Question

A factory production line is manufacturing bolts using three machines A, B and C. Out of the total output, machines A is responsible for 25%, machine B for 35% and machine C for the rest. It is known from the previous experience with the machines that 5% of the output from machine A is defective, 4% from machine B and 2% from machine C. A bolt is chosen at random from the production line and found to be defective. What is the probability that it can from (a) machine A?
(a) 0.632
(b) 0.362
(c) 0.487
(d) None of these

Answer 1

Hint: First, we have noted down the data which is given to us in mathematical form. Then be careful while noting the defective bolts from the total output which will be written as for machine A $P\left( \dfrac{D}{A} \right)=\dfrac{5}{100}$ for machine A. And, then we have to find probability of all defective bolts which will get from $P\left( D \right)=P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)+P\left( B \right)\cdot P\left( \dfrac{D}{B} \right)+P\left( C \right)\cdot P\left( \dfrac{D}{C} \right)$ . At last we have to put all the values in
$P\left( \dfrac{A}{D} \right)=\dfrac{P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)}{P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)+P\left( B \right)\cdot P\left( \dfrac{D}{B} \right)+P\left( C \right)\cdot P\left( \dfrac{D}{C} \right)}$ for machine A and get the answer.

Complete step by step solution:
Here, we are given a probability of bolt manufacture for machine A, B and C. To find how much machine C produces, is found by adding production of machine A and B and subtracting it from 100%.
We get machine C as \[100\%-\left( 25+35 \right)\%=100\%-60\%=40\%\] .
So, in mathematical form it is written as $P\left( A \right)=\dfrac{25}{100}$ , $P\left( B \right)=\dfrac{35}{100}$ , $P\left( C \right)=\dfrac{40}{100}$ .
Now, out of this output the defective bolt produced from machine A, B and C as given as $P\left( \dfrac{D}{A} \right)=\dfrac{5}{100}$ , $P\left( \dfrac{D}{B} \right)=\dfrac{4}{100}$ , $P\left( \dfrac{D}{C} \right)=\dfrac{2}{100}$ respectively. We have to find probabilities that the bolt was manufactured by machine A, B and C i.e. $P\left( \dfrac{A}{D} \right),P\left( \dfrac{B}{D} \right),P\left( \dfrac{C}{D} \right)$ respectively.
So, first we will find probability of defective bolt, which is given as:
$P\left( D \right)=P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)+P\left( B \right)\cdot P\left( \dfrac{D}{B} \right)+P\left( C \right)\cdot P\left( \dfrac{D}{C} \right)$ ……………………..(1)
Now, to find $P\left( \dfrac{A}{D} \right)$ the equation will be as given below:
$\Rightarrow P\left( \dfrac{A}{D} \right)=\dfrac{P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)}{P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)+P\left( B \right)\cdot P\left( \dfrac{D}{B} \right)+P\left( C \right)\cdot P\left( \dfrac{D}{C} \right)}$

We have all the values with us so, just substituting we get
$\Rightarrow P\left( \dfrac{A}{D} \right)=\dfrac{\dfrac{25}{100}\cdot \dfrac{5}{100}}{\dfrac{25}{100}\cdot \dfrac{5}{100}+\dfrac{35}{100}\cdot \dfrac{4}{100}+\dfrac{40}{100}\cdot \dfrac{2}{100}}$
On multiplying the values and then adding the denominator part, we get
$\Rightarrow P\left( \dfrac{A}{D} \right)=\dfrac{\dfrac{25}{100}\cdot \dfrac{5}{100}}{\dfrac{125}{10000}+\dfrac{140}{10000}+\dfrac{80}{10000}}$
$\Rightarrow P\left( \dfrac{A}{D} \right)=\dfrac{\dfrac{125}{10000}}{\dfrac{345}{10000}}$
Cancelling the denominator part as it is common in so, we have
$\Rightarrow P\left( \dfrac{A}{D} \right)=\dfrac{125}{345}=\dfrac{25}{69}$ ……………………………(2)
On dividing it, we get answer in decimal form as
$\Rightarrow P\left( \dfrac{A}{D} \right)=\dfrac{25}{69}=0.362$
Thus, option (b) is the correct answer.

Note: Students make mistakes in writing mathematical form of the given probability. Let’s take here that it is told, of the total of their output 5% bolt are defective of machine A. So, instead of writing $P\left( \dfrac{D}{A} \right)=\dfrac{5}{100}$, students write $P\left( \dfrac{A}{D} \right)=\dfrac{5}{100}$ and then substitute this value in the formula given as $\Rightarrow P\left( \dfrac{A}{D} \right)=\dfrac{P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)}{P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)+P\left( B \right)\cdot P\left( \dfrac{D}{B} \right)+P\left( C \right)\cdot P\left( \dfrac{D}{C} \right)}$ . So, from this they will find a probability of $P\left( \dfrac{D}{A} \right)$ which will be wrong. So, try to understand which data is given in the question and what you have to find from it. Do not make notation mistakes.