Assume that the overall probability of contracting breast ca

Assume that the overall probability of contracting breast cancer in a 45 year old American woman is 0.1% on average (or one in a thousand). A typical diagnostic test is a mammo- graphic scane. Assume also that a mammograph scan reading is 80% sensitive on average and 95% specific on average. Here, “sensitive” means among patients with cancer, the prob- ability that the test is positive and “specific” means among patients without cancer, the probability that the test is negative.

a) What is P(T)? Use the law of total probability here and explain what the law is and how exactly you’re using it to solve this problem.

b)  if a woman is scanned and tests positive, what is the probability she has cancer?

c) Can you explain why it’s so low?

f) What is the ratio of P(C | T) ? What does this ratio mean? What does your P(C | TC)

answer suggest? Is it possible these scans aren’t such a terrible diagnostic tool after all?

Solution

P(T) means probability of test.

P(C)= Probability of contracting with breast cancer = 0.1% = 0.001

P(C^c ) = 0.99% = 0.0099

P(T|C) = 0.8 P(T|C^c) = 0.95

P(T) = `P(TnC) + P(TnC^c) = P(T|C) * P(C) + P(T|C^c) * P(C^c) = 0.95 * 0.001 + 0.8 * 0.0099 = 0.00095 + 0.00792

= 0.00887

b) P(Women has cancer | She is scaned and test positive) = P(C| T) = P(CnT) /P(T) =0.00095 / 0.00887 = 0.107102

c) The probability is low as probability of detecting cancer is very low

d) P(T) = `P(TnC) + P(TnC^c) = P(T,C) + P(T|C^c) * P(C^c) as P(TnC^c) = P(T|C^c) * P(C^c)