This question asks for the probability that a spinner result is either a multiple of 3 or an even number from sections 1-8. The multiples of 3 are {3, 6}, and the even numbers are {2, 4, 6, 8}. The union of these sets is {2, 3, 4, 6, 8}, which contains 5 numbers (note that 6 appears in both sets but is counted only once). Therefore, P(multiple of 3 OR even) = 5/8. A common error is adding 2 + 4 = 6 favorable outcomes by counting 6 twice. When finding probabilities involving 'or', always identify the overlap to avoid double-counting.

The question asks for the probability that the number on a randomly selected card from 1 to 10 is not a multiple of 3. The sample space is 10 cards, with multiples of 3 being 3,6,9 (3 outcomes). The favorable outcomes are the remaining 7 cards (1,2,4,5,7,8,10). Thus, P(not multiple of 3) = 7/10, or equivalently 1 - 3/10. A common error is miscounting multiples, like including 10 or forgetting 9. Carefully list the sample space and count favorable outcomes to avoid such mistakes.

The question asks for the probability of landing on a letter that is not B on a spinner with 8 equal sections: 2 A, 3 B, 2 C, 1 D. The sample space is 8 sections, with 3 B sections. The favorable outcomes are the 5 non-B sections (2 A + 2 C + 1 D). Thus, P(not B) = $\frac{5}{8}$, or 1 - $\frac{3}{8}$. A common error is treating letters as equally likely instead of sections, like assuming $\frac{3}{4}$ letters are not B. Emphasize counting actual outcomes in the sample space, not just unique labels.

The question asks for the probability of rolling an even number or a number greater than 4 on a fair six-sided die. The sample space is 6 outcomes: ${1,2,3,4,5,6}$, with even numbers ${2,4,6}$ (3 outcomes) and numbers >4 ${5,6}$ (2 outcomes). The favorable outcomes are the union ${2,4,5,6}$ (4 outcomes), calculated as $3 + 2 - 1 = 4$ using inclusion-exclusion. Thus, P = $4/6 = 2/3$. A common error is forgetting to subtract the overlap (6), leading to $5/6$. Emphasize listing all outcomes to identify the exact sample space and overlaps in 'or' problems.

This is a conditional probability question asking for P(blue | not green). We have 15 marbles total: 6 red, 5 blue, and 4 green, so 11 marbles are not green (6 red + 5 blue). Given that the marble is not green, our sample space reduces to these 11 marbles, of which 5 are blue. Therefore, P(blue | not green) = 5/11. A common mistake is calculating P(blue) = 5/15 without considering the condition. For conditional probability, always identify the reduced sample space first, then count favorable outcomes within that space.

This asks for the probability that a spinner result is a multiple of 3 OR an even number. The multiples of 3 from 1-8 are {3, 6}, and the even numbers are {2, 4, 6, 8}. The union of these sets is {2, 3, 4, 6, 8}, which contains 5 outcomes. Therefore, P(multiple of 3 OR even) = 5/8. Note that 6 appears in both sets but is counted only once in the union. A common error is adding 2 + 4 = 6 favorable outcomes without recognizing the overlap. For OR probabilities with overlapping events, use the inclusion-exclusion principle or count the union directly.

This asks for the probability of getting at least one head in 3 coin flips. The complement of "at least one head" is "no heads" (all tails). With a fair coin, P(tails on one flip) = 1/2, so P(all tails in 3 flips) = (1/2)³ = 1/8. Therefore, P(at least one head) = 1 - 1/8 = 7/8. The complement approach is often easier than counting all favorable outcomes (HHH, HHT, HTH, HTT, THH, THT, TTH). For "at least one" problems, consider using the complement rule: P(at least one) = 1 - P(none).

This question asks for a conditional probability: P(Heart|Face card). In a standard deck, there are 12 face cards total (3 per suit × 4 suits). Among these 12 face cards, 3 are hearts (jack, queen, king of hearts). Using the conditional probability formula, P(Heart|Face card) = 3/12 = 1/4. A common error is using 13 (all hearts) or 52 (all cards) in the calculation instead of restricting to the 12 face cards. When solving conditional probability with cards, always identify your restricted sample space first.

This problem is best solved using the complement rule: P(at least one head) = 1 - P(no heads). With 3 coin flips, P(no heads) means getting tails on all three flips: (1/2) × (1/2) × (1/2) = 1/8. Therefore, P(at least one head) = 1 - 1/8 = 7/8. This approach is much simpler than calculating P(exactly 1 head) + P(exactly 2 heads) + P(exactly 3 heads). The complement rule is particularly useful for "at least one" problems because it reduces multiple calculations to just one. Always consider using the complement when you see phrases like "at least one" or "at least some."

This question asks for the probability that a randomly chosen bead is NOT black. The jar contains 5 black beads and 7 white beads, for a total of 12 beads. The favorable outcomes are the white beads (not black), of which there are 7. Therefore, P(not black) = 7/12. A common error would be calculating P(black) = 5/12 instead of P(not black). When a question asks for the probability of NOT an event, count all outcomes except those in the specified event.