is miscounting the multiples of 3 or forgetting to subtract from 1. When finding "not" probabilities, using the complement rule P(not A) = 1 - P(A) is often easier than counting directly.

This problem asks for the probability of drawing two beads of different colors from a jar with 5 black and 7 white beads (12 total). There are two ways this can happen: black then white, or white then black. P(black then white) = (5/12) × (7/11) = 35/132, and P(white then black) = (7/12) × (5/11) = 35/132. The total probability is 35/132 + 35/132 = 70/132 = 35/66. A common mistake is calculating only one ordering instead of both. When finding probabilities of "different" outcomes, consider all possible orderings.

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 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(\text{exactly 2 heads}) = 3 \times \frac{1}{8} = \frac{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 answering all 4 multiple-choice questions incorrectly when guessing randomly. Each question has 5 choices with 1 correct answer, so $P(\text{incorrect}) = \frac{4}{5}$ for each question. Since the questions are independent, $P(\text{all 4 incorrect}) = \left(\frac{4}{5}\right)^4 = \frac{256}{625}$. This matches answer C. A common error would be calculating $\left(\frac{1}{5}\right)^4 = \frac{1}{625}$, 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 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.