Question

In a bolt factory, Machines A, B and C manufacture respectively 25, 35 and 40 percent of the total. Out of their total output 5, 4 and 2 percent are defective. A bolt drawn from the procedure at random is found to be defective. What is the probability that it is manufactured by
(a) machine A
(b) machine C?

Answer 1

Let A,B and C be the events of drawing a bolt produced by Machine A,B and C respectively.Then

$p(A) = \frac{25}{100} = \frac{1}{4} \\ \\ p(B) = \frac{35}{100} = \frac{7}{20} \\ \\ p(C) = \frac{40}{100} = \frac{2}{5}$
let E be the event of drawing a defective bolt. So
$p\bigg( \frac{E}{A} \bigg) = \frac{5}{100} = \frac{1}{20} \\ \\ p\bigg( \frac{E}{B}\bigg)= \frac{4}{100} = \frac{1}{25} \\ \\ p\bigg( \frac{E}{C}\bigg) = \frac{2}{100} = \frac{1}{50}$
(a) Probability that the defective bolt is produced from machine A :

apply Bay's theorem of Probability

$\frac{p(A)p( \frac{E}{A} )}{ p(A)p( \frac{E}{A} ) +p(B)p( \frac{E}{B}) +p(C)p( \frac{E}{C})} \\ \\ = \frac{ \frac{1}{4} \times \frac{1}{20} }{\frac{1}{4} \times \frac{1}{20} + \frac{7}{20} \times \frac{1}{25} + \frac{2}{5} \times \frac{1}{50}} \\ \\ = \frac{1}{80} \times \frac{2000}{69} \\ \\ = \frac{25}{69}$
Hence the probability that defective bolt produced from machine A is 25/69.

(b) Probability that the defective bolt is produced from machine C :

apply Bay's theorem of Probability

$\frac{p(C)p( \frac{E}{C} )}{ p(A)p( \frac{E}{A} ) +p(B)p( \frac{E}{B}) +p(C)p( \frac{E}{C})} \\ \\ = \frac{ \frac{2}{5} \times \frac{1}{50} }{\frac{1}{4} \times \frac{1}{20} + \frac{7}{20} \times \frac{1}{25} + \frac{2}{5} \times \frac{1}{50}} \\ \\ = \frac{1}{125} \times \frac{2000}{69} \\ \\ = \frac{16}{69}$
Hence the probability that defective bolt produced from machine C is 16/69.

Hope it helps you