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How do you compute probabilities and conditional probabilities, especially from a two-way frequency table?

Probability and conditional probability: compute simple probability as favorable over total, read probabilities from two-way frequency tables, and compute conditional probability given a restricted group.

A focused answer to the Digital SAT Problem-Solving and Data Analysis skill of probability and conditional probability: simple probability, reading two-way frequency tables, and computing conditional probability within a restricted row or column.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this skill is asking
  2. The core ideas
  3. A worked two-way table reading
  4. Why the denominator changes
  5. Reasonableness and the 0-to-1 scale
  6. Expected counts from a probability

What this skill is asking

Probability measures how likely an event is, on a scale from 00 (impossible) to 11 (certain). The Digital SAT (Problem-Solving and Data Analysis domain) tests simple probability (favorable over total), reading probabilities from a two-way frequency table, and conditional probability, where the question restricts attention to a particular row or column of the table before computing.

The core ideas

Probability questions hinge on choosing the correct denominator.

A worked two-way table reading

The denominator is everything.

Why the denominator changes

The difference between a simple and a conditional probability is which total sits in the denominator. A simple probability uses the whole group (the grand total). A conditional probability, signalled by "given that", "of the", or "among the", shrinks the group to a single row or column, and that subgroup's total becomes the new denominator. Reading the phrasing carefully to spot the restriction, then locating the correct subtotal in the table, prevents the most common error of dividing by the grand total when the question has restricted the population.

Reasonableness and the 0-to-1 scale

Every probability must land between 00 and 11, so a quick sanity check is that your fraction is in that range and matches intuition: a likely event is near 11, a rare one near 00. The complement rule, P(not A)=1P(A)P(\text{not } A) = 1 - P(A), is handy when "at least one" or "not" appears, since it is often easier to compute the opposite event and subtract. On two-way table questions, after computing, confirm that your numerator is a subset of your denominator (favorable outcomes are among the total you divided by); if the numerator exceeds the denominator, you have mixed up the groups.

Expected counts from a probability

The SAT also runs probability the other way: given a probability or a rate, predict a count for a larger group. If 310\frac{3}{10} of marbles are blue and a factory makes 40004000 marbles, you expect about 310×4000=1200\frac{3}{10} \times 4000 = 1200 blue ones; if a survey finds that 40%40\% of a representative sample recycle, you estimate 40%40\% of the whole population recycle. This is the same proportional reasoning used for ratios: a probability is a ratio of favorable to total, so scaling it up to a population multiplies the probability by the population size. These "how many would you expect" questions reward recognising that a probability or a survey percentage can be applied to a new, larger group, provided that group is similar to the one the probability came from. As always, check the result is reasonable: an expected count cannot exceed the size of the group it is drawn from.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Digital SAT Math (style)1 marksA bag has 5 red, 3 blue, and 2 green marbles. If one marble is drawn at random, what is the probability it is blue? (A) 15\tfrac{1}{5} (B) 310\tfrac{3}{10} (C) 12\tfrac{1}{2} (D) 37\tfrac{3}{7}
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The correct answer is (B), 310\frac{3}{10}.

Probability is favorable outcomes over total outcomes. There are 33 blue marbles out of 5+3+2=105 + 3 + 2 = 10 total, so P(blue)=310P(\text{blue}) = \frac{3}{10}.

Digital SAT Math (style)2 marksA survey of 200 students records sport (yes/no) by grade. Of the 120 juniors, 90 play a sport. If a junior is chosen at random, what is the probability they play a sport? (A) 90200\tfrac{90}{200} (B) 120200\tfrac{120}{200} (C) 90120\tfrac{90}{120} (D) 30120\tfrac{30}{120}
Show worked answer →

The correct answer is (C), 90120=34\frac{90}{120} = \frac{3}{4}.

This is a conditional probability: the condition "a junior is chosen" restricts the group to the 120120 juniors, so that becomes the denominator. Of those, 9090 play a sport, giving 90120\frac{90}{120}. Using the full 200200 (choice A) ignores the condition.

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