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11 11 votes

In a factory, two machines $M1$ and $M2$ manufacture $60\%$ and $40\%$ of the autocomponents respectively. Out of the total production, $2\%$ of $M1$ and $3\%$ of $M2$ are found to be defective. If a randomly drawn autocomponent from the combined lot is found defective, what is the probability that it was manufactured by $M2$?

  1. $0.35$
  2. $0.45$
  3. $0.5$
  4. $0.4$

3 Answers

Best answer
14 14 votes

$(C)$ $0.5$

Let $P(M_i)$ denote the probability that the component is manufactured by machine $M_i$, and $P(def)$ denote the probability that the component is defective.

We have to find $P(M_2|def)$.

$P(M_2\mid def) = P(M_2 \cap def) / P(def) = \dfrac{0.4\times0.03}{(0.4\times 0.03) + (0.6\times0.02)} = 0.5$

Drawing probability tree diagram for such questions makes them easier to solve. Please refer link1 and link2 for more details.

• edited by
5 5 votes

Let there be 100 auto components. 60 of them are manufactured by M1 and 40 by M2. Total defective components will be 2.4 in total, 1.2 from each of M1 and M2 respectively. Then . .

The Red Crossed Out Arrow are none of our concern. Our concern is on defective components (2.4). Out of all defective components, probability that it was manufactured by M2 is = (1.2/2.4) = 0.5

1 1 vote

Based on Bayes theorem ->

prob for manufacturing 

M1     M2

.6        .4

prob for defective autocomponents

M1      M2

.02       .03

prob that a randomly chosen autocomponet is found defective is:

.6*.02+.03*.4

prob that it was produced by M2=> .03*.4/.6*.02+.03*.4   =>.5

so C is the answer

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