Question

City council consists of nine men and six women. Three representatives are chosen
at random to form a committee. What is the probability that:
a. Two women and a man are chosen?
b. Two women are chosen if Mr. Jim must be on the committee?

Answer 1

There are total 9 men and 6 women

The formula used for computing the combination is given by

_nC_r = $\frac {n!}{r!*(n-r)!}$

where ,

n = total number of items in the set

r = number of items to be selected from the set

Answer a)

Probability that 2 women & 1 man are randomly selected from a total of 15 people =

$\frac {6C2*9C1}{15C3} = \frac{\frac {6!*9!}{2!*(6-2)!*1!*(9-1)!}}{\frac {15!}{3!*(15-3!)}}$

$= \frac {27}{91}=0.2967$

Answer b)

In this case 1 person (i.e Mr Jim) is fixed in the committee of 3 for city council hence the total number of people in the selection criteria becomes 15 - 1 = 14.

Probability of randomly selecting 2 women from a total of 14 people =$\frac {6C2}{14C2}=\frac {\frac{6!}{2!*(6-2!)}}{\frac {14!}{2!*(14-2)!}}=\frac {15}{91}=0.1648$