Question

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?

• A. $\frac{17}{69}$
• B. $\frac{28}{69}$
• C. $\frac{35}{69}$
• D. None of these

Answer 1

Answer: (B)

$\frac{28}{69}$

Answer 2

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 ($E$) $=\frac{25}{100}$

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

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

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) = P($E$1)P$(\frac{D}{E_1})$+$\frac{P(E_2)P\frac{D}{E_2}}{P(E_2)P\frac{D}{E_2}}$+P($E$3)P$(\frac{D}{E_3})$

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

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

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

P(B|D) = $\frac{140}{345}$

P(B|D) = $\frac{28}{69}$

Hence, option B is the correct answer.The correct option is (B): $\frac{28}{69}$