ACT 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 ACT 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. $\frac{19}{66}$ (correct answer)
2. $\frac{1}{3}$
3. $\frac{19}{144}$
4. $\frac{47}{132}$

Explanation: The correct answer is A (19/66). Without replacement, compute P(both same color) for each color separately, then add. P(both red) = (5/12) × (4/11) = 20/132. P(both blue) = (4/12) × (3/11) = 12/132. P(both yellow) = (3/12) × (2/11) = 6/132. Total = (20 + 12 + 6)/132 = 38/132 = 19/66. B (1/3) estimates loosely from 3 colors. C (19/144) treats the draws as independent (with replacement), using 12² = 144 as the denominator. D (47/132) overcounts, possibly including the same pair twice. The key is using without replacement: after the first draw, there are only 11 marbles left, and one fewer marble of that color.

Question 2

1. 0.15
2. 0.2
3. 0.8 (correct answer)
4. 1

Explanation: This is a probability question testing the addition rule for mutually exclusive events. Choice C (0.80) is correct — for mutually exclusive events (which cannot both occur), P(A or B) = P(A) + P(B) = 0.5 + 0.3 = 0.8. Choice A (0.15) multiplies the probabilities: P(A) × P(B) = 0.5 × 0.3 = 0.15 — this is the formula for P(A and B) when events are INDEPENDENT, not the formula for P(A or B) when events are mutually exclusive. Choice B (0.20) subtracts: 0.5 − 0.3 = 0.2. Choice D (1.00) assumes mutually exclusive events together cover the entire sample space — but two mutually exclusive events can have probabilities that sum to less than 1 (there can be other outcomes). Pro tip: Mutually exclusive means the events cannot happen at the same time — like rolling a 2 and rolling a 5 on the same die. For mutually exclusive events: P(A or B) = P(A) + P(B). For non-mutually-exclusive events: P(A or B) = P(A) + P(B) − P(A and B). The simpler formula here is a gift — just add.

Question 3

1. $1.00 (correct answer)
2. $1.50
3. $2.00
4. $3.00

Explanation: This is an expected value question testing the weighted average of outcomes. Choice A ($1.00) is correct — P(even) = P(odd) = 1/2. Expected value = (win amount × P(win)) + (loss amount × P(loss)) = ($4 × 0.5) + (−$2 × 0.5) = $2.00 − $1.00 = $1.00. Note: the loss is entered as a negative value. Choice B ($1.50) likely comes from computing (4 × 0.5) + (2 × 0.5) = 2 + 1 = 3, then dividing by 2: 3/2 = $1.50 — treating the loss as positive and then halving. Choice C ($2.00) computes only the winning term: 4 × 0.5 = 2, ignoring the loss entirely. Choice D ($3.00) adds the outcomes instead of weighting: 4 + (−2) = 2... or adds the expected gain and expected loss as positives: 2 + 1 = 3. Pro tip: Expected value = Σ(outcome × probability). For each possible outcome, multiply its value by its probability, then sum all products. Losses must be entered as negative values. A positive EV means the game favors the player; negative EV means it favors the house. Here EV = $1.00 means a player would average $1 profit per roll in the long run.

Question 4

1. 1/6
2. 5/6
3. 1/3 (correct answer)
4. 2/3

Explanation: Probability is the count of favorable outcomes over the 6 equally likely faces of the die. The numbers less than or equal to 2 are 1 and 2, so there are 2 favorable outcomes and the probability is 2/6, which reduces to 1/3. The value 1/6 counts only a single face rather than both, and 2/3 counts the four faces greater than 2, which is the complement of what was asked. The value 5/6 is the complement of a single face, so it answers "not a specific number" instead of "at most 2." Read the inequality carefully to see whether the boundary value is included, then reduce the fraction before comparing it to the choices.

