Let events A and B be independent. What is the probability , in terms of P(A) an

Question

Let events A and B be independent. What is the probability , in terms of P(A) and P(B), that exactly one of the events A and B occur?
a. P(A) + P(B) -2P(A)P(B)
b. P(A) + P(B) - P(A)P(B)
c. P(A) + P(B)
d. P(A)P(B)
e. min{P(A), P(B)}

I started this problem, by saying that the probability exactly one happens is P(A union B') + P(A' union B).
Then I used the rule of addition to say that this equals P(A) + P(B') - P(A intersect B') + P(A') + P(B) - P(A' intersect B)
But P(A) + P(A') = 1 and P(B) + P(B') = 1,
So that gave me 2 - P(A intersect B') - P(A' intersect B)
Then using independence this equals 2 -P(A)P(B') -P(A')P(B).
Using the compliment rule this equals 2 -P(A)[1-P(B)] -[1-P(A)]P(B).
Which simplifies to 2 -P(A) -P(B) +2P(A)P(B)
Clearly this isn't one of the choices. Where did I go wrong?

Explanation / Answer

P(A n B') + P(A' n B) =[P(A) - P(A n B)] + [P(B) - P(A n B)] =[P(A) - P(A)P(B)] + [P(B) - P(A)P(B)] (since A and B are independent) =P(A) +P(B) - 2P(A)P(B) So answer: a. P(A) + P(B) -2P(A)P(B)

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