Question

research has shown that approximately 1 woman in 400 carries a mutation of a particular gene. about 63% of women with this mutation develop skin cancer. find the probability that a randomly selected woman will carry the mutation of this gene and will develop skin cancer. the probability that a randomly selected woman will carry the gene mutation and develop skin cancer is (round to four decimal places as needed.)

Explanation:

Step1: Find the probability of carrying the mutation

The probability that a woman carries the mutation is \(P(\text{mutation})=\frac{1}{400}= 0.0025\)

Step2: Use the conditional - probability formula

The formula for conditional probability is \(P(A\cap B)=P(A)\times P(B|A)\). Let \(A\) be the event that a woman has the mutation and \(B\) be the event that a woman develops skin cancer. We know that \(P(A)=\frac{1}{400}\) and \(P(B|A) = 0.63\)

Then \(P(A\cap B)=P(A)\times P(B|A)\)

Substitute \(P(A)=0.0025\) and \(P(B|A) = 0.63\) into the formula:

\(P(A\cap B)=0.0025\times0.63\)

Step3: Calculate the product

\(0.0025\times0.63=(2.5\times10^{- 3})\times(6.3\times10^{-1})\)

\(=(2.5\times6.3)\times10^{-3 - 1}\)

\(=15.75\times10^{-4}=0.001575\approx0.0016\)

Answer:

\(0.0016\)