. If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P A( ) ∩ = B 16, P B( ) ∩ = C 14 , P A( ) ∩ = C 18 , P A( ) ∪ ∪ B C = 910 , P A( ) ∩ ∩ B C = 115 , then find PA PB ( ), ( ) and P C( )?
Answers
Given : A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A
To find : A , B & C
Solution:
Let say P(A) = x
then P(B) = 2x
& P(C) = 3x
P ( A ∩ B ) = 0.16
P ( B ∩ C ) = 0.14
P ( A ∩ C ) = 0.18
P ( A U B U C) = 0.91
P ( A ∩ B ∩ C) = 0.115
P ( A U B U C) = P(A) + P(B) + P(C) - P ( A ∩ B ) - P ( B ∩ C ) - P ( A ∩ C ) + P ( A ∩ B ∩ C)
=> 0.91 = x + 2x + 3x - 0.16 -0.14 - 0.18 + 0.115
=> 6x = 1.275
=> x = 0.2125
P(A) = 0.2125
P(B) = 0.425
P(C) = 0.6375
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