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?

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

Infinity Learn · Accepted Answer

A: Doctor finds a $$rashB1$$: Child has measlesS: Sick $$childrenP(S/F)=0.9B2$$: Child has flu  ⇒$$PB2=9/10P(S/M)=0.10PA/B1=0.95P(R/M)=0.95PA/B2=0.08P(R/F)=0.08PB1/A=0.1$$×$$0.950.1$$×$$0.95+0.9$$×$$0.08=0.0950.095+0.072=0.0950.167=95167$$

Detailed Solution

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