An Undergraduate Study Committee of 6 members at a major university is to be formed from a pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is the probability that precisely half of the members will be women?

The probability that precisely half of the members will be women is;

P(3 women) = 0.1213

This question will be solved by hypergeometric distribution which has the formula;

P(x) = [S_C_s × (N - S)_C_(n - s)]/(NC_n)

where;

S is success from population

s is success from sample

N is population size

n is sample size

We are give;

s = 3 women (which is precisely half of the members selected)

S = 6 women

N = 24 men and women

n = 6 people selected

Thus;

P(3 women) = (⁶C₃ * ⁽¹⁸⁾C₍₃₎)/(²⁴C₆)

P(3 women) = (20 * 816)/134596

P(3 women) = 0.1213

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5/33649= approx 0.00015

Step-by-step explanation:

Total number of outcomes are  C24 6= 24!/(24-6)!/6!=19*20*21*22*23*24/(2*3*4*5*6)= 19*14*22*23

Half of the Committee =3 persons. That mens that number of the women in Commettee=3.  3 women from 6 can be elected C6  3  ways ( outputs)=

6!/3!/3!=4*5*6*/2/3=20

So the probability that 3 members of the commettee are women  is

P(women=3)= 20/(19*14*22*23)=5/(77*19*23)=5/33649=approx 0.00015

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step-by-step explanation: