Question

Suppose that, of women who undergo routine screening, 1/101 have breast cancer. Of the women who undergo screening and do have breast cancer, 80% will have a positive test. Of the women who undergo screening but don't have breast cancer, only 5% have a positive test. A woman of this age undergoes routine screening and has a positive test. What is the probability that she has breast cancer

Answer 1

Answer:

The probability that she has breast cancer if she gives a positive test result is P=0,16.

Step-by-step explanation:

The information we have is:

P(C)=1/101≈0.01

P(H)=100/101≈0.99

P(P|C)=0.80 (probability of having a positive result given that the patient has cancer)

P(P|H)=0.05 (probability of having a positive result given that the patient hasn't cancer)

where the events are:

C: have cancer

H: not have cancer

P: positive test

We need to calculate P(C|P), the probability of having cancer given a positive test result.

According to the Bayes theorem, we have:

$P(C|P)=\frac{P(C)*P(P|C)}{P(C)*P(P|C)+P(H)*P(P|H)}\\\\P(C|P)=\frac{0.01*0.80}{0.01*0.8+0.99*0.05}=\frac{0.008}{0.008+0.050}=\frac{0.008}{0.050}= 0.16$

The probability that she has breast cancer if she gives a positive test result is P=0,16.