The chances of X, Y and Z becoming managers of a certain company are 4:2:3. The probabilities that bonus scheme will be introduced if X, Y and Z become managers are 0.3, 0.5 and 0.4 respectively. If the bonus scheme has been introduced what is the probability that Z is appointed as the manager?
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Let event E₁ : person X becomes manager
event E₂ : person Y becomes manager
event E₃ : person Z becomes manager
Given, ratio of becoming manager of a certain company 4 : 2 : 3
so, P(E₁) = 4/9 , P(E₂) = 2/9 and P(E₃) = 3/9
Let event A : bonus scheme introduced
so, P(A/E₁) = 0.3, P(A/E₂) = 0.5 and P(A/E₃) = 0.4
We know,
P(A) = P(E₁).P(A/E₁) + P(E₂).P(A/E₂) + P(E₃).P(A/E₃)
= 4/9 × 0.3 + 2/9 × 0.5 + 3/9 × 0.4
= 17/45
Now, we have to find probability of Z is appointed as manager
Therefore, required probability = P(Person Z becomes manager under the condition that bonus scheme is introduced)
So, P(E₃/A) = P(A/E₃)/P(A)
= 3/9 × 0.4/17/45
= 6/17
event E₂ : person Y becomes manager
event E₃ : person Z becomes manager
Given, ratio of becoming manager of a certain company 4 : 2 : 3
so, P(E₁) = 4/9 , P(E₂) = 2/9 and P(E₃) = 3/9
Let event A : bonus scheme introduced
so, P(A/E₁) = 0.3, P(A/E₂) = 0.5 and P(A/E₃) = 0.4
We know,
P(A) = P(E₁).P(A/E₁) + P(E₂).P(A/E₂) + P(E₃).P(A/E₃)
= 4/9 × 0.3 + 2/9 × 0.5 + 3/9 × 0.4
= 17/45
Now, we have to find probability of Z is appointed as manager
Therefore, required probability = P(Person Z becomes manager under the condition that bonus scheme is introduced)
So, P(E₃/A) = P(A/E₃)/P(A)
= 3/9 × 0.4/17/45
= 6/17
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