# Probability that exactly 2 events occur and probability that neither occurs

A, B and C are events such that: $Pr(A) = 0.4$

$Pr(B) = 0.7$

$Pr(C) = 0.3$

$Pr(A \cup B) = 0.8$

$Pr(B \cap C) = 0.2$

$Pr[C \cap (A \cup B)] = 0.2$

$Pr[B \cap (A \cup C)] = 0.4$

(a) Find the probability that exactly two of $A$, $B$ and $C$ occur

(b) Find the probability that none of $A$ ,$B$, $C$ occur.

I was thinking for (a) that we could somehow use the formula:

$[(A \cap B) \cap \bar C ]$ $\cup$ $[(A \cap C) \cap \bar B]$ $\cup$ $[(B \cap C) \cap \bar A]$

and for (b) using the formula

$( \bar A \cap \bar B \cap \bar C)$

but im not sure how.

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Hint: you have the information to compute $A \cap B$ from lines 1,2,4 and $B \cup C$ from lines 2,3,5-what are they? Keep working that way and you will get there.
I have worked out $A \cap B$ and $B \cup C$ to be 0.3 ad 0.8 respectfully but im not sure where to go from there? – Matt Nov 16 '12 at 4:10