SAT Math Quiz: Probability

What this quiz covers

This quiz focuses on Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for SAT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\frac{25}{144}$
2. $\frac{49}{144}$
3. $\frac{95}{144}$ (correct answer)
4. $\frac{119}{144}$

Explanation: 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.

Question 2

1. $\frac{8}{25}$ (correct answer)
2. $\frac{4}{25}$
3. $\frac{1}{25}$
4. $\frac{9}{25}$

Explanation: 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.

Question 3

1. $\frac{1}{4}$
2. $\frac{3}{8}$
3. $\frac{1}{2}$ (correct answer)
4. $\frac{5}{8}$

Explanation: 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.

Question 4

1. $\tfrac{5}{8}$ (correct answer)
2. $\tfrac{1}{3}$
3. $\tfrac{5}{12}$
4. $\tfrac{2}{5}$

Explanation: 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.

Question 5

1. $\frac{30}{30}$
2. $\frac{5}{30}$
3. $\frac{25}{30}$ (correct answer)
4. $\frac{31}{30}$

Explanation: 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.

Question 6

1. $\tfrac{9}{64}$ (correct answer)
2. $\tfrac{3}{32}$
3. $\tfrac{27}{64}$
4. $\tfrac{9}{16}$

Explanation: This asks for the probability of getting exactly 2 correct out of 3 questions when guessing randomly among 4 choices. This is a binomial probability with $n=3$, $k=2$, $p=\frac{1}{4}$. Using the binomial formula: $P(X=2) = C(3,2) \times \left(\frac{1}{4}\right)^2 \times \left(\frac{3}{4}\right)^1 = 3 \times \frac{1}{16} \times \frac{3}{4} = 3 \times \frac{3}{64} = \frac{9}{64}.$ A common error is forgetting the binomial coefficient $C(3,2)=3$, which accounts for the different ways to get 2 correct. When finding exact counts in repeated trials, use the binomial distribution.

Question 7

1. $\frac{32}{40}$
2. $\frac{20}{40}$
3. $\frac{26}{40}$ (correct answer)
4. $\frac{6}{40}$

Explanation: This problem uses the inclusion-exclusion principle for P(French OR Spanish). Given: 18 study French, 14 study Spanish, 6 study both, out of 40 students. Using P(F or S) = P(F) + P(S) - P(F and S): P = 18/40 + 14/40 - 6/40 = 26/40. The 6 students studying both languages would be counted twice if we just added 18 + 14, so we subtract 6. A helpful strategy is to draw a Venn diagram to visualize the overlap between the two groups.

Question 8

1. $\frac{10}{20}$
2. $\frac{8}{20}$
3. $\frac{7}{20}$
4. $\frac{9}{20}$ (correct answer)

Explanation: This problem asks for the probability that a number from 1 to 20 is a multiple of 3 OR a multiple of 5. Multiples of 3: {3, 6, 9, 12, 15, 18} = 6 numbers; Multiples of 5: {5, 10, 15, 20} = 4 numbers; Multiples of both (15) = 1 number. Using $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$: $P = \frac{6}{20} + \frac{4}{20} - \frac{1}{20} = \frac{9}{20}$. The common error is forgetting to subtract the overlap (15 is both a multiple of 3 and 5). When dealing with 'or' problems, always check for elements that satisfy both conditions.

Question 9

1. $\frac{3}{10}$
2. $\frac{7}{10}$ (correct answer)
3. $\frac{3}{7}$
4. $\frac{2}{10}$

Explanation: This problem asks for the probability of NOT drawing a blue marble. The total number of marbles is 5 red + 3 blue + 2 green = 10 marbles. The number of non-blue marbles is 5 red + 2 green = 7 marbles. Therefore, P(not blue) = 7/10. A common error is confusing the complement with the total, thinking P(not blue) = 3/10 (the probability of blue). Remember that P(not A) = 1 - P(A), or you can count the favorable outcomes directly as we did here.

Question 10

1. $\tfrac{3}{52}$ (correct answer)
2. $\tfrac{13}{52}$
3. $\tfrac{12}{52}$
4. $\tfrac{3}{13}$

Explanation: This question asks for the probability of drawing a card that is both a heart AND a face card from a standard 52-card deck. A standard deck has 13 hearts and 12 face cards total (3 per suit: J, Q, K), but we need cards that satisfy both conditions. The heart suit contains exactly 3 face cards: jack of hearts, queen of hearts, and king of hearts. Therefore, $P(\text{heart AND face card}) = \tfrac{3}{52}$. A common error is multiplying $P(\text{heart}) \times P(\text{face card}) = \tfrac{13}{52} \times \tfrac{12}{52}$, which assumes independence and gives the wrong answer. For AND probabilities, count only the outcomes that satisfy both conditions simultaneously.

