Question 932887: Let A and B be two events, with probabilities P(A) = 0,6, and P(B) =
0,2. Calculate the following probabilities: P(A' ), P(AUB), P(A intersection B), P(AIB),
P(BIA), if:
(1) A and B are mutually exclusive (disjoint).
(2) A and B are independent.
(B) If A, and B are two events, where P(A) = k, P(AUB) = 0,5, P(B) = 0,33:
(1) Find k, if A and B are independent.
(2) Find k, if A and B are mutually exclusive (disjoint).
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! P(A) = 0.6, and P(B) =0.2
(1) A and B are mutually exclusive (disjoint).
P(A' ) = .4
P(AUB) = .8
P(AᑎB) = 0
P(A|B) = .6
(2) A and B are independent.
P(A' ) = .4
P(AUB) = .8
P(AᑎB) = .4*.8
P(A|B) = .6
.........
P(A) = k, P(AUB) = 0.5, P(B) = 0.33
mutually exclusive (disjoint).
P(AUB) = P(A) + P(B)
.5 = k + .33
.17 = k
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