Math miscellaneousHard
Question
If A & B are two exhaustive events such that P(A ∪ 
) = 2P( ∩ B) = P(B), then-If A & B are two exhaustive events such that P(A ∪ 
) = 2P( ∩ B) = P(B), then-
) = 2P( ∩ B) = P(B), then-If A & B are two exhaustive events such that P(A ∪ ) = 2P( ∩ B) = P(B), then-Options
A.P(A) = 
B.P(A) = 
C.
D.P(A ∪ (
∩ B)) = 1
∩ B)) = 1Solution
∵ P(A ∪ 
) = P(B)
⇒ P(A) + P() - P(A ∩ ) = P(B)
⇒ P(A) + 1 - P(B) - P(A) + P(A ∩ B) = P(B)
⇒ 1 - P(B) + P (A ∩ B) = P(B) ......(1)
A and B are exhaustive events than
P(A) + P(B) - P(A ∩ B) = 1 ...(ii)
from (i) & (ii) P(A) = P(B)
2 [P(B) - P(A ∩ B)] = P (B)
⇒ P(A) =
⇒
P(A ∪ (
∩B)) = P(A) + P(B) - P(A ∩ B) = 1
) = P(B)⇒ P(A) + P(
) - P(A ∩ ) = P(B) ⇒ P(A) + 1 - P(B) - P(A) + P(A ∩ B) = P(B)
⇒ 1 - P(B) + P (A ∩ B) = P(B) ......(1)
A and B are exhaustive events than
P(A) + P(B) - P(A ∩ B) = 1 ...(ii)
from (i) & (ii) P(A) = P(B)
2 [P(B) - P(A ∩ B)] = P (B)
⇒ P(A) =
⇒
P(A ∪ (
∩B)) = P(A) + P(B) - P(A ∩ B) = 1Create a free account to view solution
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