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Question
In an automobile factory, certain parts are to be fixed into the chassis in a section before it moves into another section. On a given day, one of the three persons A, B, and C carries out this task. A has a 45% chance, B has a 35% chance and C has a 20% chance of doing the task.
The probability that A, B, and C will take more than the allotted time is `(1)/(6), (1)/(10), and (1)/(20)` respectively. If it is found that the time taken is more than the allotted time, what is the probability that A has done the task?
Solution
P(A) = 0.45, P (B) = 0.35, P (C) = 0.20
P (A/E) = `(1)/(6), P(B/E) = (1)/(10), P (C/E) = (1)/(20)`
P(E/A) = `(P(A). P(A/E))/(P(A). P(A/E) + P(B). P(B/E) +P(C). P(C/E)`
= `(0.45 xx 1/6)/(0.45 xx 1/6 + 0.35 xx 1/10 + 0.20 xx 1/20)`
= `0.075/0.12`
P(E/A) = 0.625
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