Question

A manufacturer has three machine operators A, B and C. The first operator A produces $1\%$ defective items, whereas the other two operators B and C produce $5\%$ and $7\%$ defective items respectively. A is on the job for $50\%$ of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Answer 1

Hint : To solve this question, we will start with find the individual probability of events, i.e., probability of items produced by operator A, B and C. Then the probability of defective items produced by each operator A, B and C. And hence, we will put the values in the probability formula.

Complete step-by-step answer :
We have been given that a manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. It is given that A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, we need to find the probability that it was produced by A.
Let the event that items were produced by operator A be A, by operator B be B, by operator C be C, and item produced defective be D.
P(A) $=$ Probability of item is produced by operator A.
$= 50\% \\\ = \dfrac{{50}}{{100}} \\\ = 0.5 \\\$
P(B) $=$ Probability of item is produced by operator B.
$= 30\% \\\ = \dfrac{{30}}{{100}} \\\ = 0.3 \\\$

P(C) $=$ Probability of item is produced by operator C.
$= 20\% \\\ = \dfrac{{20}}{{100}} \\\ = 0.2 \\\$

P(D|A) $=$ Probability of a defective item produced by operator A.
$= 1\% \\\ = \dfrac{1}{{100}} \\\ = 0.01 \\\$
P(D|B) $=$ Probability of a defective item produced by operator B.

$$= 5\% \\\ = \dfrac{5}{{100}} \\\ = 0.05 \\\$$

P(D|C) $=$ Probability of a defective item produced by operator C.

$$= 7\% \\\ = \dfrac{7}{{100}} \\\ = 0.07 \\\$$

We need to find the probability that the item produced by A is defective, i.e.,
$P\left( {A|D} \right){\text{ }} = \;\dfrac{{P\left( A \right).P\left( {D|A} \right)}}{{P\left( A \right).P\left( {D|A} \right) + P\left( B \right).P\left( {D|B} \right) + P\left( C \right).P\left( {D|C} \right)}}$
On putting the values in the above formula, we get
$P\left( {A|D} \right){\text{ }} = \;\dfrac{{0.5 \times 0.01}}{{(0.5) \times (0.01) + \left( {0.3} \right) \times \left( {0.05} \right) + \left( {0.2} \right) \times \left( {0.07} \right)}}$
$= \dfrac{{0.005}}{{0.005 + 0.015 + 0.014}}$
$= \dfrac{{0.005}}{{0.034}}$
$= \dfrac{5}{{34}}$
Thus, the probability that it was produced by A is $\dfrac{5}{{34}}.$

Note : Students should carefully obtain the individual probability of the event, because if one value gets wrong, there are chances of getting the whole answer wrong. So, students are supposed to carefully put the values of individual probability of events in the formula mentioned above.