Problem 3 If a person has cancer, he/she might have 95% chance of testing positi

Question

Problem 3

If a person has cancer, he/she might have 95% chance of testing positive for cancer. In this case, the probability of A = {test is positive} given that B = {having cancer} occurred is 95%, i.e,. P(A|B) = 95%. In addition, if a person does not have cancer, he/she might have 10% chance of testing positive, i.e., P(A|Bc) = 10%. Suppose, we know that the population proportion of having cancer is 0.04%, P(B) = 0.04%.

1. What’s the probability of a person having cancer and testing positive? Compute P(A and B).

2. What’s the probability of a person not having cancer but testing positive? Compute P(A and Bc).

3. If we know a person’s test is positive, what’s the probability that he/she indeed has cancer? Compute P(B|A).

Hint: P(B|A) = P(A and B) / [P(A and B) + P(A and Bc)]

Explanation / Answer

1)probability of a person having cancer and testing positive =P(B)*P(A|B)=0.0004*0.95=0.00038

2) probability of a person not having cancer but testing positive =P(Bc)*P(A|Bc) =0.9996*0.10=0.09996

3) probability that he/she indeed has cancer given tested positive =0.00038/(0.00038+0.09996) =0.003787

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