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ACT Math

ACT Math Help: Probability

Review real example questions for Probability in ACT Math.

Question 1

A deck of cards is shuffled. What is the probability of drawing a card that is either an Ace or a diamond?

  1. 1/52
  2. 1/13
  3. 1/4
  4. 4/13
Explanation: A standard deck has 4 Aces and 13 diamonds, but the Ace of Diamonds creates overlap. Using inclusion-exclusion: P(Ace or diamond) = P(Ace) + P(diamond) - P(Ace and diamond) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13. The overlap of 1/52 must be subtracted to avoid double-counting the Ace of Diamonds. Choice C (1/4) equals 13/52, which ignores the Aces entirely.

Question 2

A box contains 4 red and 6 black balls. What is the probability of picking a red ball?

  1. 2/3
  2. 1/3
  3. 3/5
  4. 2/5
Explanation: The box contains 4 + 6 = 10 balls total, with 4 red balls as favorable outcomes. Using the probability rule P(event) = favorable / total, we get P(red) = 4/10 = 2/5. Choice B (1/3) would be incorrect, and choice A (2/3) represents the probability of picking a black ball instead of red.

Question 3

A jar contains exactly 5 red marbles, 4 blue marbles, and 3 yellow marbles. If two marbles are drawn at random without replacement, what is the probability that both marbles drawn will be the same color?

  1. 1966\frac{19}{66}6619​
  2. 13\frac{1}{3}31​
  3. 19144\frac{19}{144}14419​
  4. 47132\frac{47}{132}13247​
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 4

Events A and B are mutually exclusive. P(A)=0.5P(A) = 0.5P(A)=0.5 and P(B)=0.3P(B) = 0.3P(B)=0.3. What is P(A or B)P(A \text{ or } B)P(A or B)?

  1. 0.15
  2. 0.2
  3. 0.8
  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 5

In a carnival game, rolling an even number wins $4.00 and rolling odd loses $2.00. What is the expected value of a single roll?

  1. $1.00
  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)iscorrect—P(even)=P(odd)=1/2.Expectedvalue=(winamount×P(win))+(lossamount×P(loss))=(1.00) is correct — P(even) = P(odd) = 1/2. Expected value = (win amount × P(win)) + (loss amount × P(loss)) = (1.00)iscorrect—P(even)=P(odd)=1/2.Expectedvalue=(winamount×P(win))+(lossamount×P(loss))=(4 × 0.5) + (−2×0.5)=2 × 0.5) = 2×0.5)=2.00 − 1.00=1.00 = 1.00=1.00. Note: the loss is entered as a negative value. Choice B (1.50)likelycomesfromcomputing(4×0.5)+(2×0.5)=2+1=3,thendividingby2:3/2=1.50) likely comes from computing (4 × 0.5) + (2 × 0.5) = 2 + 1 = 3, then dividing by 2: 3/2 = 1.50)likelycomesfromcomputing(4×0.5)+(2×0.5)=2+1=3,thendividingby2:3/2=1.50 — treating the loss as positive and then halving. Choice C (2.00)computesonlythewinningterm:4×0.5=2,ignoringthelossentirely.ChoiceD(2.00) computes only the winning term: 4 × 0.5 = 2, ignoring the loss entirely. Choice D (2.00)computesonlythewinningterm:4×0.5=2,ignoringthelossentirely.ChoiceD(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 = 1.00meansaplayerwouldaverage1.00 means a player would average 1.00meansaplayerwouldaverage1 profit per roll in the long run.

Question 6

A local arcade has a prize wheel with 4 colored sections. The table shows the results over 200 spins. To make each color exactly 25%, which change is the best fix?

  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.
  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

Whole numbers 1 through 20 are placed in a hat. One slip is drawn at random. What is the probability that the number is a multiple of 3?

  1. 320\frac{3}{20}203​
  2. 15\frac{1}{5}51​
  3. 310\frac{3}{10}103​
  4. 13\frac{1}{3}31​
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

A fair coin is flipped and a fair six-sided die is rolled. What is the probability of getting heads and rolling an even number? (Assume the events are independent.)

  1. 13\tfrac{1}{3}31​
  2. 112\tfrac{1}{12}121​
  3. 14\tfrac{1}{4}41​
  4. 23\tfrac{2}{3}32​
Explanation: The sample space involves two independent events: flipping a coin (2 outcomes) and rolling a die (6 outcomes). The favorable outcomes are getting heads (probability 1/21/21/2) AND rolling an even number (2, 4, or 6, so probability 3/6=1/23/6 = 1/23/6=1/2). Since the events are independent, we multiply: P(heads and even)=P(heads)×P(even)=12×12=14P(\text{heads and even}) = P(\text{heads}) \times P(\text{even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}P(heads and even)=P(heads)×P(even)=21​×21​=41​. Choice B incorrectly uses 1/12, possibly by multiplying 1/2×1/61/2 \times 1/61/2×1/6 instead of recognizing there are 3 even numbers on a die.

Question 9

If an event occurs with a probability of 0.25, what is the probability of the event not occurring?

  1. 0.25
  2. 0.50
  3. 0.75
  4. 0.85
Explanation: For any event with probability P(event), the complement rule states that P(not event) = 1 - P(event). Given that P(event) = 0.25, we calculate P(not event) = 1 - 0.25 = 0.75. This represents the probability that the event does not occur. Choice A (0.25) incorrectly gives the original probability instead of its complement.

Question 10

A jar contains 4 white balls and 6 black balls. Two balls are drawn without replacement. What is the probability that both balls are white?

  1. 25\frac{2}{5}52​
  2. 425\frac{4}{25}254​
  3. 215\frac{2}{15}152​
  4. 16\frac{1}{6}61​
Explanation: The sample space initially has 10 balls (4 white, 6 black). For the first draw, P(white) = 4/10. After drawing one white ball without replacement, 9 balls remain with 3 white. P(second white | first white) = 3/9 = 1/3. P(both white) = (4/10) × (3/9) = 12/90 = 2/15.