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Math: Problem-Solving and Data Analysis, 3 original questions

SAT Probability Practice: Questions and Answers

Calculate simple and conditional probability and count independent outcomes in three original SAT-style questions, with explained fractions and distractor checks.

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The short answer

Probability compares favorable outcomes with the relevant set of possible outcomes when those outcomes are equally likely. A condition changes that relevant set: given that a person is a senior, use all seniors as the denominator rather than the whole table. For independent devices, list pairs or multiply outcome counts before counting favorable pairs. Keep overlap and independence assumptions explicit. Verify that a probability lies between zero and one, and simplify a fraction only after choosing the correct numerator and denominator.

Method at a glance

  1. 1Identify the event and condition
  2. 2Count within the correct group
  3. 3Build combined outcomes carefully
  4. 4Check the fraction

Try the 3 questionsRead the full method

Step-by-step method

Use these steps in order, then apply them to the questions below.

  1. Step 1: Identify the event and condition

    Underline what must happen and any given-that restriction. The condition defines the group from which the outcome is selected.

  2. Step 2: Count within the correct group

    Use favorable outcomes over all equally likely outcomes in that group. For a table, total the appropriate row or column first.

  3. Step 3: Build combined outcomes carefully

    For independent equally likely choices, multiply the individual outcome counts. List favorable ordered pairs without omitting or duplicating them.

  4. Step 4: Check the fraction

    The favorable count cannot exceed the eligible total. Confirm that a simplified fraction has the same value as the original ratio.

Try 3 SAT-style questions

These are original SAT-style questions written by PeakSAT, not official College Board questions. Choose an answer and check it to see why each choice is right or wrong, or open the worked answer.

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Question 1 of 3

A bag contains 5 red counters, 3 blue counters and 2 green counters. One counter is chosen at random, with each counter equally likely. What is the probability it is blue?

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Correct answer

Choice C: 3/10

Explanation

There are 5 + 3 + 2 = 10 counters in total and 3 favorable blue counters. The probability is 3/10. The denominator includes all eligible counters, not just the nonblue ones.

Why each choice is right or wrong

  1. Choice A

    7 is the number of nonblue counters, not the total.

  2. Choice B

    7/10 is the probability of choosing a nonblue counter.

  3. Choice C

    Correct answer

    Correct: 3 blue counters out of 10 total gives 3/10.

  4. Choice D

    This is greater than 1 and cannot be a probability.

Question 2 of 3

A school survey gives these counts: Club members: 18 seniors, 12 juniors Nonmembers: 22 seniors, 28 juniors A student is chosen at random from the seniors. What is the probability that the student is a club member?

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Correct answer

Choice A: 9/20

Explanation

The condition restricts selection to seniors. There are 18 + 22 = 40 seniors, of whom 18 are club members. The probability is 18/40 = 9/20. Juniors do not belong in this denominator.

Why each choice is right or wrong

  1. Choice A

    Correct answer

    Correct: 18 club-member seniors divided by 40 seniors equals 9/20.

  2. Choice B

    This uses all 80 surveyed students as the denominator.

  3. Choice C

    18/30 is the probability of being a senior given club membership, which reverses the condition.

  4. Choice D

    This fraction does not use the 18 favorable seniors and 40 eligible seniors.

Question 3 of 3

Two independent fair spinners are spun once. The first has labels 1, 2 and 3; the second has labels 1, 2, 3 and 4. Each label on its spinner is equally likely. What is the probability that the labels sum to 5?

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View worked answer for question 3

Correct answer

Choice D: 1/4

Explanation

There are 3 × 4 = 12 equally likely ordered pairs. The favorable pairs are (1, 4), (2, 3) and (3, 2), so there are 3 favorable outcomes. The probability is 3/12 = 1/4.

Why each choice is right or wrong

  1. Choice A

    This would require 4 favorable pairs; only 3 sum to 5.

  2. Choice B

    This would require 6 favorable pairs, twice the actual count.

  3. Choice C

    This adds the spinner outcome counts 3 + 4 instead of multiplying them.

  4. Choice D

    Correct answer

    Correct: 3 favorable pairs out of 12 equals 1/4.

Common mistakes

  • Using the entire table as the denominator after a condition is given. Restrict the denominator to the eligible group.

  • Assuming outcome totals are equally likely without a stated fair or random mechanism. Read the selection rules.

  • Adding the counts from two independent devices instead of multiplying. Combined outcomes are pairs, not isolated labels.

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Common questions

What does given that change in a probability question?

It restricts the possible outcomes. Count both favorable and total outcomes within that condition.

Can I multiply probabilities for every pair of events?

Not automatically. Independence permits multiplying the individual probabilities. Otherwise use the appropriate conditional probability.

Why can I not divide favorable labels by all labels across two spinners?

The combined outcome is a pair of labels. Count the possible pairs rather than adding the labels on the separate devices.

Sources

  1. College Board: Math section overviewsatsuite.collegeboard.org
  2. College Board: Student Question Bank Math skillssatsuite.collegeboard.org

Sources are linked for reference; they do not endorse PeakSAT. How we prepare our guides