Question 1041391: In the general population, one woman in ten will develop breast cancer. Research has shown that 1 women in 550 carries a mutation of the BRCA gene. Nine out of 10 women with this mutation develop breast cancer.
A) Find the probability that a randomly selected women will develop breast cancer given that she has a mutation of the BRCA gene. Round to one decimal place as needed.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! If I read the question as it is written, we know that probability. Given that a woman has a mutation, she has a 90% chance of developing breast cancer. What we do't know is give that a woman has a breast cancer, what is the probability that she has a mutation?
Let P(Mutation)=B
P(Cancer)=A
p(B|A)=p(B)(P(A/B)/p(A)=(1/550)(9/10)/(1/10), and that is 9/550=0.02
Set up as a table
=============BRCA+======BRCA-====Total
CA+------------90--------5410--------- 5500
CA-------------10--------49490-------49500
Total--------100---------54900-------55000 (start with a total that will give integers and of that total 1/550 have BRCA+, or 100, and that goes as shown.)Then the CA+ and CA- can be filled in for BRCA, and the final two spaces can be filled.
To find P(BRCA+/CA+), given there is a cancer, what is the probability the woman is BRCA+, we find it is 90/5500=0.0164 or 0.02.
Tables are a good way to do these if one can't remember Bayes' Theorem. Note again that I changed the question. As it is written, the answer is known.
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