# A doctor is called to see a sick child. The doctor has prior information that 80% of the sick children in that area have the flu, while the other 20% are sick with measles. Assume that there is no ot - Mathematics and Statistics

Sum

A doctor is called to see a sick child. The doctor has prior information that 80% of the sick children in that area have the flu, while the other 20% are sick with measles. Assume that there is no other disease in that area. A well-known symptom of measles is rash. From the past records, it is known that, chances of having rashes given that sick child is suffering from measles is 0.95. However occasionally children with flu also develop rash, whose chance are 0.08. Upon examining the child, the doctor finds a rash. What is the probability that child is suffering from measles?

#### Solution

Let E1 ≡ the event that child has flu

E2 ≡ the event that child has measles

It is given that E1, E2 are mutually exclusive and exhaustive.

Also P(E1) = 80% = 80/100 = 4/5,

P(E2) = 20% = 20/100 = 1/5

Let R ≡ the event that child has rash

Since chances of having rashes, if the child is suffering from measles is 0.95 and the chances of having rashes if the child has flue is 0.08, we have,

"P"("E"_2/"R") = 0.95 and "P"("R"/"E"_1) = 0.08

By Baye's Theorem, the required probability

= "P"("E"_2/"R")

= ("P"("E"_2)*"P"("R"/"E"_2))/("P"("E"_1)*"P"("R"/"E"_1) + "P"("E"_2)*"P"("R"/"E"_2))

= (1/5 xx 0.95)/(4/5 xx 0.08 + 1/5 xx 0.95)

= 0.95/(0.32 + 0.95)

= 95/127

Is there an error in this question or solution?