Question

Assume that a factory has two machines $A$ and $B$. Past records shows that machine $A$ produces 60% of the items of output and machine $B$ produces 40% of the items. Further, 2% of the items produced by machine $A$ were defective and only 1% produced by machine $B$ were defective. If a detective item is drawn at random, what is the probability that it was produced by machine $A$?

Answer 1

Consider the problem

Let, $E_{1}and{E}_2$ be the respective events of items produced by machine $A$ and $B$.

And let $x$ be the event that the produced item was found to be defective.

Therefore,

Probability of items produced by machine $A,P$ $\left(E_1\right)$

$=60\%=\frac{3}{5}$

Probability of items produced by machine $B,P$ $\left(E_2\right)$

$=40\%=\frac{2}{5}$

And,

Probability that machine $A$ produced defective items, $P\left(\frac{x}{E_1}\right)$

$=2\%=\frac{2}{100}$

Probability that machine $B$ produced defective items, $P\left(\frac{x}{E_2}\right)$

$=1\%=\frac{1}{100}$

So, the probability that randomly selected items was from machine $A$ is given by $P\left(\frac{E_1}{x}\right)$

Now, Apply Bayes' Theorem

$\begin{array}{c}P\left(\frac{E_1}{x}\right)=\frac{P\left(E_1\right)P\left(\frac{x}{E_1}\right)}{P\left(E_1\right)P\left(\frac{x}{E_1}\right)+P\left(E_2\right)P\left(\frac{x}{E_2}\right)}\\ =\frac{\frac{3}{5}\times \frac{2}{100}}{\frac{2}{5}\times \frac{1}{100}+\frac{3}{5}\times \frac{2}{100}}\\ =\frac{6}{2+5}\\ =\frac{6}{11}\end{array}$

Hence, the required probability of machine $A$ is $\frac{6}{11}$