Prob and StatsProbability RulesS-CP.A.4

Conditional Probability from Two-Way Tables

Divide by the row or column total that matches the condition.

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In Algebra 1 you read two-way tables with row and column percents. Those percents are conditional probabilities: the row or column you restrict to is the condition.

Key idea

In a two-way table, divide the cell by the total of the condition: $$P(\text{row} \mid \text{column}) = \dfrac{\text{cell}}{\text{column total}}, \qquad P(\text{column} \mid \text{row}) = \dfrac{\text{cell}}{\text{row total}}.$$ With no condition, divide by the grand total.

Why it works

A condition picks out one row or one column of the table: those are the only individuals we are now choosing from. The cell where the row and column meet holds the ones that satisfy both. So the denominator is always the total of whatever you were "given."
200 students who took a certification exam.
Took review classDid notTotal
Passed4590135
Did not pass56065
Total50150200

Worked example: Given the column

Find the probability that a student passed, given that the student took the review class.
  1. The condition is "took the review class": use that column, 50 students.
  2. Of those, 45 passed.
  3. $P(\text{passed} \mid \text{review}) = \dfrac{45}{50} = 0.9$.

Answer: $0.9$

Worked example: Given the row

Find the probability that a student took the review class, given that the student passed.
  1. The condition is "passed": use that row, 135 students.
  2. Of those, 45 took the class.
  3. $P(\text{review} \mid \text{passed}) = \dfrac{45}{135} = \dfrac{1}{3}$.

Answer: $\frac{1}{3}$

Worked example: No condition

Find the probability that a randomly chosen student took the review class and passed.
  1. No "given," so all 200 students are in play.
  2. $P = \dfrac{45}{200} = 0.225$.

Answer: $0.225$

Common mistake

Dividing by the grand total when the question gives a condition. "Of the students who took the review class, what fraction passed?" has 50 in the denominator, not 200. Check: find the word "given" (or "of those who," "among") and divide by that group's total.
Comparing $P(\text{passed} \mid \text{review}) = 0.9$ with $$P(\text{passed} \mid \text{no review}) = 90/150 = 0.6$$ shows an association between the class and passing. (This is observational data, so it does not prove the class causes passing.)

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Sources and further reading

  • OpenStax, Introductory Statistics 2e, section 3.4 Contingency Tables
  • Common Core State Standards for Mathematics, S-CP.A.4 (Construct and interpret two-way frequency tables)
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