Conditional probability from a two-way table
For P(A | B), divide by the count in B, not by the grand total.
Worked example
Two-way table: passed (P) and attended lectures (A). What is P(P | A)?
| A | not A | |
|---|---|---|
| P | 10 | 23 |
| not P | 54 | 48 |
Worked solution
- 1.P(P | A) = number (P and A) / number A = 10 / (10 + 54) = 0.1563
Typical mistake: Dividing by the grand total instead of the total of the condition.
Your turn
Three problems of exactly this type, each time with new numbers. Comma or point; fractions like 1/6 work too.
Task 1 of 3Conditional probability
Task
Two-way table: passed (P) and attended lectures (A). What is P(P | A)?
| A | not A | |
|---|---|---|
| P | 16 | 58 |
| not P | 57 | 29 |
P (4 decimal places)
Comma or point; fractions like 1/6 work too.
Formulas for this topic
- P(A | B) = P(A ∩ B) / P(B)
- Total probability: P(B) = P(B | A) · P(A) + P(B | Ā) · P(Ā)
- Bayes: P(A | B) = P(B | A) · P(A) / P(B)
Typical mistake: Dividing by the grand total instead of the total of the condition.
Frequently asked questions
What is the typical mistake in "Conditional probability from a two-way table"?
Dividing by the grand total instead of the total of the condition.
Why is P(ill | positive) often small?
For rare diseases there are far more healthy people, so even a few false positives outnumber the true positives.