Question 5

1. 1/8
2. 1/2 (correct answer)
3. 5/8
4. 3/8

Explanation: Probability is favorable outcomes divided by total equally likely outcomes. The spinner has 8 equal sections, and the even numbers among 1 through 8 are 2, 4, 6, and 8, so there are 4 favorable outcomes out of 8, which reduces to 1/2. The value 3/8 comes from counting only three even numbers and forgetting 8, and 5/8 comes from over-counting by including an odd section. The value 1/8 is the chance of landing on one specific section, not on any of the four even ones. Count the favorable outcomes explicitly by listing them, then reduce the fraction to lowest terms before matching it to the choices.

Question 6

1. Increase the area of the Red section by decreasing the area of the Green section.
2. Decrease the area of the Blue section by increasing the area of the Yellow section. (correct answer)
3. Decrease the area of the Green section by increasing the area of the Yellow section.
4. Increase the area of the Blue section, and it does not matter which section decreases.

Explanation: This is an experimental probability and data analysis question testing whether students can identify which change moves results toward a target. Choice B (decrease Blue, increase Yellow) is correct — the target is 25% per color, which means 50 spins each over 200 trials. Current results: Red=50 (25%, already correct), Blue=80 (40%, too high), Green=40 (20%, too low), Yellow=30 (15%, too low). The only change that moves a too-high color down and a too-low color up is decreasing Blue and increasing Yellow. Choice A increases Red (already at target) and decreases Green (already too low) — making two things worse. Choice C decreases Green (already too low) — moving in the wrong direction. Choice D increases Blue (already too high) — the opposite of what's needed. Pro tip: Before evaluating the answer choices, calculate what each color's current share is and label each one as "too high," "correct," or "too low." The right answer must fix all problems simultaneously — if an answer makes any color worse, eliminate it immediately.

Question 7

1. $\frac{3}{20}$
2. $\frac{1}{5}$
3. $\frac{3}{10}$ (correct answer)
4. $\frac{1}{3}$

Explanation: This is a probability question testing careful counting within a bounded set. Choice C (3/10) is correct — list the multiples of 3 between 1 and 20: 3, 6, 9, 12, 15, 18 — exactly 6 numbers. P = 6/20 = 3/10. Choice A (3/20) uses a numerator of 3, perhaps counting only the first three multiples (3, 6, 9) and stopping, or confusing the divisor with the count. Choice B (1/5) uses a numerator of 4, suggesting the student counted 4 multiples or computed ⌊20/3⌋ with an error. Choice D (1/3) applies a shortcut: "dividing by 3 means a 1/3 chance" — this would be true if the numbers went from 1 to infinity, but within the bounded set 1–20, there are exactly 6 multiples of 3 out of 20 numbers, not 20/3 ÷ 20 = 1/3. Pro tip: For "multiples of n from 1 to N" problems, the count is ⌊N/n⌋ — divide and take the integer part. Here: ⌊20/3⌋ = 6. Always list them to verify, especially near the boundary (18 qualifies, 21 does not).

Question 8

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

Explanation: The sample space for a six-sided die is $\{1, 2, 3, 4, 5, 6\}$. The favorable outcomes are $\{1, 2, 3\}$ = 3 outcomes. Since these are mutually exclusive events (rolling exactly one number), $P(1 \text{ or } 2 \text{ or } 3) = \frac{3}{6} = \frac{1}{2}$. This represents half of all possible outcomes.

Question 9

1. $\frac{8}{25}$ (correct answer)
2. $\frac{8}{17}$
3. $\frac{12}{25}$
4. $\frac{1}{3}$

Explanation: The correct answer is A (8/25). First find the total number of marbles: 12 + 8 + 5 = 25. The probability of drawing a blue marble is the number of blue marbles divided by the total: 8/25. B (8/17) uses only the red and green marbles as the denominator (12 + 5 = 17), omitting blue from the total — a subtle error. C (12/25) gives the probability of drawing a red marble instead of blue. D (1/3) treats all three colors as equally likely, ignoring their different quantities. On probability questions, always identify the correct sample space (denominator) before computing the ratio.

