A doctor is called to see a sick child. The doctor knows (prior to the visit) that 90% of the sick children in that neighborhood are sick with the flu, denoted by F, while 10% are sick with the measles, denoted by M. A well known symptom of measles is a rash, denoted by R. The probability of having a rash for a child sick with the measles is 0.95. However, occasionally children with the flu also develop a rash, with conditional probability 0.08. Upon examination the child, the doctor finds a rash. Then what is the probability that the child has the measles?

Moderate

Solution

A: Doctor finds a rash
B₁: Child has measles
S: Sick children
$P (S / F) = 0.9$
B₂: Child has flu $\Rightarrow P (B_{2}) = 9 / 10$
$\begin{array}{l} P (S / M) = 0.10 \\ P (A / B_{1}) = 0.95 \\ P (R / M) = 0.95 \\ P (A / B_{2}) = 0.08 \\ P (R / F) = 0.08 \end{array}$
$\begin{array}{l} P (B_{1} / A) = \frac{0.1 \times 0.95}{0.1 \times 0.95 + 0.9 \times 0.08} \\ = \frac{0.095}{0.095 + 0.072} \\ = \frac{0.095}{0.167} = \frac{95}{167} \end{array}$

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