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A new medical test has been designed to detect the presence of the mysterious Brainlesserian disease. Among those who have the disease, the probability that the disease will be detected by the new test is 0.9. However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.06. It is estimated that 16 % of the population who take this test have the disease. If the test administered to an individual is positive, what is the probability that the person actually has the disease?

Answer :

MrsWilliams18
Let D denote the event that a particular person has the disease. P (D) = 0.16. Let N indicate the event that the person doesn't have the disease. 
Let M denote the event that the medical test turns positive. We need to compute P (D l M) . 

We know that P (D)  = 0.16, P (N) = 0.84. And, P (D l M) = 0.9, P ( D l N) = 0.06.

When we plug this information into the Bayes' Theorem equation, we get the following: 

[tex]P \frac{(D∩M)}{P(M)} [/tex]


= [tex] \frac{0.9x0.16}{0.9x0.16+0.06x0.84} [/tex]


= 0.74

This means that the probability the person actually has the disease is 0.74, or 74%. 

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