The same, as probabilities

P(D \cap +) = 0.02 \times 0.95 = 0.019

P(D' \cap +) = 0.98 \times 0.1 = 0.098

P(+) = 0.019 + 0.098 = 0.117

P(D \mid +) = \frac{0.019}{0.117} = 0.162

19 true positives and 98 false ones, out of 1000

Conditional probability

The test is positive. How worried should you be?

A test catches 95% of people with a disease. It seems obvious that a positive result means you almost certainly have it. Count people, and it isn't so.

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Two in a hundred

Here are 1000 people. The disease affects 2% of them: the 20 filled dots.

Everyone is tested

The test is positive for 95% of people with the disease: 19 of the 20, now ringed. It is also wrongly positive for 10% of the 980 without it: 98 more rings.

Given a positive

Now suppose all you know is that a person tested positive. Everyone else is ruled out. Only the 117 ringed people count, and 19 of them have the disease:

P(D \mid +) = \frac{19}{117} = 0.162

Why so low?

The disease is rare, so the healthy group is huge, and 10% of a huge group swamps 95% of a small one: 98 false positives to 19 true ones. The test's 95% is $P(+ \mid D)$ , not $P(D \mid +)$ .

Chapter 2

Given what you know, count only what's left.

Two-way tables, Venn diagrams, tree diagrams read both ways, independence, criticising probability models, and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›