According to government data, 75% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women are randomly selected: a. What is the probability that exactly 2 of them have never been married? b. That at most 2 of them have never been married? c. That at least 13 of them have been married?

Answer:We are given that According to government data, 75% of employed women have never been married.So, Probability of success = 0.75So, Probability of failure = 1-0.75 = 0.25If 15 employed women are randomly selected: a. What is the probability that exactly 2 of them have never been married?We will use binomial Formula : [tex]P(X=r) =^nC_r p^r q^{n-r}[/tex]At x = 2 [tex]P(X=r) =^{15}C_2 (0.75)^2 (0.25^{15-2}[/tex] [tex]P(X=2) =\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{13}[/tex] [tex]P(X=2) =8.8009 \times 10^{-7}[/tex] b. That at most 2 of them have never been married?At most two means at x = 0 ,1 , 2So,  [tex]P(X=r) =^{15}C_0 (0.75)^0 (0.25^{15-0}+^{15}C_1 (0.75)^1 (0.25^{15-1}+^{15}C_2 (0.75)^2 (0.25^{15-2}[/tex]  [tex]P(X=r) =(0.75)^0 (0.25^{15-0}+15 (0.75)^1 (0.25^{15-1}+\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{15-2})[/tex] [tex]P(X=r) =9.9439 \times 10^{-6}[/tex]c. That at least 13 of them have been married?P(x=13)+P(x=14)+P(x=15) [tex]={15}C_{13}(0.75)^{13} (0.25^{15-13})+{15}C_{14} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})[/tex] [tex]=\frac{15!}{13!(15-13)!}(0.75)^{13} (0.25^{15-13})+\frac{15!}{14!(15-14)!} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})[/tex] [tex]=0.2360[/tex]