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 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 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 generated password contains the letter A or the digit 3 (or both), where passwords are one letter from {A, B, C, D} followed by one digit from {1, 2, 3}, with total 12 equally likely passwords. The relevant outcomes are the 6 favorable passwords: A1, A2, A3, B3, C3, D3. The probability is number of favorable / total = 6/12 = 1/2. Alternatively, using inclusion-exclusion: P(A) + P(3) - P(A and 3) = (3/12) + (4/12) - (1/12) = 6/12 = 1/2. A common error is overlooking the overlap, leading to 3/12 + 4/12 = 7/12. Carefully count the sample space and use inclusion-exclusion for 'or' to avoid double-counting.

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.

This question asks for the probability of getting tails on a fair coin flip and then a number greater than 4 on a standard six-sided die roll. The relevant outcomes are tails (1 out of 2 for the coin) and 5 or 6 (2 out of 6 for the die), with independent events. The probability is P(tails) × P(greater than 4) = (1/2) × (2/6) = (1/2) × (1/3) = 1/6. Since the events are independent, multiplying the individual probabilities gives the joint probability. A common error is to add the probabilities instead of multiplying, resulting in 1/2 + 1/3 = 5/6, which is incorrect for 'and' events. For independent events, always multiply to find the probability of both occurring.

This question asks for the probability that two tiles drawn with replacement from a bag with 5 black and 3 white tiles are the same color. The total tiles are 8, with each draw independent. The probability is P(both black) + P(both white) = $(5/8)^2$ + $(3/8)^2$ = 25/64 + 9/64 = 34/64 = 17/32. This considers the two mutually exclusive ways to get the same color. A common error is treating it as without replacement, calculating (5/8)(4/7) + (3/8)(2/7) = 15/32. Since replacement is specified, maintain the same sample space for each draw.

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 of not landing on an even number when spinning a spinner with 8 equal sections numbered 1 through 8. The relevant outcomes are the odd numbers: 1, 3, 5, 7, which are 4 out of 8 sections. The probability is 4/8 = 1/2. This is also 1 - P(even), where even numbers are 2, 4, 6, 8 (4/8). A common error is miscounting the even or odd numbers, perhaps thinking 8 is not even. Carefully list the sample space to ensure accurate counts of favorable outcomes.