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)?

Respuesta :

Answer:

[tex]P(X = 1) = 0.1470[/tex]

Step-by-step 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.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem

We want to find P(X = 1), so [tex]x = 1[/tex].

There are 5 woman, so [tex]n = 5[/tex]

There is a 51% probability that a woman with this gene has cancer, so [tex]p = 0.51[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 1) = C_{5,1}.(0.51)^{1}.(0.49)^{4} = 0.1470[/tex]