These Digital SAT questions give you a two-way table, a frequency table or a short scenario and ask for a probability, a conditional probability, or a missing count. Almost every one comes down to choosing the right denominator.
Probability and conditional probability is one of seven skills in College Board’s Problem-Solving and Data Analysis domain, next to percentages, ratios and rates and four data skills. The domain is about 15% of SAT Math, or 5 to 7 questions, shared among all seven. College Board doesn’t publish a count for any single skill.
The Math section is two adaptive modules of 22 questions, 35 minutes each, with the Desmos calculator built in. About a quarter of the questions are student-produced response (SPR): you type the answer, with no choices to check against. College Board’s framework describes this skill as follows:
“Use one- and two-way tables, area models, and other representations to find relative frequency, probabilities, and conditional probabilities.”
The numbers can come from “a data display showing frequencies for a single variable, a two-way table, an area model, or a description of a situation.” And now and then the question runs in reverse: “Infrequently, given a probability, determine an unknown number.” Section 05 covers that. Skills Insight, College Board’s guide to score bands, ties these statements to Math scores:
Look at the gap: a probability from a two-way table is a 370–410 skill, a conditional one a 610–670 skill. The division is identical. Picking the right total is the hard part, and this guide drills it.
Every SAT probability question reduces to the same fraction: P(event) = favorable outcomes / total outcomes. A bag of 20 marbles with 8 red ones gives P(red) = 8/20 = 2/5. The only extra work on most SAT questions is figuring out which numbers count as “favorable” and which count as “total” — and a two-way table organizes exactly those counts for you.
Relative frequency is computed the same way: a category’s count over the total. Answers can be fractions, decimals or percents, so be ready to convert: 3/5 = 0.6 = 60%.
A survey of 250 students recorded whether each plays a sport and whether each plays an instrument:
| Plays instrument | No instrument | Total | |
|---|---|---|---|
| Plays a sport | 60 | 90 | 150 |
| No sport | 50 | 50 | 100 |
| Total | 110 | 140 | 250 |
Row and column totals are already favorable/total pairs waiting to be divided. Walk through pulling a conditional probability out of this table below.
Using the table above, what is P(plays a sport)?
A two-way table has three kinds of numbers: row totals and column totals (each the total for one category, ignoring the other), single cells (the count for both categories together), and the grand total (everyone). Before dividing anything, decide which of these four numbers the question is actually asking for.
Using the sports table, what value answers each question?
Using the sports table, what is P(no sport)?
Many tables arrive half-filled. Every row and column still has to add up, so each blank is one subtraction away. Here’s a survey of 200 adults with only four numbers given:
| Drinks coffee | No coffee | Total | |
|---|---|---|---|
| Morning person | 30 | ? | 80 |
| Night person | ? | ? | ? |
| Total | 110 | ? | 200 |
Here’s the finished table. Sections 04 and 07 use it too.
| Drinks coffee | No coffee | Total | |
|---|---|---|---|
| Morning person | 30 | 50 | 80 |
| Night person | 80 | 40 | 120 |
| Total | 110 | 90 | 200 |
P(A | B) means “the probability of A, given that B already happened.” In a two-way table, that means you divide the joint cell by the row or column total for B — never by the grand total. Restrict your attention to just B's row (or column) first, then find A's share of it.
The condition hides in several phrasings: given that, of the, among, chosen from those who. Each means “restrict to this group first.” If none appears and one person is picked from everyone, divide by the grand total.
Use the finished coffee table. Pick the denominator before you compute.
Using the sports table, what is P(plays an instrument | no sport)?
At a clinic, 30% of patients are under 18. Of those under 18, 40% got a flu shot, and so did 60% of patients 18 or older. A patient who got a shot is chosen at random. What is the probability that the patient is under 18? Turn the percents into counts with a round total.
An area model draws the same thing as a square split 0.3 and 0.7, then by shot rate. The pieces 0.3 × 0.4 = 0.12 and 0.7 × 0.6 = 0.42 are probabilities, and 0.12 / 0.54 = 2/9 again.
