Question

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?

Answer 1

Answer:

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)

Answer 2

Answer:

The probability that all the officers are women is $\frac{56}{1140}$

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

= $\frac{8!}{3!(8-3)!} / \frac{20!}{3!(20-3)!}$

=$\frac{8!}{3!5!} /\frac{20!}{3!17!}$

=$\frac{56}{1140}$

Therefore the probability that all the officers are women is $\frac{56}{1140}$