Question

In a bolt factory, machines A, B and C manufacture 25%, 35%, 40% respectively. Of the total of their output 5%, 4% and 2% are defective. A bolt is drawn and is found to be defective. What are the probabilities that it was manufactured by the machines A, B and C?
(a) $\dfrac{25}{69},\dfrac{28}{69},\dfrac{16}{69}$
(b) $\dfrac{28}{69},\dfrac{25}{69},\dfrac{16}{69}$
(c) $\dfrac{25}{69},\dfrac{16}{69},\dfrac{28}{69}$
(d) $\dfrac{16}{69},\dfrac{28}{69},\dfrac{25}{69}$

Answer 1

Hint: First, we have to note 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 all the three machines and get the answers.

Complete step-by-step answer:
Here, we are given probability of bolt manufacture for machine A, B and C. 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)
Similarly, we will find $P\left( \dfrac{B}{D} \right)$ the equation will be as given below:
$\Rightarrow P\left( \dfrac{B}{D} \right)=\dfrac{P\left( B \right)\cdot P\left( \dfrac{D}{B} \right)}{P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)+P\left( B \right)\cdot P\left( \dfrac{D}{B} \right)+P\left( C \right)\cdot P\left( \dfrac{D}{C} \right)}$

We have all the values with us so, just substituting we get
$\Rightarrow P\left( \dfrac{B}{D} \right)=\dfrac{\dfrac{35}{100}\cdot \dfrac{4}{100}}{\dfrac{25}{100}\cdot \dfrac{5}{100}+\dfrac{35}{100}\cdot \dfrac{4}{100}+\dfrac{40}{100}\cdot \dfrac{2}{100}}$
On multiplying the values and then adding the denominator part, we get
$\Rightarrow P\left( \dfrac{B}{D} \right)=\dfrac{\dfrac{35}{100}\cdot \dfrac{4}{100}}{\dfrac{125}{10000}+\dfrac{140}{10000}+\dfrac{80}{10000}}$

$\Rightarrow P\left( \dfrac{B}{D} \right)=\dfrac{\dfrac{140}{10000}}{\dfrac{345}{10000}}$
Cancelling the denominator part as it is common in so, we have
$\Rightarrow P\left( \dfrac{B}{D} \right)=\dfrac{140}{345}=\dfrac{28}{69}$ ……………………………(3)
Similarly, we will find $P\left( \dfrac{C}{D} \right)$ the equation will be as given below:
$\Rightarrow P\left( \dfrac{C}{D} \right)=\dfrac{P\left( C \right)\cdot P\left( \dfrac{D}{C} \right)}{P\left( A \right)\cdot P\left( \dfrac{D}{A} \right)+P\left( B \right)\cdot P\left( \dfrac{D}{B} \right)+P\left( C \right)\cdot P\left( \dfrac{D}{C} \right)}$

We have all the values with us so, just substituting we get
$\Rightarrow P\left( \dfrac{C}{D} \right)=\dfrac{\dfrac{40}{100}\cdot \dfrac{2}{100}}{\dfrac{25}{100}\cdot \dfrac{5}{100}+\dfrac{35}{100}\cdot \dfrac{4}{100}+\dfrac{40}{100}\cdot \dfrac{2}{100}}$
On multiplying the values and then adding the denominator part, we get
$\Rightarrow P\left( \dfrac{C}{D} \right)=\dfrac{\dfrac{40}{100}\cdot \dfrac{2}{100}}{\dfrac{125}{10000}+\dfrac{140}{10000}+\dfrac{80}{10000}}$
$\Rightarrow P\left( \dfrac{C}{D} \right)=\dfrac{\dfrac{80}{10000}}{\dfrac{345}{10000}}$
Cancelling the denominator part as it is common in so, we have
$\Rightarrow P\left( \dfrac{C}{D} \right)=\dfrac{80}{345}=\dfrac{16}{69}$ ……………………………(4)
Thus, the bolt was manufactured by machine A, B and C i.e. $P\left( \dfrac{A}{D} \right),P\left( \dfrac{B}{D} \right),P\left( \dfrac{C}{D} \right)$ is $\dfrac{25}{69},\dfrac{28}{69},\dfrac{16}{69}$ respectively.
Hence, option (a) is correct.

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 od 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.