Question 11

1. $\tfrac{3}{12}\cdot\tfrac{9}{11}$
2. $\tfrac{9}{12}\cdot\tfrac{8}{11}$
3. $\tfrac{3}{12}\cdot\tfrac{9}{12}$
4. $2\cdot\tfrac{3}{12}\cdot\tfrac{9}{11}$ (correct answer)

Explanation: This question asks for the probability of selecting exactly one defective bulb when drawing two bulbs without replacement from a box with 3 defective and 9 nondefective bulbs. Exactly one defective means either (defective first, nondefective second) OR (nondefective first, defective second). For the first case: P(D then N) = $\frac{3}{12} \times \frac{9}{11}$. For the second case: P(N then D) = $\frac{9}{12} \times \frac{3}{11}$. Adding these: P(exactly one defective) = $\frac{3}{12} \times \frac{9}{11} + \frac{9}{12} \times \frac{3}{11} = 2 \times \left( \frac{3}{12} \times \frac{9}{11} \right)$. The factor of 2 accounts for both possible orders. A common error is calculating only one order or forgetting that the probabilities change without replacement.

Question 12

1. $\frac{9}{25}$
2. $\frac{17}{29}$
3. $\frac{51}{145}$ (correct answer)
4. $\frac{18}{30}$

Explanation: 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.

Question 13

1. $\tfrac{4}{10}$
2. $\tfrac{21}{25}$
3. $\tfrac{9}{25}$
4. $\tfrac{16}{25}$ (correct answer)

Explanation: 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.

Question 14

1. $\tfrac{40}{40}$
2. $\tfrac{28}{40}$ (correct answer)
3. $\tfrac{12}{40}$
4. $\tfrac{34}{40}$

Explanation: 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.

Question 15

1. $\frac{3}{10}$
2. $\frac{7}{10}$ (correct answer)
3. $\frac{2}{5}$
4. $\frac{1}{2}$

Explanation: 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.

Question 16

1. $\frac{2}{3}$ (correct answer)
2. $\frac{5}{6}$
3. $\frac{7}{12}$
4. $\frac{3}{4}$

Explanation: This question asks for $P(\text{multiple of 3 OR prime})$ 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, $P(\text{multiple of 3 OR prime}) = \frac{8}{12} = \frac{2}{3}$. Be careful not to double-count numbers that satisfy both conditions when dealing with OR probabilities.

Question 17

1. $\frac{3}{8}$
2. $\frac{5}{8}$ (correct answer)
3. $\frac{6}{8}$
4. $\frac{7}{8}$

Explanation: We need to find the probability of landing on an even number OR a number greater than 6 on a spinner with sections 1-8. The even numbers are ${2, 4, 6, 8}$ (4 outcomes) and numbers greater than 6 are ${7, 8}$ (2 outcomes). Since 8 appears in both sets, we must use $P(\text{even or } >6) = P(\text{even}) + P(>6) - P(\text{even and } >6) = \frac{4}{8} + \frac{2}{8} - \frac{1}{8} = \frac{5}{8}$. Alternatively, we can count the favorable outcomes directly: ${2, 4, 6, 7, 8}$ gives 5 outcomes out of 8 total. The key error to avoid is double-counting the number 8.

Question 18

1. $1/3$ (correct answer)
2. $1/2$
3. $2/3$
4. $1/6$

Explanation: Outcomes greater than 4 are 5 and 6, so the probability is $2/6 = 1/3$. The other options count the wrong number of favorable or total outcomes.

Question 19

1. $\frac{g+2}{g+f+1}$ (correct answer)
2. $\frac{g+2}{g+f}$
3. $\frac{g+2}{g+f+2}$
4. $\frac{g}{g+f}$

Explanation: To find the chances of picking a guard, our formula should be the number of guards divided by the total number of players. We start with $g$ guards and $f$ forwards, and then add 2 guards and subtract one forward. So after the roster changes, we have $g+2$ guards, and we have $g+2+f-1$ total players. Since the probability of an outcome is the number of options that give you that outcome (guards) divided by the number of total options (all players), our probability is then $\frac{g+2}{g+f+1}$.

Question 20

1. $\frac{5}{8}$ (correct answer)
2. $\frac{3}{8}$
3. $\frac{1}{2}$
4. $\frac{5}{16}$

Explanation: This question asks for the probability of drawing a red ball from a box containing red and blue balls. The favorable outcomes are the 5 red balls, and the total possible outcomes are all 8 balls (5 red + 3 blue). Using the probability formula $P(\text{red}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{5}{8}$.