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Question
In a factory, machine A produces 30% of total output, machine B produces 25% and the machine C produces the remaining output. The defective items produced by machines A, B and C are 1%,1.2%, 2% respectively. An item is picked at random from a day's output and found to be defective. Find the probability that it was produced by machine B?
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Solution
Let E1 = choosing machine A
E2 = choosing machine B
E3 = choosing machine C
A = Producing a defective output
Given, P(E1) = 30% = `30/100` = 0.3
P(E2) = 25% = `25/100` = 0.25
P(E3) = [100 – (30 + 25)]% = 45%
= `45/100` = 0.45
and `P(A/E_1)` = P(Producing defective output from machine A)
= 1% = `1/100` = 0.01
`P(A/E_2)` = P(Producing defective output from machine B)
= 1.2% = `1.2/100` = 0.12
`P(A/E_3)` = P(Producing defective output from machine C)
= 2% = `2/100` = 0.02
Required probability = `P(E_2/A)`
= P(The found defective item is produced by machine B)
Using Bayes' theorem,
`P(E_2/A) = (P(E_2).P(A/E_2))/(P(E_1).P(A/E_1) + P(E_2).P(A/E_2) + P(E_3).P(A/E_3))`
= `(0.25 xx 0.12)/((0.3 xx 0.01) + (0.25 xx 0.12) + (0.45 xx 0.02))`
= `300/(300 + 300 + 900)`
= `300/1500`
= `1/5`
Thus, required probability is `1/5`.
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