Suppose we have three events,
Suppose we have three events, A, B, and C such that:
– A and B are independent
– B and C are independent
– P[AUBUC]=0.90
-P[A]= 0.20
– P[C]= 0.60
Compute P [C | AUB]
Answer:
P(AUBUC) = P(A) + P(B) + P(C) – P(A 
B) – P(A C) – P(B C) + P(A B C)
– A and B are independent
– B and C are independent
It means, A and C are also independent
So,
P(A B) = P(A)*P(B)
P(A C) = P(A)*P(C)
P(B C) = P(B)*P(C)
P(A B C) = P(A)*P(B)*P(C)
So, as per the formula-
0.9 = 0.2 + 0.6 + P(B) – 0.2*P(B) – 0.6*P(B) – 0.2*0.6 +0.2*0.6*P(B)
0.9 = 0.68 + 0.32*P(B)
P(B) = 0.6875
So, P [C | AUB] = P(A B C)/P(AUB)
P(A B C) = P(A)*P(B)*P(C) =0.2*0.6*0.6875 = 0.0825
P(AUB) = P(A) + P(B) – P(AB) = 0.2 + 0.6875 -0.2*0.6875 = 0.75
P [C | AUB] = 0.0825/0.75 = 0.11
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