Why a Positive Medical Test Does Not Always Mean You Are Ill

A factory has three machines, and a defective bolt turns up in a quality check. Which machine most likely made it? Questions like this run in reverse — from an observed result back to its cause — and the theorem of total probability together with Bayes' theorem answers them precisely.

This lesson covers total probability over a partition, Bayes' theorem, and applied problems such as diagnostic tests and defective items.

**If events $E_1,E_2,\dots ,E_n$ form a partition of the sample space — they are mutually exclusive and together cover every outcome — then for any event A, $P(A)=\sum P(E_i)P(A|E_i)$.

Why it works.** A splits into the pieces $A\cap E_1$, $A\cap E_2$ and so on, which do not overlap, and each piece has probability $P(E_i)P(A|E_i)$ by the multiplication theorem.

Worked example. Machines $E_1$, $E_2$ and $E_3$ make 50%, 30% and 20% of a factory's bolts, with defect rates of 2%, 3% and 4%. The probability that a randomly chosen bolt is defective is

$$P(D)=0.5\times 0.02+0.3\times 0.03+0.2\times 0.04=0.010+0.009+0.008=0.027$$

Worked example 2. Bag I has 3 red and 2 black balls; bag II has 2 red and 6 black balls. A fair coin decides which bag is used. Then

$$P(\text{red})=\frac{1}{2}\times \frac{3}{5}+\frac{1}{2}\times \frac{2}{8}=0.3+0.125=0.425$$

An everyday example. The chance that a parcel arrives late, when it may travel with one of three courier companies that have different delay rates, is found by total probability.

The substance. The cases must form a genuine partition — if two cases overlap, the same outcomes are counted twice and the total can exceed 1.

**Bayes' theorem gives the probability of a cause $E_i$ once the result A is known: $P(E_i|A)=\frac{P(E_i)P(A|E_i)}{\sum P(E_j)P(A|E_j)}$ — the share of the total probability of A that comes through $E_i$.

Terms.** $P(E_i)$ is the prior probability, known before the result; $P(E_i|A)$ is the posterior probability, updated after it.

Worked example. A bolt from the factory above is found to be defective. The chance that it came from each machine is

- $P(E_1|D)=\frac{0.010}{0.027}\approx 0.370$
- $P(E_2|D)=\frac{0.009}{0.027}\approx 0.333$
- $P(E_3|D)=\frac{0.008}{0.027}\approx 0.296$

The three posteriors add up to 1, as they must.

Worked example 2. In the two-bag problem, a red ball is drawn. The probability that it came from bag I is $\frac{0.3}{0.425}\approx 0.706$.

An everyday example. A teacher who sees an unusually perfect homework sheet and weighs whether it was copied is reasoning backwards from evidence, just as Bayes' theorem does.

The substance. **Machine $E_1$ has the lowest defect rate yet is the likeliest source** — because it makes the most bolts, the prior matters as much as the defect rate.

Set up the partition, such as ill or healthy, list each prior and each conditional probability of the observed result, find the total probability of that result, and divide the branch you want by that total.

Worked example (diagnostic test). A disease affects 1% of a population. A test detects it in 95% of people who have it, but also gives a positive result for 5% of healthy people.

- $P(+)=0.01\times 0.95+0.99\times 0.05=0.0095+0.0495=0.059$
- $P(\text{ill}|+)=\frac{0.0095}{0.059}\approx 0.161$

So only about 16 in every 100 people who test positive actually have the disease.

Worked example (defective items). A shop buys 60% of its bulbs from supplier A and 40% from supplier B, with defect rates of 3% and 5%. A bulb is found faulty.

- $P(F)=0.6\times 0.03+0.4\times 0.05=0.018+0.020=0.038$
- $P(B|F)=\frac{0.020}{0.038}\approx 0.526$ and $P(A|F)\approx 0.474$

An everyday example. A health screening camp in a village uses exactly this reasoning to explain why a positive screening result is followed by a confirmatory test.

The substance. When a condition is rare, most positive results are false positives — the few healthy people who wrongly test positive outnumber the ill, because the healthy group is so much larger.