Statistics 3.2--Probability

Question

Research has shown that approximately 1 woman in 500 carries a mutation of a particular gene. About 67% of women with this mutation develop esophageal cancer. Find the probability that a randomly selected woman will carry the mutation of this gene and will develop esophageal cancer.

Answer 1

Hi Riley,

You have two events of interest here. First, the woman has the genetic mutation. Second, she develops esophageal cancer. Designate:

A= Woman has genetic mutation.

B= Woman develops esophageal cancer.

Now, we are interested in the probability that any randomly selected woman will both have the mutation and develop the cancer. Formula for this is:

P(B given A) = P(A and B)/P(A)

Now, all we need to do is identify the relevant probabilities and solve for the one we want, which is P(A and B), the woman both having the gene and developing esophageal cancer.

P(B given A) = 0.67 (probability that woman develops cancer given she has the genetic mutation)

P(A) = 1/500 (probability that woman will have the gene)

Thus:

0.67 = P(A and B)/(1/500)

Multiply both sides by (1/500) to isolate P(A and B)

P(A and B) = 0.67/500

P(A and B) = 0.00134

I hope this helps.