Question

Research has shown that approximately 1 woman in 500 Carrie’s a mutation of particular gene. About 48% of women with this mutation develop colon cancer find the probability that a randomly selected woman will carry the mutation of this gene and will develop colon cancer

Answer 1

To find the probability that a randomly selected woman will carry the mutation of this gene and develop colon cancer, we need to multiply the probabilities of the two events: carrying the gene mutation and developing colon cancer.

Let's denote the probability of carrying the gene mutation as P(M) and the probability of developing colon cancer given the gene mutation as P(C|M). Given the information provided, P(M) can be calculated as 1 in 500, which is equivalent to 1/500.

The probability of developing colon cancer given the gene mutation is stated as 48%, which can be expressed as 0.48.

To find the probability of both events occurring, we multiply P(M) and P(C|M):

P(M and C) = P(M) * P(C|M) = (1/500) * 0.48

Therefore, the probability that a randomly selected woman will carry the mutation of this gene and develop colon cancer is (1/500) * 0.48, which can be calculated as a decimal or percentage based on your preference.