Question

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')?

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|K^c) = ?
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|K^c) = P(E₃ ∩ K^c) / P(K^c)
P(K^c) = 1 - P(K) = 1 - 0.0615 = 0.9385
P(E₃∩K^c) = P(E₃)P(K^c|E₃) = P(E₃) (1 - P(K|E₃)) = 0.3 x (1 - 0.01) = 0.297
P(E₃|K^c) = 0.297 / 0.9385 = 0.3165