Answer:
Let's assume that the probability of having a male or female member is 0.5 or 50%.
There are different ways to approach this problem, but one possible method is to use the binomial distribution.
First, let's calculate the probability of having exactly 3 females, which is the midpoint between 0 and 7 females.
P(3 females) = (7 choose 3) * 0.5^7 = 0.2734
This means that there is a 27.34% chance of having exactly 3 females in the committee.
Next, let's calculate the probability of having 4, 5, 6, or 7 females, which would satisfy the condition of having more females than males.
P(4 or more females) = P(4 females) + P(5 females) + P(6 females) + P(7 females)
P(4 females) = (7 choose 4) * 0.5^7 = 0.2734
P(5 females) = (7 choose 5) * 0.5^7 = 0.1641
P(6 females) = (7 choose 6) * 0.5^7 = 0.0547
P(7 females) = (7 choose 7) * 0.5^7 = 0.0078
Therefore,
P(4 or more females) = 0.2734 + 0.1641 + 0.0547 + 0.0078 = 0.5
This means that there is a 50% chance of having more females than males in the committee, given that the probability of having a male or female member is equal.