How a Positive Medical Test Can Still Mean You Are Probably Healthy

A factory's defective bolt could have come from any of three machines; a positive test could be a true or a false alarm. The theorem of total probability finds the chance of the effect, and Bayes' theorem turns the question around to ask which cause is most likely.

This part covers partitions and total probability, applying total probability, Bayes' theorem, and interpreting results in real problems.

**Events $E_1,E_2,\dots ,E_n$ partition a sample space if they are pairwise disjoint, together cover the whole space, and each has non-zero probability; then for any event $A$, the theorem of total probability gives $P(A)=\sum P(E_i)P(A|E_i)$.

Conditions for a partition:

- Pairwise disjoint** — $E_i\cap E_j=\varphi$ for $i\ne j$
- Exhaustive — $E_1\cup E_2\cup \cdots \cup E_n=S$
- Non-zero — $P(E_i)>0$ for every $i$

The theorem:

$$P(A)=P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+\cdots +P(E_n)P(A|E_n)$$

Why it holds. $A$ splits into the disjoint pieces $A\cap E_1,A\cap E_2,\dots$, and each piece has probability $P(E_i)P(A|E_i)$ by the multiplication theorem.

Worked example — spotting a partition. For a die, $\{1,2\}$, $\{3,4,5\}$ and $\{6\}$ form a partition; $\{1,2\}$ and $\{2,3\}$ cannot be part of one, because they overlap.

An everyday example. Grouping a school's students by house, with every student in exactly one house, is a partition — the chance that a random student plays chess can be worked out house by house and added.

The substance. **An event and its complement, $E$ and $E^{\prime }$, always form a partition** — the most common two-case use of the theorem.

Identify the possible causes that form a partition, write each cause's probability and the probability of the event under that cause, multiply in pairs, and add.

Worked example 1 — two bags. Bag I has $3$ red and $4$ black balls; Bag II has $5$ red and $6$ black. A bag is chosen at random and one ball is drawn. The probability that it is red:

$$P(R)=\frac{1}{2}\cdot \frac{3}{7}+\frac{1}{2}\cdot \frac{5}{11}=\frac{33+35}{154}=\frac{34}{77}$$

Worked example 2 — three machines. Machines A, B and C make $50$, $30$ and $20$ per cent of a factory's bolts, and $2$, $3$ and $4$ per cent of their bolts respectively are defective. The probability that a random bolt is defective:

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

Setting it out. A tree diagram, with one branch per cause and a second branch for the event, keeps the products organised.

An everyday example. The chance of reaching school late combines the chances of taking the bus, an auto or a bicycle with the chance of being late on each.

The substance. **The cause probabilities must add up to $1$** — if they do not, the causes do not form a partition.

**Bayes' theorem gives the probability of a particular cause $E_i$ after event $A$ has been observed, $P(E_i|A)=\frac{P(E_i)P(A|E_i)}{{\sum }_{j}P(E_j)P(A|E_j)}$, turning prior probabilities into posterior probabilities.

Derivation.** By the multiplication theorem $P(E_i\cap A)=P(E_i)P(A|E_i)$, and by total probability the denominator is $P(A)$, so

$$P(E_i|A)=\frac{P(E_i\cap A)}{P(A)}=\frac{P(E_i)P(A|E_i)}{{\sum }_{j=1}^{n}P(E_j)P(A|E_j)}$$

Terms:

- Prior — $P(E_i)$, the probability of a cause before the evidence
- Likelihood — $P(A|E_i)$, how probable the evidence is under that cause
- Posterior — $P(E_i|A)$, the updated probability after the evidence

Worked example — two bags. A red ball has been drawn. The probability it came from Bag I:

$$P(\text{I}|R)=\frac{\frac{1}{2}\cdot \frac{3}{7}}{\frac{34}{77}}=\frac{3}{14}\cdot \frac{77}{34}=\frac{33}{68}$$

Worked example — three machines. A bolt is found to be defective. The probability that machine B made it:

$$P(B|D)=\frac{0.3\times 0.03}{0.027}=\frac{0.009}{0.027}=\frac{1}{3}$$

An everyday example. A mechanic hearing a strange engine noise weighs how common each fault is against how likely each fault is to cause that noise — Bayes' theorem in practice.

The substance. **The posterior probabilities of all the causes add up to $1$**, which is a useful check on the arithmetic.

Translate the situation into causes with prior probabilities and an observed result with conditional probabilities, apply Bayes' theorem, and read the posterior as the updated belief — which can differ sharply from intuition when a cause is rare.

Worked example — a screening test. Suppose a disease affects $0.1$ per cent of a population. A test detects it in $99$ per cent of people who have it, but also wrongly shows positive for $1$ per cent of healthy people. A person tests positive. What is the probability that they have the disease?

$$P(D)=0.001,P(+|D)=0.99,P(D^{\prime })=0.999,P(+|D^{\prime })=0.01$$

$$P(D|+)=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.01}=\frac{0.00099}{0.01098}\approx 0.09$$

Interpreting it. Even after a positive result, the probability of disease is only about **$0.09$, because the disease is rare and false positives among the many healthy people outnumber true positives.

What changes the answer. A higher prior — for example, for someone who already has symptoms — or a second independent positive test raises the posterior sharply.

An everyday example. A smoke alarm beeping while toast browns in the kitchen is far more likely a false alarm than a fire — a rare cause stays unlikely even when the alarm sounds.

The substance. A test's accuracy alone does not give the chance you have the condition** — the prior probability matters just as much.