Question 1101679
since the pool is quite large and the probability of hiring a qualified woman is the same as hiring a qualified man, then:


p = .5
q = .5
n = 22
x = 0 to 3


this is a binomial probability type problem.


the standard form is p(x) = p^x * q^(n-1) * c(n,x)


for x = 0 to 3, the formula becomes:


p(0) = .5^0 * .5^22 * c(22,0)
p(1) = .5^1 * .5^21 * c(22,1)
p(2) = .5^2 * .5^20 * c(22,2)
p(3) = .5^3 * .5^19 * c(22,3)


results are:


sum of p(x) for x equals 0 to 3 is equal to 4.27723 * 10^-4 rounded to 9 decimal places.


the decimal equivalent to this is .000427723 rounded to 9 decimal places.


rounded to 8 decimal places, this is equal to .00042772.


c(n,x) is the number of ways you can draw a set of x elements out of a set of n elements.


the formula is c(n,x) = n! / (x! * (n-x)!)


for example:


c(22,3) is equal to 22! / 3! * 19!)
this is equal to (22 * 21 * 20 * 19!) / (3! * 19!)
the 19! in the numerator and denominator cancel out to get:
(22 * 21 * 20) / (3!)
since 3! is equal to 3*2*1 = 6, this then becomes equal to:
(22 * 21 * 20) / 6
the end result is that c(22,3) is equal to 1540
p(3) is therefore equal to .5^3 * .5^19 * 1540 which is equal to 0.000367165 rounded to 9 decimal places, as shown in the attached excel spreadsheet printout.


<img src = "http://theo.x10hosting.com/2017/111601.jpg" alt="$$$">
<img src = "http://theo.x10hosting.com/2017/111602.jpg" alt="$$$">


the probability of getting less than or equal to 3 women is exceedingly small.
it is much smaller than .5 which one would expect, given that women are equally as qualified as men.