Question

According to government data, 44% 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

Solution :

Given :

P (never married) = P = 0.44

Sample size, n = 15

Let $X \sim \text{Binomial}$ (n = 15, P = 0.44)

P(X = x) = $^nC_xP^x(1-P)^{n-x}; x=0, 1,....n$

a). The probability that exactly two of them have been married is

P(X=2) = $^{15}C_2(0.44)^2(1-0.44)^{15-2}$

          $=\frac{15!}{2!(15-2)!}(0.44)^2(0.56)^{13}$

          $=\frac{15\times14}{2}\times (0.44)^2(0.56)^{13}$

        = 0.0108

b). That at most two of them have never been married.

$P(X \leq 2)= P(X=0)+P(X+1)+P(X=2)$

              $=^{15}C_0(0.44)^0(1-044)^{15-0}+^{15}C_1(0.44)^1(1-0.44)^{15-1}+^{15}C_2(0.44)^2(1-0.44)^{15-2}$

$=(1)(1)(0.56)^{15}+(15)(0.44)(0.56)^{14}+(105)(0.44)^2(0.56)^{13}$

= 0.000167+0.001969+0.010828

= 0.012964

= 0.0130

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

P(married) = 1 - P(never married)

                 = 1 - 0.44

                = 0.56

$P(X \geq 13)= P(X=13)+P(X=14)+P(X+15)$

                 $=^{15}C_{13}(0.56)^{13}(1-0.56)^{15-13}+^{15}C_{14}(0.56)^{14}(1-0.56)^{15-14}+^{15}C_{15}(0.56)^{15}(1-0.56)^{15-15}$$=(105)(0.56)^{13}(0.44)^{2}+(15)(0.56)^{14}(0.44)+(1)(0.56)^{15}(1)$

$=0.010828+0.001969+0.000167$

= 0.012964

=0.0130