BTGmoderatorDC wrote:Let S be a set of outcomes and let A and B be events with outcomes in S. Let ∼B denote the set of all outcomes in S that are not in B and let P(A) denote the probability that event A occurs. What is the value of P(A) ?
(1) P(A ⋃ B) = 0.7
(2) P(A ⋃∼B) = 0.9
OA C
Source: Official Guide
Let's take each statement one by one.
(1) P(A ⋃ B) = 0.7
P(A ⋃ B) = P(A) + P(B) - P(A & B) = 0.7. Can't get the value of P(A). Insufficient.
(2) P(A ⋃∼B) = 0.9
P(A ⋃ ~B) = P(A) + P(~B) - P(A & ~B) = 0.9. Can't get the value of P(A). Insufficient.
(1) and (2) together
From P(A ⋃∼B) = 0.9, we can deduce that only P(B) = 1 - 0.9 = 0.1. Note that the sum of probabilities of all events = 1.
From P(A ⋃ B) = 0.7, we can rewrite P(A ⋃ B) as P(A ⋃ B) = P(A) + Only P(B)
=> 0.7 = P(A) + 0.1 => P(A) = 0.7 - 0.1 = 0.6. Sufficient.
The correct answer:
C
Hope this helps!
-Jay
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