PSAT Data — two-way tables, relative frequency and conditional probability

Probability on the PSAT/NMSQT is mostly probability read from data. College Board's description of Problem-Solving and Data Analysis asks students to use various representations of data to find relative frequency, probabilities and conditional probabilities, and the representation that carries most of that work is the two-way table. Every question in this material prints its own full table, with row and column totals, so you practice on the object the test actually uses.

The twelve quiz questions move through the table one idea at a time. A simple probability uses the grand total. A conditional probability restricts to one row, another to one column, and a third, an age group given a preference for print, offers the reversed division as a trap. A joint probability asks for one cell out of everything, with the conditional version of the same cell among the options. An OR question makes you subtract the overlap, and the option that forgets to do so is larger than one, which is a useful alarm. Two questions compare groups: the flu rates of people with and without a shot, where each rate needs its own row total, and two factory shifts where the shift with fewer defective parts turns out to have the higher defect rate. One table gives proportions of all respondents rather than counts, and the rule does not change. The rest fill a missing cell from the totals, convert a row to percents, and scale a relative frequency up to predict a count in a batch of 2,000.

No question claims that one variable causes another; the tables describe, and the questions ask only what they show.

The flashcards cover the vocabulary and the three rules that matter: which total goes in the denominator, why the word "given" moves you into a row or a column, and the formula for "or".

The printable written work has eight open problems stated in sentences: the same gym data asked both ways round, an OR probability with the double count explained, a two-way table built from a paragraph of facts, two hospitals whose complication counts mislead, a conditional probability from percents of the whole, an expected number of failures, row percents compared, and the joint, conditional and reversed probabilities of one cell set side by side.

The oral exam asks one question at a time and wants you to say which total you are dividing by and why.

All tables are invented for practice, and nothing is taken from any published test.

  • Read counts, row totals, column totals and the grand total of a two-way table
  • Compute simple, joint and conditional probabilities from a table
  • Choose the denominator from the condition and avoid the reversed conditional
  • Compute P(A or B) by subtracting the overlap
  • Compare groups using rates (relative frequencies) rather than raw counts
  • Work with a table of proportions of the whole
  • Complete a table from partial information
  • Predict an expected count from a relative frequency

Practice material written by Zestly, based on the PSAT/NMSQT Math skill "Probability and conditional probability" as described in College Board's Fall 2026 PSAT/NMSQT Student Guide and the Assessment Framework for the Digital SAT Suite (version 3.01, August 2024). All tables are invented. Zestly is not affiliated with College Board.

Sample question

A survey of 300 adults asked for their main news source. | | Online | TV | Print | Total | |---|---|---|---|---| | Age 18-35 | 80 | 20 | 20 | 120 | | Age 36+ | 60 | 90 | 30 | 180 | | Total | 140 | 110 | 50 | 300 | If one of the 300 adults is chosen at random, what is the probability that the person prefers online news OR is age 36 or older?

See the answer

About 0.87

Add the two groups and subtract the overlap counted twice: $140 + 180 - 60 = 260$ adults, and $\frac{260}{300} \approx 0.87$. Equivalently, the only adults who are in neither group are the 18-35 year olds who prefer TV or print, $20 + 20 = 40$, and $1 - \frac{40}{300} \approx 0.87$. About 1.07 adds 140 and 180 without removing the 60 counted twice, and a probability can never exceed 1. The value 0.2 is the overlap alone, $\frac{60}{300}$. About 0.47 counts only the online readers.

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