MATH SOLVE

4 months ago

Q:
# 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)?

Accepted Solution

A:

Answer:[tex]P(X = 0) = C_{5,0}.(0.51)^{0}.(0.49)^{5} = 0.0345[/tex][tex]P(X = 1) = C_{5,1}.(0.51)^{1}.(0.49)^{4} = 0.1657[/tex][tex]P(X = 2) = C_{5,2}.(0.51)^{2}.(0.49)^{3} = 0.3185[/tex][tex]P(X = 3) = C_{5,3}.(0.51)^{3}.(0.49)^{2} = 0.3060[/tex][tex]P(X = 4) = C_{5,4}.(0.51)^{4}.(0.49)^{1} = 0.1470[/tex][tex]P(X = 5) = C_{5,5}.(0.51)^{5}.(0.49)^{0} = 0.0282[/tex]Step-by-step explanation:We want the probability mass function of X.For each women with the gene selected, there are only two possible outcomes. Either they have breast or ovarian cancer, or they do not. The women are chosen at random, which means that the probability of any of them having breast or ovarian cancer is independent from other women. So we use the binomial probability distribution to solve this problem.Binomial probability distributionThe 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 combinations 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.51% of them contract breast or ovarian cancer by the age of 50. Let 5 women be selected in this manner, and let X = the number of women who have breast or ovarian cancer. This means that [tex]p = 0.51, n = 5[/tex]What is P(X)?[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex][tex]P(X = 0) = C_{5,0}.(0.51)^{0}.(0.49)^{5} = 0.0345[/tex][tex]P(X = 1) = C_{5,1}.(0.51)^{1}.(0.49)^{4} = 0.1657[/tex][tex]P(X = 2) = C_{5,2}.(0.51)^{2}.(0.49)^{3} = 0.3185[/tex][tex]P(X = 3) = C_{5,3}.(0.51)^{3}.(0.49)^{2} = 0.3060[/tex][tex]P(X = 4) = C_{5,4}.(0.51)^{4}.(0.49)^{1} = 0.1470[/tex][tex]P(X = 5) = C_{5,5}.(0.51)^{5}.(0.49)^{0} = 0.0282[/tex]