A committee consists of eight women and twelve men. Three committee members will be chosen as officers. What is the probability that all the officers are women?

The probability that all the officers are women is

Explanation:

Given a committee consisting of 8 women and 12 men, from which 3 committee members are to be chosen as officers from a committee.  

The total number of committee members is 8 + 12 = 20 members

C(n,r) = which represents the number of combinations of taking r objects out of n.

The probability that all the officers are women = C(8,3) / C(20,3)

=

=

=

Therefore the probability that all the officers are women is

0.049 (approximately)

Explanation:

Solution:

Since there is 8 women and 12 men, from which 3 are chosen as officers.

Use C(n,r)=n! ÷ (r!(n-r)!) to represent number of combinations of taking r objects out of n.

Therefore;

Ways to choose 3 women out of 8 = C(8, 3)

Ways to choose from 20 members = C(20,3)

Probability of choosing 3 women out of 20

= C(8,3)÷ C(20,3)

Substituting our values for n=8 and r=3 we get

8! ÷ 3!(8-3)!

Substituting 8-3 with 5

8! / 3!×5!

Expanding factorial and cancelling out

8×7×6×5! ÷ 3!×5!

Multiple out

336÷6 = 56

Substituting our values for n=8 and k=3 we get

20! ÷ 3!(20-3)!

Substituting 20-3 with 17

20! ÷ 3!×17!

Expanding factorial and cancelling out

20x19x18x17! ÷ 3!×17!

Multiple out

6840÷6

= 1140

= 56/1140

= 3/17

=0.049 (approximately)