1) Let A and B be two events. Suppose that P(A), P(B) and P(AB) are given. What is the probability that neither A nor B will occur?
I know the ANSWER is 1 - P(A) - P(B) + P(AB) but I'm not sure how to get it.
2) In a bridge game, each of the four players gets 13 random cards. What is the probability that every player has an ace?
I don't have the answer for this one, but am confused nonetheless on how to solve it.
Any input would be greatly appreciated.
October 6th 2005, 08:05 AM
Right I will give it a go at explaining the first one, though I am a bit rusty on probability. :)
1) If you have 2 events A and B there are 4 things that can happen. Just A. Just B. Both A and B. Or even neither! As one of these 4 things have to happen, they must add up to 1.
So P(A)+P(B)+P(A n B)+P(Neither) = 1
Rearranging gives : P(Neither) = 1 - P(A) - P(B) - P(A n B)