Question 51

In a factory where toys are manufactured, machines A, B and C produce 25%, 35% and 40% of the total toys, respectively. Of their output, 5%, 4% and 2% respectively, are defective toys. If a toy drawn at random is found to be defective, what is the probability that it is manufactured on machine B?

Solution

Let $$E_{1}$$, $$E_{2}$$ and $$E_{3}$$ be the events that the toys are produced by the machines A, B and C respectively.

Probability of toys manufactured by A i.e P($$E_{1}$$) = $$\frac{25}{100}$$

Probability of toys manufactured by B i.e P($$E_{2}$$) = $$\frac{35}{100}$$

Probability of toys manufactured by C i.e P($$E_{3}$$) = $$\frac{40}{100}$$

Let D be the event of the toy being defective.

Now, P$$(\frac{D}{E_{1}})$$ = $$\frac{5}{100}$$; P$$(\frac{D}{E_{2}})$$ = $$\frac{4}{100}$$; P$$(\frac{D}{E_{3}})$$ = $$\frac{2}{100}$$

Probability that the toy drawn at random is defective and is manufactured on machine B,

P(B|D) = $$\frac{P(E_{2})P(D/E_{2})}{P(E_{1})P(D/E_{1}) + P(E_{2})P(D/E_{2}) + P(E_{3})P(D/E_{3})}$$

P(B|D) = $$\frac{(35/100)(4/100)}{(25/100)(5/100) + (35/100)(4/100) + (40/100)(2/100)}$$

P(B|D) = $$\frac{(35 \times 4)}{(25 \times 5) + (35 \times 4) + (40 \times 2)}$$

P(B|D) = $$\frac{140}{125 + 140 + 80}$$

P(B|D) = $$\frac{140}{345}$$

P(B|D) = $$\frac{28}{69}$$

Hence, option B is the correct answer.


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