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Bayes: flu, measles and rash

Short answer

By Bayes' theorem: P(Measles|Rash) = (0.95×0.10) / (0.95×0.10 + 0.08×0.90) = 0.095/0.167 ≈ 0.569. Probability is about 56.9%.

A doctor is called to see a sick child. The doctor has prior information that 90% of sick children in that neighborhood have the flu, while the other 10% are sick with measles. Let F stand for an event of a child being sick with flu and M stand for an event of a child being sick with measles. Assume for simplicity that there are no other maladies in that neighborhood. A well-known (and common) symptom of measles is a rash and has probability of 0.95. However, very occasionally, children with flu also develop rash and has probability of 0.08. Upon examining the child, the doctor finds a rash. What is the probability that the child has measles?

Given: P(F) = 0.90, P(M) = 0.10, P(R | M) = 0.95, P(R | F) = 0.08. We need P(M | R) by Bayes' theorem.

Bayes' theorem

P(M | R) = P(R | M)·P(M) / P(R)

Step 1: Total probability of a rash

Only flu and measles exist, so P(R) = P(R | M)·P(M) + P(R | F)·P(F) = (0.95 × 0.10) + (0.08 × 0.90) = 0.095 + 0.072 = 0.167

Step 2: Posterior

P(M | R) = 0.095 / 0.167 = 0.5689

Check

P(F | R) = 0.072 / 0.167 = 0.4311, and 0.5689 + 0.4311 = 1.

Answer: P(child has measles | rash) ≈ 0.569 (about 56.9%).

Note that measles was only 10% likely before the examination, but the rash raises it to over 50% because a rash is far more likely with measles (0.95) than with flu (0.08).

✓ Answer checked by Claude Opus 5.5 (high effort)

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