# 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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