This question asks for the probability that at least one of two socks is black when drawing without replacement from 4 black and 6 white socks (10 total). It's easier to calculate the complement: P(both white) = (6/10) × (5/9) = 30/90 = 1/3. Therefore, P(at least one black) = 1 - 1/3 = 2/3. A common error is trying to directly calculate P(one black) + P(both black) without accounting for the different arrangements, when using the complement is much simpler.

This question asks for the probability of two independent events both occurring: rolling an even number AND getting heads. A standard die has three even numbers (2, 4, 6) out of six possible outcomes, so P(even) = 3/6 = 1/2. A fair coin has P(heads) = 1/2. Since these events are independent, we multiply: P(even AND heads) = (1/2) × (1/2) = 1/4. A common mistake is adding the probabilities instead of multiplying them for AND events.

This question asks for the probability that both marbles drawn are blue when drawing without replacement from a jar with 5 red, 3 blue, and 2 green marbles (10 total). For the first draw, the probability of getting blue is 3/10. After removing one blue marble, there are 2 blue marbles left out of 9 total marbles, so the probability of getting blue on the second draw is 2/9. The probability of both events occurring is (3/10) × (2/9) = 6/90 = 1/15. A common error would be to use 3/10 for both draws, forgetting that without replacement changes the second probability.

This question asks for P(multiple of 3 OR prime) from tickets numbered 1-12. Multiples of 3 are: 3, 6, 9, 12 (4 numbers). Prime numbers from 1-12 are: 2, 3, 5, 7, 11 (5 numbers). Number 3 appears in both sets, so using inclusion-exclusion: favorable outcomes = 4 + 5 - 1 = 8. Therefore, P(multiple of 3 OR prime) = 8/12 = 2/3. Be careful not to double-count numbers that satisfy both conditions when dealing with OR probabilities.

The question is asking for the probability that the student gets at least one question correct by guessing. The relevant outcomes are the possible guess results, but since independent, easier to compute complement. The probability of getting a question wrong is 3/4, so P(all wrong) = $(3/4)^5$ = 243/1024. Then P(at least one correct) = 1 - 243/1024 = 781/1024. A common error is to calculate P(exactly one) or something instead of at least one. Another error is to use $(1/4)^5$ for all correct. Strategy: For 'at least one,' consider the complement 'none' which is often easier to calculate.

The question is asking for the conditional probability that the student has a library card given that the student has a part-time job. The relevant outcomes are the students with a part-time job, with favorable being those who also have a library card, from the description. From the text, there are 10 students with both, and 6 with job but no library, so total with job = 16, favorable = 10. The conditional probability is 10/16 = 5/8. A common error is to use the total students or wrong numbers. Strategy: Construct a table from the given information to visualize the counts.

The question is asking for the probability that both marbles drawn are red, without replacement. The relevant outcomes are all possible ways to draw 2 marbles from 12, with favorable being drawing 2 red from 6 red. The probability can be calculated as P(first red) * P(second red | first red) = (6/12) * (5/11) = (1/2) * (5/11) = 5/22. Alternatively, using combinations, number of ways to choose 2 red is C(6,2) = 15, total ways C(12,2) = 66, so 15/66 = 5/22. A key error is to forget that it's without replacement and use $(6/12)^2$ = 1/4 instead. A good strategy is to determine if replacement is mentioned; if not, assume without for marbles.

The question is asking for the probability of getting heads on the coin and a number greater than 4 on the die. The relevant outcomes are the 2 coin outcomes and 6 die outcomes, total 12 equally likely, favorable are heads with 5 or 6, so 2 favorable. The probability is P(heads and >4) = P(heads) * P(>4) since independent = (1/2) * (2/6) = (1/2)*(1/3) = 1/6. There are 2 favorable out of 12, 2/12 = 1/6. A key error is to think >4 is 4,5,6 or something, but >4 is 5 and 6. Strategy: List all possible outcomes for small sample spaces to verify.

This question asks for the probability that a randomly chosen student from a class of 30 takes Spanish or French, where 18 take Spanish, 12 take French, and 5 take both. The relevant outcomes are the students taking at least one language, which is 18 + 12 - 5 = 25. The probability is 25/30. This uses inclusion-exclusion to avoid double-counting the 5 who take both. A common error is adding without subtracting, giving 30/30, which overcounts. Always subtract the intersection when finding the union in 'or' probabilities.

This question asks for the probability that both marbles drawn without replacement from a jar with 6 red, 4 blue, and 2 green marbles are blue. The total number of marbles is 12, with 4 blue marbles. The probability is calculated as P(first blue) × P(second blue | first blue) = (4/12) × (3/11) = (1/3) × (3/11) = 1/11. This accounts for the reduced sample space after the first draw, as there are now 11 marbles left with 3 blue. A common error is to treat it as with replacement, calculating $(4/12)^2$ = 1/9, which overestimates the probability. When draws are without replacement, always adjust the total and favorable outcomes for each successive draw.