Answer :
Answer:
P(at most four) = 0.00960541
Step-by-step explanation:
For each employee hired, there are only two possible outcomes. Either it is a women, or it is not. The probability of an employee being a women is independent of any other employee, which means that the binomial probability distribution is used 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.
19 new employees.
This means that [tex]n = 19[/tex]
Approximately equal number of qualified men as qualified women.
This means that [tex]p = 0.5[/tex]
Probability of getting four or fewer women when 19 people are hired
This is:
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{19,0}.(0.5)^{0}.(0.5)^{19} = 0.00000191[/tex]
[tex]P(X = 1) = C_{19,1}.(0.5)^{1}.(0.5)^{18} = 0.00003624[/tex]
[tex]P(X = 2) = C_{19,2}.(0.5)^{2}.(0.5)^{17} = 0.00032616[/tex]
[tex]P(X = 3) = C_{19,3}.(0.5)^{3}.(0.5)^{16} = 0.00184822[/tex]
[tex]P(X = 4) = C_{19,4}.(0.5)^{4}.(0.5)^{15} = 0.00739288[/tex]
Then
[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.00000191 + 0.00003624 + 0.00032616 + 0.00184822 + 0.00739288 = 0.00960541[/tex]
So
P(at most four) = 0.00960541