Question

A factory has three machines A,B and C, which produce $100,200and300$items of a particular type daily. The machines produce $2\%,3\%and5\%$ defective items respectively. One day when the production was over, an item was picked up randomly and it was found to be defective. Find the probability that it was produced by machine A.

Answer 1

Let $E_{1,}{E}_2,E_3$ is machine $A,B,C$

$D=$defective items

total production $100+200+300=600$

$\begin{array}{c}p\left(E_1\right)=\frac{100}{600}\\ p\left(E_1\right)=\frac{1}{6}\\ p\left(E_2\right)=\frac{200}{600}=\frac{1}{3}\\ p\left(E_3\right)=\frac{300}{600}=\frac{1}{2}\end{array}$

now

$\begin{array}{c}p\left(\frac{D}{E_1}\right)=0.02\\ p\left(\frac{D}{E_2}\right)=0.03\\ p\left(\frac{D}{E_3}\right)=0.05\end{array}$

probability

$\begin{array}{c}P\left(\frac{E_1}{D}\right)=\frac{p\left(E_1\right)p\left(\frac{D}{E_1}\right)}{p\left(E_1\right)p\left(\frac{D}{E_1}\right)+p\left(E_2\right)p\left(\frac{D}{E_2}\right)+p\left(E_3\right)p\left(\frac{D}{E_3}\right)}\\ =\frac{\frac{1}{6}\times 0.02}{\frac{1}{6}\times 0.02+\frac{1}{3}\times 0.03+\frac{1}{2}\times 0.05}\\ =\frac{2}{23}\end{array}$