Question

research has shown that approximately 1 woman in 700 carries a mutation of a particular gene. about 45% 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. the probability that a randomly selected woman will carry the gene mutation and develop colon cancer is (round to four decimal places as needed.)

Explanation:

Step1: Calculate the probability of a woman having the gene mutation

The probability that a woman has the gene - mutation is $\frac{1}{700}\approx0.001429$.

Step2: Calculate the probability of a woman with the mutation developing colon cancer

The probability that a woman with the mutation develops colon cancer is $0.45$.

Step3: Calculate the joint - probability

We use the multiplication rule for independent events (in the context of conditional probability). The probability that a randomly selected woman has the gene mutation and develops colon cancer is the product of the probability of having the mutation and the probability of developing cancer given the mutation. So $P=\frac{1}{700}\times0.45$.
$P = 0.001429\times0.45=0.00064305\approx0.0006$

Answer:

$0.0006$