Question

In a factory machine A produces $30\%$ of the total output, machine B produces $25\%$ and machine C produces the remaining output. If defective items produced by machine A, B, and C are $1\%$, $12\%$, $2\%$ respectively. Three machines working together produced 10000 items in a day. An item is drawn at random from a day’s output and found to be defective. Find the probability that it was produced by machine B.

Answer 1

Hint: First of we will find the probability of items produced by machines A, B and C to be defective. Then by using the conditional probability we will find the probability of an item drawn randomly was produced by machine B using the formula $P\left( \dfrac{B}{D} \right)=\dfrac{P\left( B \right)\times P\left( \dfrac{D}{B} \right)}{P\left( B \right)\times P\left( \dfrac{D}{B} \right)+P\left( A \right)\times P\left( \dfrac{D}{A} \right)+P\left( C \right)\times P\left( \dfrac{D}{C} \right)}$ .

Complete step-by-step answer:
In question we are given that machine A produces $30\%$ of the total output, machine B produced $25\%$ and the machine C produced the remaining output this can be shown mathematically as,
$P\left( A \right)=30\%=\dfrac{30}{100}=0.3$
$P\left( B \right)=25\%=\dfrac{25}{100}=0.25$
$P\left( C \right)=100-\left( 30+25 \right)=\dfrac{45}{100}=0.45$
Now, let’s consider D as defective product and defective items produced by A, B and C are $1\%$, $12\%$, $2\%$ respectively. This, can be shown mathematically as,
$P\left( \dfrac{D}{A} \right)=1\%=\dfrac{1}{100}=0.01$
$P\left( \dfrac{D}{B} \right)=12\%=\dfrac{12}{100}=0.12$
$P\left( \dfrac{D}{C} \right)=2\%=\dfrac{2}{100}=0.02$
Now, to find the probability of product that is defective is produced by B can be given by the formula,
$P\left( \dfrac{B}{D} \right)=\dfrac{P\left( B \right)\times P\left( \dfrac{D}{B} \right)}{P\left( B \right)\times P\left( \dfrac{D}{B} \right)+P\left( A \right)\times P\left( \dfrac{D}{A} \right)+P\left( C \right)\times P\left( \dfrac{D}{C} \right)}$
Now, substituting the values in the equation we will get,
$P\left( \dfrac{B}{D} \right)=\dfrac{0.25\times 0.12}{\left( 0.25\times 0.12 \right)+\left( 0.3\times 0.01 \right)+\left( 0.45\times 0.02 \right)}$
$\Rightarrow P\left( \dfrac{B}{D} \right)=\dfrac{0.03}{0.03+0.003+0.009}$
$\Rightarrow P\left( \dfrac{B}{D} \right)=\dfrac{3\times {{10}^{-3}}}{\left( 30+3+9 \right)\times {{10}^{-3}}}=\dfrac{30}{15}$
$\Rightarrow P\left( \dfrac{B}{D} \right)=0.71$
Thus, we can say that the probability of a product drawn is defective and it is produced by machine B is $0.71$.

Note: This problem can also be solved by taking different variables but that will consume more time so it is less preferable. Students might make mistake using the conditional probability and they might use $P\left( \dfrac{D}{B} \right)$ instead of $P\left( \dfrac{B}{D} \right)$ and due to that whole problem might go wrong so students should be careful while solving such sums.