Probability Finite Help!!

Question

Assume that there are 16 board members: 11 females, and 5 males including Neville. There are 3 tasks to be assigned. Note that assigning the same people different tasks constitutes a different assignment.

(1) Find the probability that both males and females are given a task.?

(2) Find the probability that Neville and at least one female are given tasks.?

Answer 1

1) You're looking for the probability of 2 females + 1 male or 2 males and 1 female.  Since there is no replacement (if 1 person gets picked, they can't get picked again) we use a hypogeometric distribution to solve.

 

P(2F + 1M) + P(2M + 1F) = [C(11,2)*C(5,1)/C(16,3)] + [C(11,1)*C(5,2)/C(16,3)] = .6875

 

2) P(Neville + 1F + 1M not Neville) * P(Neville + 2F + 0M)

= [C(1,1) * C(11,1)*C(4,1)/C(15,3)] * [C(1,1)*C(11,2)*C(4,0)/C(15,3)] = .218