Question

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 a conditional probability of 0.08. Upon examination of the child, the doctor finds a rash. Then what is the probability that the child has measles?

• (A) $\frac{96}{169}$
• (B) $\frac{95}{167}$Your Answer
• (C) $\frac{98}{165}$
• (D) $\frac{92}{163}$

Answer 1

The Correct Option is B

Explanation:
$A$: Doctor finds a rash$B_1$: Child has measles$S$: Sick children$P(\frac{S}{F})=0.9$$B_2$: Child has flu$P(B2)=\frac{9}{10}$$P(\frac{S}{M})=0.10$$P(\frac{A}{B_1})=0.95$$P(\frac{R}{M})=0.95$$P(\frac{A}{B_2})=0.08$$P(\frac{R}{F})=0.08$$P(\frac{B_1}{A})=\frac{0.1\times 0.95}{0.1\times 0.95+0.9\times 0.08}$$=\frac{[0.095]}{[0.095+0.072]}$$=\frac{0.095}{0.167}$$=\frac{95}{167}$Hence, the correct option is (B).