Question

Among women who have the BCRS1/2 gene, 51% of them contract breast or ovarian cancer by the age of 50. Imagine an experiment in which a 50 year old woman with this gene is chosen at random, and it is recorded whether or not she has breast or ovarian cancer. Let 5 women be selected in this manner, and let X = the number of women who have breast or ovarian cancer. What is P(X=1)?

Answer 1

`P(X = 1) = 0.1470`

Explanation:

There can only be two outcomes. Either a woman has breast or ovarian cancer, or she hasn't. So we can solve this problem by the binomial probability formula.

Binomial probability

Th binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinatios of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In this problem

We want to find P(X = 1), so
`x = 1`.

There are 5 woman, so
`n = 5`

There is a 51% probability that a woman with this gene has cancer, so
`p = 0.51`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 1) = C_(5,1).(0.51)^(1).(0.49)^(4) = 0.1470`