Probability and conditional probability

Official Digital SAT skill — Problem-Solving and Data Analysis domain.

What this question tests

Probability and conditional probability questions on the Digital SAT test whether you can translate a real-world description into counts and then turn those counts into a fraction. You’ll often see two-way tables (like seniors vs. drama club) or short scenarios describing overlapping groups, and the key move is choosing the correct denominator for the probability you’re asked to find. Sometimes you compute a basic probability, like the chance a randomly chosen student is in a club, which uses the entire population as the denominator. Other times you compute a conditional probability, which means you restrict attention to a subgroup first (the condition), and then compute the chance of another event within that subgroup. This skill is tested because it reveals whether you can reason precisely about “of all” versus “of those who,” which is a core data literacy habit in statistics and everyday decision-making.

What to know

A probability is a ratio of favorable outcomes to total outcomes, so write it as $P(A)=\frac{\text{number of outcomes in }A}{\text{total number of outcomes}}$ when all outcomes are equally likely.
Conditional probability means the sample space is restricted to the condition first, so use $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ or, in count form, $P(A\mid B)=\frac{\text{count of }A\text{ and }B}{\text{count of }B}$.
The phrase “given B” tells you the denominator must be the total number of cases satisfying $B$, because you are only considering that subgroup.
The intersection $A\cap B$ means both events happen, which corresponds to the overlap cell in a two-way table if you have one.
A two-way table can be navigated by rows and columns: pick the row/column for the condition to find the conditional denominator, then use the overlapping cell for the numerator.
Probabilities are numbers between 0 and 1, and if a fraction seems too large or too small, re-check whether you used the correct subgroup as the denominator.

How to approach it

First, identify the exact event being asked for and the condition, because the wording determines which counts belong in the numerator and which group sets the denominator.
Next, locate the condition group in the table (or in the scenario) and treat that group as your entire universe, because conditional probability is “within this subgroup.”
Then find the overlap between the event and the condition, because that overlap count is the favorable outcomes inside the restricted universe.
After that, write the probability as a fraction with the overlap count on top and the condition-group count on the bottom, because this directly matches $P(A\mid B)$.
If the question is not conditional, use the total population as the denominator, because basic probability is about the whole sample space.
Finally, do a quick sanity check by estimating whether your answer is plausible from the table proportions, because this catches swapped denominators and intersection mistakes.

Common traps

Wrong denominator (using the whole table instead of the condition subgroup) happens because students see “randomly chosen” and default to the total, so avoid it by explicitly asking “of which group?” before writing the fraction.
Swapping condition and event happens because both are mentioned close together, so avoid it by translating “given X” into “denominator is X” before you compute.
Confusing $P(A\cap B)$ with $P(A\mid B)$ happens because both use the overlap, so avoid it by remembering the conditional needs the condition’s total as the denominator, not the grand total.
Using a row total when the condition is a column (or vice versa) happens because table structure is easy to skim incorrectly, so avoid it by tracing the row/column labels with your finger and confirming the subgroup count.
Picking a distractor that looks familiar (like a reduced fraction from another part of the table) happens under time pressure, so avoid it by re-deriving the numerator and denominator from the wording rather than matching patterns.

Tips & shortcuts

Circle or underline the word “given” and write the denominator next to it before you look at answer choices.
In a two-way table, compute the condition-group total first, then find the overlap, because that order reduces denominator errors.
If two choices share the same numerator, compare denominators: the conditional one must match the condition-group size, not the grand total.

Worked example

In a group of $200$ students, $120$ are enrolled in math, $90$ are enrolled in science, and $50$ are enrolled in both math and science. If one student is chosen at random, what is the probability that the student is enrolled in science, given that the student is not enrolled in math?

A. $\frac{50}{90}$ ✓ (correct answer)
B. $\frac{1}{2}$
C. $\frac{40}{200}$
D. $\frac{40}{120}$

Why: To find $P(\text{science}\mid \text{not math})$, first determine the conditioned group. The number not enrolled in math is $200-120=80$. The number enrolled in science but not math is $90-50=40$ (subtract the $50$ who take both). Therefore, $P(\text{science}\mid \text{not math})=\frac{40}{80}=\frac{1}{2}$. The correct choice is $\frac{1}{2}$.

Use the Practice Questions for this skill to drill it, then attempt a Timed Practice Test.