Question

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 1

We are given that According to government data, 75% of employed women have never been married.

So, Probability of success = 0.75

So, Probability of failure = 1-0.75 = 0.25

If 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 :
`P(X=r) =^nC_r p^r q^(n-r)`

At x = 2

`P(X=r) =^(15)C_2 (0.75)^2 (0.25^(15-2)`

`P(X=2) =(15!)/(2!(15-2)!) (0.75)^2 (0.25^(13)`

`P(X=2) =8.8009 * 10^(-7)`

b. That at most 2 of them have never been married?

At most two means at x = 0 ,1 , 2

So,
`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)`

`P(X=r) =(0.75)^0 (0.25^(15-0)+15 (0.75)^1 (0.25^(15-1)+(15!)/(2!(15-2)!) (0.75)^2 (0.25^(15-2))`

`P(X=r) =9.9439 * 10^(-6)`

c. That at least 13 of them have been married?

P(x=13)+P(x=14)+P(x=15)

`={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))`

`=(15!)/(13!(15-13)!)(0.75)^(13) (0.25^(15-13))+(15!)/(14!(15-14)!) (0.75)^(14)(0.25^(15-14)+{15}C_(15) (0.75)^(15) (0.25^(15-15))`

`=0.2360`