SAT Math: Conditional Probability and Two-Way Tables Practice Questions
Identify the correct restricted sample space, separate joint from conditional probabilities, and check independence from counts. Try each original question before checking its worked explanation. This focused practice is free and requires no account.
Free original practice questions for study purposes. Open Exam Prep is an independent study resource and is not affiliated with, endorsed by, or sponsored by the makers of SAT.
Answer the questions below — you get instant feedback and a full explanation for each.
1. A club has 24 seniors and 16 juniors. Of the seniors, 15 arrive by bus and 9 walk. Of the juniors, 5 arrive by bus and 11 walk. If one bus rider is selected at random, what is the probability that the rider is a senior?
Explanation. Conditioning on bus riders restricts the sample space to 15 + 5 = 20 people. Fifteen are seniors, so the probability is 15/20 = 3/4. Dividing by all 40 club members would answer a different question.
2. A club has 24 seniors and 16 juniors. Nine seniors and 11 juniors walk; all other members take the bus. If one club member is selected at random, what is the probability that the member is a junior who walks?
Explanation. The event requires both being a junior and walking. There are 11 such members out of 24 + 16 = 40 total members, so the joint probability is 11/40. No condition restricts selection to walkers or juniors.
3. Of 60 respondents, 36 prefer tea and 24 prefer coffee. Eighteen tea-preferring respondents and 12 coffee-preferring respondents order online. In this data set, are beverage preference and ordering online independent?
Explanation. Among tea-preferring respondents, the online fraction is 18/36 = 1/2. Among coffee-preferring respondents it is 12/24 = 1/2, also equal to the overall online fraction 30/60. Conditioning on beverage preference does not change the online probability, so the events are independent in these data.
4. A training group has 50 participants, of whom 40 pass a test. Another group has 30 participants, of whom 12 pass. If a member of the training group is selected at random, what is the probability that the person did not pass?
Explanation. Within the training group, 50 - 40 = 10 participants did not pass. The condition fixes the denominator at 50, so the probability is 10/50 = 1/5. The other group is outside the restricted sample space.
5. A bag has 50 labeled tokens: 30 marked F and 20 marked M. Eighteen F tokens and 12 M tokens are blue. If a blue token is selected at random, what is the probability it is marked F?
Explanation. There are 18 + 12 = 30 blue tokens. Eighteen are marked F, giving conditional probability 18/30 = 3/5. The total of 50 tokens is not the denominator once selection is restricted to blue tokens.
6. In a group of 100 respondents, 60 like math and 50 like science. For a uniformly selected respondent, the events liking math and liking science are independent. How many respondents like both subjects?
Explanation. Independence gives P(math and science) = P(math)P(science) = (60/100)(50/100) = 0.30. Thus 30% of 100 respondents, or 30, like both. The stated independence assumption is needed to determine the overlap.
📘 Looking for more study material? Browse SAT prep books and study resources. Some links are affiliate links; we may earn a commission at no cost to you.
FAQ
What changes the denominator in conditional probability?
A condition restricts the selection to one group. Count all members of that group in the denominator, then count the favorable members within it.
How do I check independence?
Check whether conditioning on one event changes the probability of the other. Equivalently, the joint probability must equal the product of the individual probabilities.