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Nick Roz · Answer 1 · 2021-07-14T02:51:05+0000

Answer:

P(X = 0) = C_(5,0).(0.51)^(0).(0.49)^(5) = 0.0345

P(X = 1) = C_(5,1).(0.51)^(1).(0.49)^(4) = 0.1657

P(X = 2) = C_(5,2).(0.51)^(2).(0.49)^(3) = 0.3185

P(X = 3) = C_(5,3).(0.51)^(3).(0.49)^(2) = 0.3060

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

P(X = 5) = C_(5,5).(0.51)^(5).(0.49)^(0) = 0.0282

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 distribution

The 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 combinations 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.

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
p = 0.51, n = 5

What is P(X)?