A doctor is called to see a sick child. The doctor knows (prior to the visit) that 90% of the sick children in that neighbourhood 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?

A doctor is called to see a sick child. The doctor knows (prior to the visit) that 90% of the sick children in that neighbourhood 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? Correct Answer 95/167

Calculation:

⇒ A: Doctor finds a rash

⇒ B1 : Child has measles

⇒ S: Sick children

⇒ P(S/F) = 0.9

⇒ B2 : Child has flu

⇒ P (B2) = 9/10

⇒ P(S/M) = 0.10

⇒ P(A/B1) = 0.95

⇒ P(R/M) = 0.95

⇒ P (A/B2) = 0.08

⇒ P (R/F) = 0.08

⇒ P (B1/A) = (0.1 × 0.95)/ $\frac{0.1 \times 0.95}{0.1 \times 0.95 + 0.9 \times 0.08}$

⇒ /

⇒ 0.095/0.167

⇒ 95/167