Sometimes the question gives the probability and asks for a count. If the total is known, multiply: with 480 students and P(junior) = 0.35, there are 0.35 × 480 = 168 juniors. If the unknown is part of the total, write the fraction and solve. Below, x seniors ride the bus. If a bus rider is chosen at random, the probability that the student is a senior is 0.6. What is x?
| Rides the bus | Doesn't ride | Total | |
|---|---|---|---|
| Junior | 18 | 22 | 40 |
| Senior | x | 30 | x + 30 |
| Total | x + 18 | 52 | x + 70 |
Desmos tip: graph y = x/(x + 18) and y = 0.6 and click the crossing: (27, 0.6). It’s a fast check on SPR questions, where no choices catch an algebra slip.
Use the bus table, but drop the 0.6. Suppose instead that if one student is chosen at random from everyone in the table, the probability that the student is a senior who doesn't ride the bus is 0.25. What is x?
For two independent events (a coin flip and a die roll — neither affects the other), multiply: P(A and B) = P(A) × P(B). For mutually exclusive “or” events (a single die roll can't show both a 2 and a 5), add: P(A or B) = P(A) + P(B). If the events can overlap, subtract the overlap once: P(A or B) = P(A) + P(B) − P(A and B).
The most common error is using the wrong operation — adding independent events instead of multiplying them, or forgetting to subtract the overlap for “or” events that can happen together.
The complement is often fastest: P(not A) = 1 − P(A). It’s also a favorite wrong choice, as several quizzes on this page show.
A fair coin is flipped and a fair six-sided die is rolled. What is P(the coin shows heads and the die shows a 5)?
Most wrong choices on probability questions come from the same few slips. Learn the list and you can often eliminate a choice on sight.
P(coffee | night) = 80/120, but P(night | coffee) = 80/110. The group after the bar supplies the denominator.P(night and coffee) picks from everyone: 80/200.Using the finished coffee table, one of the 200 adults is chosen at random. What is the probability that the person is a night person who drinks coffee?
A mix of everything above. Name the denominator before you reveal each answer.
A jar has 9 red and 6 blue marbles. What is P(red)?
Using the sports table, what is P(plays a sport and does not play an instrument)?
Using the sports table, what is P(plays a sport and plays an instrument)?
Using the sports table, what is P(plays an instrument | plays a sport)?
Using the coffee table, what is P(skips coffee | morning person)?
Using the coffee table, what is P(morning person | skips coffee)?
SPR: A bag holds 10 red tiles and some green tiles. If P(green) = 3/5, how many green tiles are there?
Two independent events have P(A) = 1/4 and P(B) = 2/5. What is P(A and B)?
Rolling a fair die once, what is P(rolling a 2 or a 5)?
The same “restrict, then divide” instinct carries over to sample-to-population estimates.
Here are three Probability questions from our problem sets, from easier to harder, in the same format as the Digital SAT. Answer each one to see the full explanation.
Questions 1–3 use this survey of 300 shoppers.
| Bought gloves | No gloves | Total | |
|---|---|---|---|
| Bought a jacket | 117 | 63 | 180 |
| No jacket | 21 | 99 | 120 |
| Total | 138 | 162 | 300 |
What is P(bought a jacket)?
What is P(bought gloves | bought a jacket)?
What is P(did not buy a jacket and did not buy gloves)?
Two independent events have P(A) = 3/8 and P(B) = 2/3. What is P(A and B)?
A fair spinner has 8 equal sections numbered 1 through 8. What is P(landing on a multiple of 3 or a multiple of 4)?
In a class survey, 6 students own both a dog and a cat. If a student who owns a cat is chosen at random, the probability that the student also owns a dog is 0.2. How many students in the class own a cat?
Related: Percentages · Two-variable data & scatterplots · Data inference & margin of error · Evaluating statistical claims · Distributions, center & spread · SAT Math overview