This question asks for the probability that at least one of two beads drawn (with replacement) is black. With 5 black and 7 white beads (12 total), P(black) = 5/12 on each draw. For "at least one black," it's easier to use the complement: P(at least one black) = 1 - P(both white). Since P(white) = 7/12 on each draw, P(both white) = (7/12) × (7/12) = 49/144. Therefore, P(at least one black) = 1 - 49/144 = 95/144. A common error is trying to add probabilities incorrectly; using the complement for "at least one" problems is often more efficient.

This conditional probability asks for P(A|B) where A is "first flip heads" and B is "exactly one head in two flips". First, identify all outcomes with exactly one head: HT and TH (event B has 2 outcomes). Among these, only HT has heads on the first flip (event A∩B has 1 outcome). Therefore, P(A|B) = P(A∩B)/P(B) = (1/4)/(2/4) = 1/2. A key insight is that given exactly one head occurs, it's equally likely to be on the first or second flip. When working with conditional probability, always clearly identify which outcomes satisfy both the condition and the event.

This question asks for the probability of selecting a hardcover book from a shelf of 20 books. The hardcover books are: 3 fiction hardcovers and 2 poetry hardcovers, giving us 5 total hardcover books. Using the probability formula P(hardcover) = favorable outcomes / total outcomes, we get P(hardcover) = 5/20 = 1/4. A common error is to only count hardcovers from one category or to miscount the total. When dealing with overlapping categories (book type and cover type), carefully identify all items that meet your criteria.

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. After removing one blue marble, 14 marbles remain with only 4 blue, so P(second blue|first blue) = 4/14. The probability of both events is P(both blue) = (5/15) × (4/14) = 20/210 = 2/21. A key error is forgetting to adjust the counts for the second draw or treating it as with replacement. When drawing without replacement, always update both the favorable outcomes and total count.

This problem asks for P(at least one black sock) when drawing 2 socks from 3 black and 7 white (10 total) without replacement. It's easier to use the complement: P(at least one black) = 1 - P(both white). P(both white) = (7/10) × (6/9) = 42/90 = 7/15. Therefore, P(at least one black) = 1 - 7/15 = 8/15. The complement approach is often simpler than calculating P(exactly one black) + P(both black). When you see "at least one," consider using the complement rule with "none."

This question asks for the probability that a randomly selected book is NOT hardcover. Total books: 90 fiction + 60 nonfiction = 150 books. Hardcover books: 30 fiction + 15 nonfiction = 45 hardcover books. Therefore, not hardcover books = 150 - 45 = 105 books. P(not hardcover) = 105/150 = 7/10. A common error is calculating P(hardcover) = 45/150 = 3/10 but forgetting to find the complement. When asked for "not" probabilities, subtract from 1 or count the complementary outcomes directly.

This question asks for the probability that at least one of two beads drawn with replacement is black, from a jar with 6 white and 4 black beads (total 10). The total outcomes for each draw are 10, and 'at least one black' is the complement of both white. The probability of both white is (6/10) × (6/10) = 36/100 = 9/25, so P(at least one black) = 1 - 9/25 = 16/25. This uses the complement because it's easier than adding P(first black and second white) + P(first white and second black) + P(both black). A common error is forgetting replacement, treating it as without and using 6/10 × 5/9 for both white. Another mistake is calculating only both black. When using complements, ensure the sample space is correctly identified for independent events.

This question asks for the probability that a randomly chosen student from a class of 40 is taking French or Spanish. There are 18 taking French, 16 taking Spanish, and 6 taking both, so the favorable outcomes for the union are 18 + 16 - 6 = 28. The probability is P(French or Spanish) = 28/40. This uses inclusion-exclusion to avoid double-counting the 6 students in both. Calculation: P(French) + P(Spanish) - P(both) = 18/40 + 16/40 - 6/40 = 28/40. A common error is adding without subtracting, getting 34/40. In Venn diagram-style problems, carefully calculate the union by subtracting the intersection from the sample space totals.

This question asks for the probability that a randomly drawn card from a 52-card deck is a heart or a face card (jack, queen, or king). The deck has 52 cards, with 13 hearts and 12 face cards, but 3 face hearts overlap. The favorable outcomes are 13 + 12 - 3 = 22 cards. Thus, P(heart or face) = 22/52. This calculation applies inclusion-exclusion: P(heart) + P(face) - P(both) = 13/52 + 12/52 - 3/52. A common error is not subtracting the overlap, getting 25/52. When dealing with unions in card problems, list the suits and ranks to accurately count the sample space and overlaps.

This question asks for the probability that the spinner lands on an even number or a number greater than 6. The spinner has 8 equal sections labeled 1 through 8, so the total sample space is 8 outcomes. The favorable outcomes are even numbers (2, 4, 6, 8; 4 outcomes) and numbers greater than 6 (7, 8; 2 outcomes), with overlap at 8, so the union has 5 unique outcomes. The probability is P(even or >6) = P(even) + P(>6) - P(both) = 4/8 + 2/8 - 1/8 = 5/8. This uses inclusion-exclusion to handle the overlap. A common error is not subtracting the overlap, leading to 6/8. When calculating unions, carefully list all favorable outcomes to avoid double-counting in the sample space.