Final answer:
The probability that a randomly chosen committee of 5 from a group of 3 men and 4 women has more women than men is 16/35, which is option B.
Step-by-step explanation:
To determine the probability that there are more women than men in a committee of 5 people, randomly chosen from a group of 3 men and 4 women, we must first understand that this is a hypergeometric distribution problem since it involves sampling without replacement from two groups.
We are interested in the cases where there are more women than men on the committee. This can happen in two scenarios: either 3 women and 2 men are selected, or all 4 women and 1 man are selected.
The total number of ways to select a committee of 5 from the 7 people is the combination C(7,5).
The number of ways to select 3 women out of 4 and 2 men out of 3 is C(4,3) x C(3,2).
The number of ways to select all 4 women and 1 man is C(4,4) x C(3,1).
The probability of having more women than men on the committee is the sum of the probabilities for each favorable scenario divided by the total number of ways to form the committee:
P(more women than men) = (C(4,3) x C(3,2) + C(4,4) x C(3,1)) / C(7,5).
Calculating the values, we get:
P(more women than men) = (4 x 3 + 1 x 3) / 21 = 15 / 21 = 5 / 7 = 16/35