Question 10

1. $5/12$
2. $1/12$
3. $1/3$
4. $7/12$ (correct answer)

Explanation: The box contains $5 + 4 + 3 = 12$ balls total. The event "drawing a yellow or green ball" includes $4 + 3 = 7$ favorable outcomes since these are mutually exclusive events. Using $P(event) = favorable / total$, we get $P(yellow or green) = 7/12$. Choice A ($5/12$) incorrectly counts only the red balls instead of yellow and green combined.

Question 11

1. $\frac{1}{3}$
2. $\frac{6}{11}$
3. $\frac{2}{3}$
4. $\frac{2}{5}$ (correct answer)

Explanation: This is a fundamental probability problem where you're finding the probability of a single event from a simple sample space. Count the total number of marbles: 5 + 6 + 4 = 15 marbles. The probability of selecting a blue marble is the number of blue marbles divided by the total: 6/15, which simplifies to 2/5. Strategy: Always reduce fractions to lowest terms on the ACT—the correct answer will be in simplest form.

Question 12

1. $\frac{2}{5}$ (correct answer)
2. $\frac{5}{36}$
3. $\frac{1}{3}$
4. $\frac{1}{18}$

Explanation: This is a conditional probability question requiring reduction of the sample space. Choice A (2/5) is correct — the condition is that the sum equals 8. First, list all outcomes summing to 8: (2,6), (3,5), (4,4), (5,3), (6,2) — exactly 5 equally likely outcomes. These become the new (conditional) sample space. Of these 5 outcomes, those containing at least one 5: (3,5) and (5,3) — exactly 2 outcomes. P = 2/5. Choice B (5/36) uses the 5 favorable "sums to 8" outcomes over the full 36-outcome sample space, ignoring the conditional. Choice C (1/3) misidentifies the denominator — using 6 possible outcomes instead of the correct 5. Choice D (1/18) counts only 2 outcomes containing a 5 but places them over the unconditioned sample space of 36: 2/36 = 1/18. Pro tip: Conditional probability ("given that...") shrinks the sample space. First identify all outcomes that satisfy the given condition — those become your new denominator. Then count how many of those meet the additional requirement for your numerator.

Question 13

1. $\frac{1}{9}$
2. $\frac{2}{21}$ (correct answer)
3. $\frac{1}{3}$
4. $\frac{4}{21}$

Explanation: This is a dependent probability question testing probability without replacement. Choice B (2/21) is correct — P(1st marble is green) = 5/15 = 1/3. After removing one green marble, P(2nd marble is green) = 4/14 = 2/7. Multiply the two probabilities: (1/3)(2/7) = 2/21. Choice A (1/9) treats the two draws as independent — computing (1/3) × (1/3) = 1/9, as if the first marble were replaced before the second draw. Choice C (1/3) gives the probability of the first marble being green only, stopping after the first draw. Choice D (4/21) uses 4/15 for the first draw instead of 5/15 — perhaps counting only 4 of the 5 green marbles. Pro tip: "Without replacement" means the denominator and possibly the numerator change for the second draw. After drawing one green marble, you have 14 total and 4 green remaining. Always update BOTH the favorable outcomes and the total before computing the second probability.

Question 14

1. $\frac{1}{4}$
2. $\frac{9}{20}$
3. $\frac{3}{4}$ (correct answer)
4. $\frac{11}{20}$

Explanation: This is a probability question testing complementary probability. Choice C (3/4) is correct — there are 6 + 5 + 9 = 20 total marbles. "Not blue" means red or green: 6 + 9 = 15 marbles. P(not blue) = 15/20 = 3/4. Choice A (1/4) inverts the logic — it represents the probability of an outcome that is NOT "not blue," essentially giving P(blue) reframed. Choice B (9/20) counts only the green marbles (9 out of 20), ignoring the 6 red marbles that are also "not blue." Choice D (11/20) adds blue + green (5 + 9 = 14)... wait — actually reflects adding blue to something incorrectly. Pro tip: For "not" probability questions, count everything EXCEPT the excluded group, then divide by the total. It's often faster than computing P(blue) and subtracting from 1.

