ACT Probability and Statistics Practice

Probability and statistics on the ACT rarely reach advanced formulas. The section tests whether you can compute averages carefully, read tables accurately, count outcomes, and spot when a conclusion overreaches the data. This lesson explains the four core skill areas and gives you eight original four-choice questions, each with a verified solution and notes on why every wrong answer fails. For the overall math format and pacing, see our enhanced ACT math guide.

Mean, median, and mode (and how a new value shifts the mean)

The mean is the arithmetic average: sum of values divided by the number of values. The median is the middle value when the data are ordered; with an even number of values, average the two middle values. The mode is the most frequent value. ACT questions often add or remove one value and ask for the new mean, so keep track of both the numerator and denominator.

Reading two-way tables

A two-way table organizes counts by two categories at once — for example, grade level and pet preference. Before you calculate, identify whether the question wants a row total, a column total, a single cell, or a fraction of the grand total. The most common error is using the right number from the wrong row or column.

Probability basics (single events, "at least one" via complement, counting principle)

For equally likely outcomes, the probability of an event is favorable outcomes divided by total possible outcomes. The complement rule saves time when a question asks for "at least one": compute the probability of the opposite event (none) and subtract from 1. The fundamental counting principle says that if one choice has m options and an independent second choice has n options, there are m × n total combinations.

What averages can't tell you (inference limits: no causation from association, small samples)

Averages and associations describe data, but they do not automatically explain why the pattern exists. A study may find that students who do X have higher average scores, but that alone does not prove X causes the higher scores — there may be confounding variables, self-selection, or chance. Likewise, a small or unrepresentative sample cannot support a sweeping generalization. On the ACT, the correct answer often states only what the data directly show, while wrong answers claim causation or make predictions for groups that were not studied.

Practice set

The eight questions below are original PrepSolution practice items written in the enhanced four-choice format. Every answer was verified with two independent methods, and every wrong option reflects a common mistake.

Question 1: Mean with an added value

Four test scores are 78, 82, 88, and 90. After a fifth score is added, the mean of the five scores is 86. What is the fifth score?
A) 84
B) 88
C) 92
D) 96

Solution: the sum of the original four scores is 78 + 82 + 88 + 90 = 338. For five scores to have a mean of 86, their total must be 86 × 5 = 430. The fifth score is 430 − 338 = 92.

Check another way: the original mean is 338 ÷ 4 = 84.5. To raise the mean to 86 across five scores, the added score must supply the original mean (84.5) plus an extra 1.5 for each of the five scores: 84.5 + 5 × 1.5 = 84.5 + 7.5 = 92.

• A) 84 fails because 338 + 84 = 422, giving a mean of 84.4, not 86.
• B) 88 fails because 338 + 88 = 426, giving a mean of 85.2.
• D) 96 fails because 338 + 96 = 434, giving a mean of 86.8.

Question 2: Median from an ordered list

What is the median of the ordered list 14, 17, 19, 21, 25, 29, 31, 35?
A) 18
B) 20
C) 23
D) 28

Solution: there are eight values, so the median is the average of the two middle values, the 4th and 5th: (21 + 25) / 2 = 46 / 2 = 23.

Check another way: four values lie at or below 21 and four at or above 25, so the median must split the data between them. The midpoint is 23.

• A) 18 fails from averaging the 3rd and 4th values, (19 + 21)/2, one position too low.
• B) 20 is the mean of the full list, not the median; outliers pull these apart.
• D) 28 fails from averaging the 5th and 6th values, (25 + 29)/2, one position too high.

Question 3: Weighted average

A chemistry class has 16 students with an average score of 78. A physics class has 24 students with an average score of 88. What is the combined average score for all 40 students?
A) 82.0
B) 83.0
C) 84.0
D) 85.0

Solution: chemistry total points = 16 × 78 = 1,248. Physics total points = 24 × 88 = 2,112. Combined total = 3,360. Combined average = 3,360 ÷ 40 = 84.

Check another way: the combined average must be between 78 and 88, weighted toward 88 because the larger class scored higher. The physics class is 24/40 = 60% of the students, so the average should be about 60% of the way from 78 to 88: 78 + 0.6(10) = 84.

• A) 82.0 fails from swapping the class sizes: (24 × 78 + 16 × 88) / 40 = 3,280 / 40.
• B) 83.0 fails from averaging the two class means without accounting for class size: (78 + 88) / 2.
• D) 85.0 fails from misapplying the weights, giving too much credit to the higher-scoring class.

Question 4: Probability from a two-way table

A survey of 120 students asked whether they prefer cats or dogs. The results are shown below.

