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Question
The chances of A, B and C becoming manager of a certain company are 5 : 3 : 2. The probabilities that the office canteen will be improved if A, B, and C become managers are 0.4, 0.5 and 0.3 respectively. If the office canteen has been improved, what is the probability that B was appointed as the manager?
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Solution
Let A1, A2, and A3 be the events of A, B, and C becoming managers of the company respectively.
Let B be the event that the office canteen will be improved.
We have to find the conditional probability P (A2/B).
Since A1, A2 and A3 are mutually exclusive and exhaustive events, applying Bayes theorem.
p(A2/B) = `("P"("A"_2) * "P"("B"/"A"_2))/("P"("A"_1) * "P"("B"/"A"_1) + "P"("A"_2) * "P"("B"/"A"_2) + "P"("A"_1) * "P"("B"/"A"_3))`
Give P(A1) = `5/10`, P(B/A1) = 0.4
P(A2) = `3/10`, P(B/A2) = 0.5
P(A3) = `2/10`, P(B/A3) = 0.3
P(A2/B) = `(3/10 xx 0.5)/(5/10 xx 0.4 + 3/10 xx 0.5 + 2/10 xx 0.3)`
= `(0.15/10)/((2.0 + 1.5 + 0.6)/10)`
P(A2/B) = `0.15/4.1`
= `15/41`
If the office canteen is improved than the probability of that B was appointed as the manager is `15/41`
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