Question

There are 6 women and 9 men eligible to be in a committee of 5. Find the expected number of women on the committee given that at least one woman must be on the committee. Round the probabilities of the distribution to four decimal places or keep them as fractions. Round the answer to two decimal places.

Answer 1

P = 0.2517

Explanation:

In this case we must calculate the probability of event, which would be the number of specific events (that is, at least one woman and the rest men, 4), then it would be to choose 1 of 6 women by 4 of 9 men divided by the number of total events, which would be to choose 5 (committee size) out of 15 (9 men + 6 women, total number of people)

P (at least one woman) = 6C1 * 9C4 / 15C5

we know that nCr = n! / (r! * (n-r)!)

replacing we have:

6C1 = 6! / (1! * (6-1)!) = 6

9C4 = 9! / (4! * (9-4)!) = 126

15C5 = 15! / (5! * (15-5)!) = 3003

Therefore it would be:

P (at least one woman) = 6 * 126/3003

P = 0.2517

That is, approximately 1 out of 4 women.