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?

Answer 1

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`.