An insurance sales representative selects three policies to review.the group of policies he can select from contains eight life policies,five automobile policies and two homeowner's policies.find the probability of selecting:

Question

I.all life policies.
II.both homeowner's policies.
III.one of each policy.

(base on probability and combinatorial analysis )

Answer 1

prob = C(8,3)xC(5,0)xC(2,0)/C(15,3)
= 56/455
= 8/65

prob(both homeowners plus one of the others)
= C(2,2) x (13,1)/C(15,3) = 13/455 = 1/35

one of each:
C(8,1) x C(5,1) x C(2,1)/C(15,3)
= 80/455
= 19/91