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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. $0.4$

could u plz put his q in numerical ability?

How both questions solving method is same?? one is giving percentage from total production...

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?

ANOTHER QUESTION IS FROM 2012

https://gateoverflow.in/2211

An automobile plant contracted to buy shock absorbers from two suppliers X and Y . X supplies 60% and Y supplies 40% of the shock absorbers. All shock absorbers are subjected to a quality test. The ones that pass the quality test are considered reliable. Of X′s shock absorbers, 96% are reliable. Of Ys shock absorbers, 72% are reliable.

$(C)$ $0.5$

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.

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

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

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