Probability Question 79
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?
Options:
A) 91/165
B) 90/163
C) 82/161
D) 95/167
Show Answer
Answer:
Correct Answer: D
Solution:
A: Doctor finds a rash $ B_1: $ Child has measles S: Sick children $ P(S/F)=0.9 $
$ B_2: $ Child has flu $ \Rightarrow P(B_2)=9/10 $
$ 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 $
$ 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} $
BETA
