Question

A factory has two machines \(A\) and \(B\).Past record shows that machine \(A\) produce \(60\%\) of the item of output and machine \(B\) produced \(40\%\) of the items.Further,\(2\%\) of the items produced by machine \(A\) and \(1\%\) produced by machine \(B\) were defective.All the items are put into one stockpile and then one item is chosen at random from this and is found to defective.What is the probability that it was produced by machine \(B\)?

Answer 1

The correct answer is: \(\frac{1}{4}\)
Given \(P(A)=\frac{60}{100},P(B)=\frac{40}{100}\)
Let D denotes a defective item:
\(∴P(D|A)=\frac{2}{100}\, and\, P(D|A)=\frac{1}{100}\)
\(P(B|D)=\frac{P(B)P(D|B)}{P(A)P(D|A)+P(B)P(D|B)}\)
\(=\frac{\frac{40}{100}×\frac{1}{100}}{\frac{60}{100}×\frac{2}{100}+\frac{40}{100}×\frac{1}{100}}\)
\(=\frac{40}{120+40}\)
\(=\frac{40}{160}\)
\(=\frac{1}{4}\)