Question 15

1. $\frac{75}{200}$
2. $\frac{75}{135}$ (correct answer)
3. $\frac{75}{100}$
4. $\frac{135}{200}$

Explanation: This is a conditional probability question testing restricted sample space. Choice B (75/135) is correct — "given that the patron bought popcorn" restricts the sample space to only the 135 popcorn buyers. Of those, 75 are children. P(child | popcorn) = 75/135. Choice A (75/200) uses the total patron count as the denominator — the student ignores the conditional "given popcorn" restriction. Choice C (75/100) uses total children (100) as the denominator — the student restricts by the wrong group. Choice D (135/200) gives the probability of buying popcorn at all — it answers "what fraction of all patrons bought popcorn?" not the conditional question asked. Pro tip: "Given that" is the key phrase for conditional probability. It always shrinks your sample space. Ask: "What group am I restricted to?" (popcorn buyers = 135). Then ask: "Within that group, what am I counting?" (children who bought popcorn = 75). The denominator is always the "given" group's total, not the overall total.

Question 16

1. $\frac{1}{5}$
2. $\frac{4}{9}$ (correct answer)
3. $\frac{3}{5}$
4. $\frac{2}{3}$

Explanation: The correct answer is B (4/9). This is a conditional probability question: P(plays sport | plays instrument). The condition restricts the sample space to the 90 students who play an instrument. Of those 90, exactly 40 also play a sport. P = 40/90 = 4/9. A (1/5) uses 40/200, the unconditional probability of playing both — ignores the given condition. C (3/5) uses 120/200, the overall probability of playing a sport with no condition. D (2/3) uses an incorrect row or column total. The critical insight: conditional probability means the denominator changes to reflect only the cases where the given condition is met.

Question 17

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

Explanation: The correct answer is B (3/8). The ratio 3:5 means 3 + 5 = 8 total parts. With 40 chips total, each part = 40 ÷ 8 = 5 chips. Red chips = 3 × 5 = 15. P(red) = 15/40 = 3/8. A (1/4) uses 10 red chips instead of 15, possibly treating the ratio as direct counts. C (1/2) ignores the ratio and assumes equal halves. D (5/8) gives the probability of drawing a blue chip instead of red — the student finds the right value but for the wrong color. Always scale the ratio to the actual total before computing probability.

Question 18

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

Explanation: The sample space for two coin flips is $\{HH, HT, TH, TT\}$ with 4 equally likely outcomes. The event "exactly one head" corresponds to outcomes $\{HT, TH\}$ = 2 favorable outcomes. $P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2}$. Choice A incorrectly calculates P(at least one head).

Question 19

1. 13/52
2. 1/4
3. 1/2 (correct answer)
4. 1/13

Explanation: A standard deck has 52 cards with 13 spades and 13 clubs, giving us 26 favorable outcomes total. Since drawing a spade and drawing a club are mutually exclusive events, P(spade or club) = P(spade) + P(club) = 13/52 + 13/52 = 26/52 = 1/2. Choice B (1/4) represents the probability of drawing cards from just one suit.

Question 20

1. 1/10
2. 2/5 (correct answer)
3. 1/5
4. 3/10

Explanation: Among numbers 1 to 10, the prime numbers are {2, 3, 5, 7}, giving us 4 favorable outcomes out of 10 total outcomes. Using the probability rule P(event) = favorable / total, we get P(prime) = 4/10 = 2/5. Choice C (1/5) incorrectly counts only 2 prime numbers, while choice D (3/10) suggests only 3 prime numbers were counted.