How to solve SAT probability and two-way table questions

Probability questions on the Digital SAT ask how likely it is that a randomly chosen person or item has some property. Most are built on a two-way table, and the hardest part is picking the right group to divide by. Conditional probability questions narrow that group with words like "given that," "of those who," or "among." Problem-Solving and Data Analysis is about 15% of the Math section, or 5 to 7 of the 44 questions. Those questions are split across seven skills, so expect about one probability question per test, sometimes two.

What the questions look like

Two-way table, whole group. A table splits a group by two categories, such as grade level and preferred format. You find the probability that a person chosen from the whole group falls in one cell, row, or column.

Two-way table, conditional. The question restricts the group first. "If a student who prefers online is chosen at random, what is the probability the student is in grade 11?" Only one column of the table matters.

Counts and expected values. You are told a probability or a rate and asked for an expected number, such as how many items out of 1,250 are expected to be defective if the probability of a defect is 0.04. That answer is $0.04 \times 1250 = 50$.

Step-by-step method

1. Find the group you are choosing from. This is the denominator. With no condition, it is the grand total. With "given that," "of those," or "among," it is the total of that row or column only.
2. Find the count that meets the condition inside that group. This is the numerator. It must be a part of the denominator group.
3. Divide, and match the form of the choices. Simplify the fraction or convert to a decimal or percent as needed.
4. For "or" questions, avoid double counting. $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$.
5. For independent events, multiply. If two events don't affect each other, $P(A \text{ and } B) = P(A) \cdot P(B)$.
6. Sanity check. Every probability is between 0 and 1, and a conditional probability uses a smaller denominator than the grand total.

Worked example

A school surveyed 200 students in grades 10 and 11 about whether they prefer reading online or in print. The results are shown.

(a) If one of the 200 students is chosen at random, what is the probability that the student prefers online?

No condition, so the denominator is the grand total:

$$P(\text{online}) = \frac{114}{200} = 0.57$$

(b) If a student who prefers online is chosen at random, what is the probability that the student is in grade 11?

The condition is "prefers online," so only the Online column counts. The denominator is 114, and the grade 11 students in that column number 66:

$$P(\text{grade 11} \mid \text{online}) = \frac{66}{114} = \frac{11}{19} \approx 0.579$$

(c) If a grade 11 student is chosen at random, what is the probability that the student prefers online?

Now the condition is "grade 11," so only the Grade 11 row counts. The denominator is 120:

$$P(\text{online} \mid \text{grade 11}) = \frac{66}{120} = \frac{11}{20} = 0.55$$

Parts (b) and (c) share the numerator 66 but have different answers, because they restrict the group in different ways.

(d) If one of the 200 students is chosen at random, what is the probability that the student is in grade 10 or prefers print?

Grade 10 has 80 students and Print has 86, but the 32 grade 10 students who prefer print are in both groups. Subtract them once:

$$P(\text{grade 10 or print}) = \frac{80 + 86 - 32}{200} = \frac{134}{200} = 0.67$$

Common traps

• Dividing by the grand total on a conditional question. In part (b), $\frac{66}{200} = 0.33$ is the probability of being both grade 11 and online. It is a likely wrong choice. When you see "given," "of those," or "among," the denominator shrinks.
• Reversing the condition. Parts (b) and (c) look alike but use different denominators. Say the condition out loud: "out of the students who prefer online" tells you to use the Online column total.
• Double counting in "or" questions. Adding 80 and 86 counts the 32 grade 10 print readers twice and gives $\frac{166}{200}$. Subtract the overlap.
• Using a row total when you need a column total. Label the row and column you are using before you divide. Totals sit next to each other in the table, and grabbing the wrong one is easy under time pressure.
• Treating table percents as counts. If a table already shows percents of each row, those percents are conditional on the row. Read the title and headers to see what each number means.