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.
Best Answer
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.