Question

In a factory which manufactures bolts, machines $A$, $B$ and $C$ manufacture respectively $25\%$, $35\%$ and $40\%$ of the bolts. Of their output $5\%$, $4\%$ and $2\%$ are respectively defective bolts. $A$ bolt is drawn at random from the total production and is found to be defective. Find the probability that it is manufactured by machine $B$.

• A. $\frac{23}{70}$
• B. $\frac{28}{69}$
• C. $\frac{25}{69}$
• D. $\frac{27}{70}$

Answer 1

Answer: (B)

Let $E_1$, $E_2$, $E_3$ and $A^{\prime }$ be the events defined as follows :
$E_1=$ bolt is manufactured by machine $A$
$E_2=$ bolt is manufactured by machine $B$
$E_3=$ bolt is manufactured by machine $C$
$A^{\prime }=$ bolt is defective.
$\therefore P\left(E_1\right)=\frac{25}{100}=\frac{1}{4}$,
$P\left(E_2\right)=\frac{35}{100}=\frac{7}{20}$
$P\left(E_3\right)=\frac{40}{100}=\frac{2}{5}$
$P\left(A^{\prime }|E_1\right)=\frac{5}{100}=\frac{1}{20}$,
$P\left(A^{\prime }|E_2\right)=\frac{4}{100}=\frac{1}{25}$
$P\left(A^{\prime }|E_3\right)=\frac{2}{100}=\frac{1}{50}$
We want to find $P(E_2|A^{\prime })$
By Bayes' Theorem,

$P\left(E_2|A^{\prime }\right)=\frac{P\left(E_2\right)\cdot P\left(A^{\prime }|E_2\right)}{P\left(E_1\right)P\left(A^{\prime }|E_1\right)+P\left(E_2\right)P\left(A^{\prime }|E_2\right)+P\left(E_3\right)P\left(A^{\prime }|E_3\right)}$
$=\frac{\frac{7}{2}\times \frac{1}{25}}{\frac{1}{4}\times \frac{1}{20}+\frac{7}{20}\times \frac{1}{25}+\frac{2}{5}\times \frac{1}{50}}$
$=\frac{\frac{7}{25}}{\frac{1}{4}+\frac{7}{25}+\frac{4}{25}}$
$=\frac{7}{25}\times \frac{100}{69}=\frac{28}{69}$