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 $A'$ 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 $C$
$A' =$ pipe is defective.
$\therefore P\left(E_{1}\right) = \frac{500}{3500} = \frac{1}{7}$,
$P\left(E_{2}\right) = \frac{1000}{3500} = \frac{2}{7}$
$P\left(E_{3}\right) = \frac{2000}{3500} = \frac{4}{7}$
$P\left(A'|E_{1}\right) = 0.005$,
$P\left(A'|E_{2}\right) = 0.008$,
$P\left(A'|E_{3}\right) = 0.010$
We want to find $P(E_1|A')$
By Bayes' Theorem,

$P\left(E_{1}|A'\right) = \frac{P\left(E_{1}\right) P\left(A' |E_{1}\right)}{P \left(E_{1}\right)P\left(A'|E_{1}\right) + P\left(E_{2}\right)P\left(A'|E_{2}\right)+ P\left(E_{3}\right)P\left(A'|E_{3}\right)}$
$=\frac{\frac{1}{7}\times0.005 }{\frac{1}{7}\times0.005 +\frac{2}{7}\times0.008 +\frac{4}{7}\times0.010 }$
$\frac{0.005}{0.005+0.016+0.040}$
$= \frac{0.005}{0.061} = \frac{5}{61}$