Answer:
0.000001 = 0.0001% probability that exactly 2 of them have never been married
Step-by-step explanation:
For each employed women, there are only two possible outcomes. Either they have already been married, or they have not. The women are chosen at random, which means that the probability of a woman having been already married is independent from other women. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The 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.
74% of employed women have never been married.
This means that [tex]p = 0.74[/tex]
15 employed women are randomly selected
This means that [tex]n = 15[/tex]
a. What is the probability that exactly 2 of them have never been married
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{15,2}.(0.74)^{2}.(0.26)^{13} = 0.000001[/tex]
0.000001 = 0.0001% probability that exactly 2 of them have never been married