Question

A firm produces steel pipes in three plants $A$, $B$ and $C$ with daily production of $500$, $1000$ and $2000$ units respectively. It is known that fractions of defective output produced by the three plants are respectively $0.005$, $0.008$ and $0.010$. $A$ pipe is selected at random from a day's total production and found to be defective. What is the probability that it came from first plant?

• A. $\frac{4}{61}$
• B. $\frac{8}{61}$
• C. $\frac{5}{61}$
• D. $\frac{9}{61}$

Answer 1

Answer: (C)

Let $E_1$, $E_2$, $E_3$ and $A^{\prime }$ be the events defined as follows :
$E_1=$ pipe is selected from plant $A$
$E_2=$ pipe is selected from plant $B$
$E_3=$ pipe is selected from plant $C$
$A^{\prime }=$ pipe is defective.
$\therefore P\left(E_1\right)=\frac{500}{3500}=\frac{1}{7}$,
$P\left(E_2\right)=\frac{1000}{3500}=\frac{2}{7}$
$P\left(E_3\right)=\frac{2000}{3500}=\frac{4}{7}$
$P\left(A^{\prime }|E_1\right)=0.005$,
$P\left(A^{\prime }|E_2\right)=0.008$,
$P\left(A^{\prime }|E_3\right)=0.010$
We want to find $P(E_1|A^{\prime })$
By Bayes' Theorem,

$P\left(E_1|A^{\prime }\right)=\frac{P\left(E_1\right)P\left(A^{\prime }|E_1\right)}{P\left(E_1\right)P\left(A^{\prime }|E_1\right)+P\left(E_2\right)P\left(A^{\prime }|E_2\right)+P\left(E_3\right)P\left(A^{\prime }|E_3\right)}$
$=\frac{\frac{1}{7}\times 0.005}{\frac{1}{7}\times 0.005+\frac{2}{7}\times 0.008+\frac{4}{7}\times 0.010}$
$\frac{0.005}{0.005+0.016+0.040}$
$=\frac{0.005}{0.061}=\frac{5}{61}$