Question

A factory has two machines $A$ and $B$. Past record shows that machine $A$ produced $60\%$ of the items of output and machine $B$ produced $40\%$ of the items. Further, $2\%$ of the items produced by machine $A$ and $1\%$ produced by machine $B$ were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. Then, the probability that it was produced by machine $B$ is

• A. $\frac{1}{5}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

Answer: (D)

Let $E_1$ : the event that the item is produced by machine $A$
and $E_2$ : the event that the item is produced by machine $B$.
Then, $E_1$ and $E_2$ are mutually exclusive and exhaustive events. Moreover,
$P\left(E_1\right)=60\%=\frac{60}{100}=\frac{3}{5}$
$\text{ and }P\left(E_2\right)=40\%=\frac{40}{100}=\frac{2}{5}$
Let $E$ : the event that the item chosen is defective,
$\therefore P\left(\frac{E}{E_1}\right)=P(\text{ machine }A\text{ produced defective items })$
$=2\%=\frac{2}{100}$
$P\left(\frac{E}{E_2}\right)=P(\text{ machine }B\text{ produced defective items })$
$=1\%=\frac{1}{100}$
The probability that the randomly selected item was from machine $B$, given that it is defective, is given by $P\left(\frac{E_2}{E}\right)$.
By using Bayes' theorem, we obtain
$P\left(\frac{E_2}{E}\right)=\frac{P\left(\frac{E}{E_2}\right)P\left(E_2\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}$
$=\frac{\frac{1}{100}\times \frac{2}{5}}{\frac{2}{100}\times \frac{3}{5}+\frac{1}{100}\times \frac{2}{5}}=\frac{\frac{2}{500}}{\frac{6}{500}+\frac{2}{500}}$
$=\frac{2}{6+2}=\frac{1}{4}$