PROBLEM 4 The proportion of people in a given community disease is 0.005. A test

Question

PROBLEM 4 The proportion of people in a given community disease is 0.005. A test is available disease, the probability that the test will be positive (i.e. indicates the presence of the who have a certain to diagnose the disease. If a person has the a person does not have the disease, the probability the test will be disease) is 0.99. If positive is 0.01 a) If a person tests positive, what is the probability that the person a b) If 10 patients test positive, what is the probability that at least 2 of them the disease? actually have the diseasé?

Explanation / Answer

P(test positive) = 0.005 * 0.99 + (1 - 0.005 ) * 0.01 = 0.0149

A) P(the actual has the disease | person test positive)

= P(person test positive | the actual has the disease) * P(actual has the disease)/P(person test positive)

= 0.99 * 0.005/0.0149 = 0.332

B) P(X = x) = nCx * px * (1 - p )n- x

P(X > 2) = 1 - P(X < 2)

= 1 - (P(X = 0) + P(X = 1))

= 1 - (10C0 * (0.0149)^0 * (0.9851)^10 + 10C1 * (0.0149)^1 * (0.9851)^9)

= 1 - 0.9908 = 0.0092

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