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A doctor is called to see a sick child. The doctor has prior information that
95% of sick children in that neighborhood have the flu, while the other 5%
are sick with measles. Let F stand for an event of a child being sick with flu
and M stand for an event of a child being sick with measles.
A well-known symptom of measles is a rash (the event of having which is
denoted by R). P(R|M) = 0.93. However, occasionally children with flu also
develop rash, so that P(R|F) = 0.09.Upon examining the child, the doctor
finds a rash. What is the probability that the child has measles?
0.57
0.35
0.65
0.20

Respuesta :

The probability that the child has measles is gotten as; 0.35

How to use Baye's Theorem?

F is the event of a child being sick with flu.

M is the event of a child being sick with measles.

A is the event that the doctor finds a rash.

B1 is the event that the child has measles

S is the sick children.

P(R|M) = 0.93.

P(R|F) = 0.09

P(S|F) = 0.95

P(S|M) = 0.05

Thus, the probability that the child has measles is;

P(M|R) = [(0.05 * 0.93)/[(0.05 * 0.93) + (0.95 * 0.09)]

P(M|R) = 0.35

Read more about Baye's Theorem at; https://brainly.com/question/16038936

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