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Q1. 4 Events: Event C. Event E1. Event E2. Event E3 P(C) = 0.15 P(E1)=P(E3)=P(E2)=0.3 p(K/E1)=0.08...

Q1. 4 Events: Event C. Event E1. Event E2. Event E3

P(C) = 0.15 P(E1)=P(E3)=P(E2)=0.3

p(K/E1)=0.08 P(K/E2)=0.04 P(K/E3)=0.01 P(k/C)=0.15

Ask: What is probability that given not K (K'=1-P(k)), and it would be one of the event in E3? P(E3/K')?

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Answer #1

P(C) = 0.15, P(E1)=P(E2)=P(E3)=0.3
P(K|E1)=0.08, P(K|E2)=0.04, P(K|E3)=0.01, P(K|C)=0.15
P(E3|Kc) = ?
First using total law of probability, we find P(K)
P(K) = P(E1)P(K|E1) + P(E2)P(K|E2) + P(E3)P(K|E3)+P(C)P(K|C)
P(K) = (0.3 x 0.08) + (0.3 x 0.04) + (0.3 x 0.01) + (0.15 x 0.15)
P(K) = 0.0615
By definition of conditional probability:
P(E3|Kc) = P(E3 ∩ Kc) / P(Kc)
P(Kc) = 1 - P(K) = 1 - 0.0615 = 0.9385
P(E3∩Kc) = P(E3)P(Kc|E3) = P(E3) (1 - P(K|E3)) = 0.3 x (1 - 0.01) = 0.297
P(E3|Kc) = 0.297 / 0.9385 = 0.3165

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