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

The Correct Option is B

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): \(\frac{28}{69}\)