If one student is selected at random from the 120 students, what is the probability that the student is a sophomore who prefers dogs?
A) 1/6
B) 1/4
C) 1/3
D) 1/2

Solution: the number of sophomores who prefer dogs is the cell where the Sophomores row and Dogs column meet: 40. The total number of students is 120. Probability = 40/120 = 1/3.

Check another way: 40/120 reduces by dividing numerator and denominator by 40, giving 1/3.

• A) 1/6 fails from using sophomores who prefer cats: 20/120 = 1/6.
• B) 1/4 fails from using freshmen who prefer dogs: 30/120 = 1/4.
• D) 1/2 fails from using all sophomores: 60/120 = 1/2.

Question 5: Complement rule — "at least one"

A bag contains only red, blue, and green marbles. The probability of selecting a red marble is 0.35, and the probability of selecting a blue marble is 0.45. If two marbles are selected with replacement, what is the probability that at least one of them is green?
A) 0.16
B) 0.20
C) 0.32
D) 0.36

Solution: P(green) = 1 − 0.35 − 0.45 = 0.20. P(not green) = 1 − 0.20 = 0.80. P(no green in two draws) = 0.80 × 0.80 = 0.64. Therefore, P(at least one green) = 1 − 0.64 = 0.36.

Check another way: P(exactly one green) = 2 × 0.20 × 0.80 = 0.32. P(two greens) = 0.20 × 0.20 = 0.04. Adding the two cases gives 0.32 + 0.04 = 0.36.

• A) 0.16 fails from computing P(two greens) instead of P(at least one green).
• B) 0.20 fails from giving the probability of a single green marble, ignoring the two draws.
• C) 0.32 fails from giving P(exactly one green) but omitting the P(two greens) case.

Question 6: Counting principle

A password consists of one uppercase letter followed by one digit. How many different passwords are possible if the letter can be any of the 26 letters and the digit can be any digit from 0 to 9?
A) 36
B) 52
C) 260
D) 520

Solution: by the fundamental counting principle, multiply the number of letter options by the number of digit options: 26 × 10 = 260.

Check another way: for each of the 26 letters there are 10 possible digits, so the total is 10 + 10 + ... (26 times) = 260.

• A) 36 fails from adding 26 + 10 instead of multiplying.
• B) 52 fails from multiplying 26 by 2, as if each letter had only two digit choices.
• D) 520 fails from using 26 × 20, perhaps counting each digit twice.

Question 7: Inference limits

A study of high school students found that students who participate in a school music program have higher average GPAs than students who do not participate in the program. Based only on this finding, which conclusion is supported?
A) Participating in the music program causes students to have higher GPAs.
B) Students who want higher GPAs should join the music program.
C) There is an association between music-program participation and higher average GPAs among the students studied.
D) Students in the music program are more motivated than students who are not in the program.

Solution: the study found a difference in average GPAs between two groups. That supports only an association. The correct answer is C.

Why the other choices overclaim: A asserts causation, which an observational comparison cannot establish. B turns the finding into a recommendation for a different population or future behavior. D introduces a new characteristic (motivation) that was not measured. Only C restates what the data actually show.

• A) overclaims by treating association as proof of causation.
• B) overclaims by giving advice the study was not designed to support.
• D) overclaims by inventing an unmeasured explanation for the observed difference.

Question 8: Percent from a two-way table

Using the same pet-preference table, what percent of all students prefer dogs?
A) 41.7%
B) 50.0%
C) 58.3%
D) 70.0%

Solution: the total number of students who prefer dogs is 70, and the total number of students surveyed is 120. The percent is (70 / 120) × 100% = 58.333...%, which rounds to 58.3%.

Check another way: 70/120 reduces to 7/12, and 7 ÷ 12 = 0.5833..., confirming 58.3%.

• A) 41.7% fails from using the students who prefer cats: 50/120 ≈ 41.7%.
• B) 50.0% fails from using one class total (60/120) instead of the dog-preference total.
• D) 70.0% fails from treating the count 70 as a percentage out of 100 rather than out of 120.

What to review next

• If averages and weighted averages were tricky, review the formula reference in our ACT math formulas guide.
• For more probability, table-reading, and counting problems, work the full free ACT practice set.
• To build these skills into a weekly schedule, use the ACT study plan.
• For the overall enhanced math section — topics, pacing, and calculator policy — see the enhanced ACT math guide.

Use these eight questions as focused practice, not as a score prediction. A short set cannot estimate your ACT scaled score; only a full official form with its own scoring key can do that.