Probability questions on the PSAT/NMSQT show up in two main formats: reading probabilities directly out of a two-way frequency table, and calculating probabilities for events described in words — sometimes a single event, sometimes a combination of two or more events happening together. Both formats rest on the same basic definition (favorable outcomes divided by total outcomes), but the PSAT tests several variations on that idea that are worth knowing cold before test day.
Two-way table questions ask about probabilities within a category — for example, "given that a person is from a certain group, what's the probability of X?" These are conditional probability questions, and the key is dividing by the total for the given condition (a row or column total), not the grand total. Compound event questions, on the other hand, ask about two things happening together — "the probability that A and B both occur" — which usually means multiplying two probabilities together, but only when the events are independent (like drawing with replacement). When a selection is made without replacement, like choosing two items from a box, the probabilities change between the first and second draw, which is a detail many students miss. Questions using "or" instead of "and" require the opposite instinct: adding two probabilities together and then subtracting any overlap, so you don't double-count outcomes that satisfy both conditions.
Below are 11 PSAT/NMSQT-style probability practice problems covering all of these variations: two-way tables, "and"/"or" compound events, probability without replacement, and multi-step conditional scenarios. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free PSAT/NMSQT Math QBank.
Questions 1–5 refer to the following information. The table below shows the number of college faculty members in three departments: biological sciences, education, and social sciences.
| Biological Sciences | Education | Social Sciences | Total | |
|---|---|---|---|---|
| Male | 10 | 26 | 19 | 55 |
| Female | 15 | 21 | 17 | 53 |
| Total | 25 | 47 | 36 | 108 |
What is the probability that a randomly chosen faculty member is a female given that she is from Biological Sciences?
"Given that she is from Biological Sciences" restricts the group to only Biological Sciences faculty — so the denominator is the Biological Sciences column total (25), not the grand total (108).
Of those 25 Biological Sciences faculty, 15 are female.
15/25 = 3/5
The correct answer is B.
What is the probability that a randomly chosen faculty member is a male or from Social Sciences?
For an "or" probability, add the two individual probabilities and then subtract the overlap, so faculty who are both male and from Social Sciences aren't counted twice:
P(male) + P(Social Sciences) − P(male and Social Sciences)
55/108 + 36/108 − 19/108 = 72/108
Simplify: 72/108 = 2/3. The correct answer is B.
What is the probability that a randomly chosen faculty member is a female from Education department or a male from Social Sciences?
Unlike the previous problem, these two groups — "female from Education" and "male from Social Sciences" — can't overlap (a person can't be both), so there's no double-counting to subtract. Just add the two counts directly.
Female Education faculty: 21. Male Social Sciences faculty: 19.
(21 + 19)/108 = 40/108
Simplify by dividing both by 4: 40/108 = 10/27. The correct answer is B.
What is the probability that a randomly chosen faculty member is from Biological Sciences given that the faculty member is a male?
"Given that the faculty member is a male" restricts the group to the Male row — the denominator is the male row total (55), not 108.
Of those 55 male faculty, 10 are from Biological Sciences.
10/55 = 2/11
The correct answer is B. Notice this is a different question — and a different answer — from Problem 1, even though both involve Biological Sciences and gender; the direction of the condition matters.
For Biological Science and Education faculties combined, 1/6 of the female and 1/4 of the male faculty members are associate professors. If a person is randomly chosen from these two departments, what is the probability that a faculty member is an associate professor?
First combine Biological Sciences and Education. Female: 15 + 21 = 36. Male: 10 + 26 = 36. Total faculty in these two departments: 36 + 36 = 72.
Find the number of associate professors in each group. Female associate professors: 36 × 1/6 = 6. Male associate professors: 36 × 1/4 = 9.
Total associate professors: 6 + 9 = 15.
Probability: 15/72. Simplify by dividing both by 3: 15/72 = 5/24. The correct answer is B.
5 down, 6 to go — keep practicing.
Explore 2,500+ Free PSAT/NMSQT Math ProblemsQuestions 6–7 refer to the following information. A bag contains 15 balls, numbered 1 through 15.
What is the probability of selecting a number that is odd or a multiple of 5?
Count each group separately, then subtract the overlap. Odd numbers from 1 to 15: 1, 3, 5, 7, 9, 11, 13, 15 — that's 8 numbers. Multiples of 5 from 1 to 15: 5, 10, 15 — that's 3 numbers.
The overlap — numbers that are both odd and a multiple of 5 — is 5 and 15, which is 2 numbers. These would be double-counted if not subtracted.
Favorable outcomes: 8 + 3 − 2 = 9.
Probability: 9/15 = 3/5. The correct answer is B.
A ball is selected at random then replaced in the bag. A second selection is then made. What is the probability that the first number is a prime number and the second number is a multiple of 3?
Because the ball is replaced before the second draw, the two selections are independent — the outcome of the first pick doesn't affect the second, so their probabilities can be multiplied directly.
Primes from 1 to 15: 2, 3, 5, 7, 11, 13 — 6 numbers, so P(prime) = 6/15.
Multiples of 3 from 1 to 15: 3, 6, 9, 12, 15 — 5 numbers, so P(multiple of 3) = 5/15.
P(both) = (6/15) × (5/15) = 30/225 = 2/15. The correct answer is B.
Product p = s · t is formed from the two sets below, in which s is a number from set S and t is a number from set T. What is the probability that the product s · t will be a positive number?
There are 4 choices for s and 3 choices for t, giving 4 × 3 = 12 total possible pairs (and products). A product is positive only when both numbers share the same sign.
In S, there are 3 negative numbers (−5, −2, −1) and 1 positive number (4). In T, there is 1 negative number (−2) and 2 positive numbers (3, 7).
Negative × negative pairs: 3 × 1 = 3. Positive × positive pairs: 1 × 2 = 2. Total positive-product pairs: 3 + 2 = 5.
Probability: 5/12. The correct answer is B.
8 down, 3 to go.
Explore 2,500+ Free PSAT/NMSQT Math ProblemsQuestions 9–10 refer to the following information. Janis is making a flight reservation for her business trip. The travel agent informs that the probability that her flight to Phoenix will arrive on schedule is 90% and the probability that her flight from Phoenix to Atlanta will arrive on schedule is 80%.
What is the probability that both flights arrive on schedule?
The two flights arriving on schedule are independent events, so the probability that both occur is found by multiplying their individual probabilities.
P(both on schedule) = 0.90 × 0.80
= 0.72
The correct answer is B.
What is the probability that her flight to Phoenix is on schedule but her flight from Phoenix to Atlanta is not?
The probability that the Phoenix-to-Atlanta flight is not on schedule is the complement of 80%: 1 − 0.80 = 0.20.
Multiply the probability that the first flight is on schedule by the probability that the second flight is not:
0.90 × 0.20 = 0.18
The correct answer is B.
In a box of 12 headlamps, 3 are defective. If you choose two headlamps without replacement, what is the probability that both headlamps are defective?
Since the first headlamp is not replaced, the total count and the defective count both change for the second draw. Find the probability of the first headlamp being defective: 3/12.
If that happens, only 2 defective headlamps remain out of 11 total for the second draw: 2/11.
Multiply the two probabilities together: (3/12) × (2/11) = 6/132
Simplify by dividing both by 6: 6/132 = 1/22. The correct answer is B.
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