Question

In the general population, 1% of the women between the ages of 40 and 50 suffer from breast cancer. A hospital testing for breast cancer has 90% accuracy when testing positive and 85% accurancy when testing negative. Answer the following questions based on the above information. Let's define the following events: + is the event that a random woman tests positive for breast cancer C is the event that a random woman has cancer.

Find the probability that a woman tests positive given that she has cancer, that is find P(+|C):

Find the probability a random woman tests positive, that is find P(+):

Find the probability that a woman has cancer given that the she has already tested positive, P(C|+):

Assume that a woman has 10%10% chance of having cancer if she tests positive first time. Find the probability that a woman has cancer who tested positive twice that is find P(C|++):

Answer 1

From given information we have

P(C) = 0.01

P(+|C) = 0.90

P(+' | C') = 0.85

By the complement rule, we have

P(C') = 1 - P(C) = 1 -0.01 = 0.99

P(+' | C) = 1 - P(+ | C) = 1 - 0.90 = 0.10

P(+ | C') = 1 - P(+' | C') = 1 - 0.85 = 0.15

The probability that a woman tests positive given that she has cancer is

P(+|C) = 0.90

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By the law of total probability, the probability a random woman tests positive is

P(+) = P(+ | C')P(C') +  P(+ | C)P(C) = 0.15 * 0.99 + 0.90 * 0.01 = 0.1575

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The probability that a woman has cancer given that the she has already tested positive is

P(C | +) = [ P(+ |C) P(C) ] / P(C) = [0.90 *0.01] / 0.01 = 0.90

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the only thing change now is

P(C) =P(C|+) = 0.10

Now

P(C | ++) = [ P(++ |C) P(C) ] / [P(++ |C) P(C)+P(++ |C') P(C')] = [0.90 *0.10] / [0.15 * 0.90 + 0.90 * 0.10 ] = 0.40