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Medical case histories indicate that different illnesses may produce identical symptoms. Suppose that a particular set of symptoms, denoted H , occurs only when any one of three illnesses, I1 , I2 , or I3, occurs. Assume that the simultaneous occurrence of more that one of these illnesses is impossible and that:
P(I1) = .01, P(I2) = .005, P(I3) = .02.
The probabilities of developing the set of symptoms H , given each of these illnesses, are known to be
P (H|I1) = .90, P (H|I2) = .95, P (H|I3) = .75
Assuming that an ill person exhibits the symptoms, H, what is the probability that the person has illness I1?

Answer :

Answer:

0.31

Step-by-step explanation:

Given an ill person exhibits  the symptoms H, find the probability that the person has illness I1.

So in other words, we need to find P (I1 | H).

We will make use of Baye's rule.

P (I1 | H) =[tex]\frac{P (H | I1)* P(I1)}{P(H)}[/tex]

P(H) can be calculated using total law of probability.

P(H) = P (H | I1)*P(I1) + P (H | I2)*P(I2) + P (H | I3)*P(I3)

       = (0.90)(0.01) + (0.95)*(0.05) + (0.75)*(0.02)

      = 0.009 + 0.00475 + 0.015 =0.02875

P (H | I1)*P(I1)  = (0.90)(0.01) = 0.009

so

P (I1 | H) =[tex]\frac{P (H | I1)* P(I1)}{P(H)}[/tex] = [tex]\frac{0.009}{0.02875}[/tex] = 0.31

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