Correct Answer - Option 4 : 4 / 7

__Concept:__

Bayes’ theorem:

\({\rm{P}}\left( {{\rm{A|B}}} \right) = \frac{{{\rm{P}}({\rm{B}}|{\rm{A}}){\rm{\;\;P}}\left( {\rm{A}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

Where A and B are events and P(B).

P(A|B) is a conditional probability: the likelihood of an event of A occurring given that B is true.

P(B|A) is a conditional probability: the likelihood of the event of B occurring given that A is true.

P(A) and P(B) are the probabilities of observing A and B respectively.

__Calculation:__

Let

A: toy manufactured by machine A.

B: toy manufactured by machine B.

C: toy manufactured by machine C.

D: defective toy.

Given:

P(A) = 0.30; P(B) = 0.40; P(C) = 0.30

Probability of defective toy manufactured by A = P(D|A) = 0.02

Probability of defective toy manufactured by B = P(D|B) = 0.03

Probability of defective toy manufactured by C = P(D|C) = 0.01

We need to determine the probability that a toy is chosen, found to be defective, and was manufactured by B = P(B|D)

Using Bayes’ theorem,

\({\rm{P}}\left( {{\rm{B|D}}} \right) = \frac{{{\rm{P}}\left( {{\rm{D|B}}} \right){\rm{P}}\left( {\rm{B}} \right)}}{{{\rm{P}}\left( {\rm{D}} \right)}}\)

P(D) = Probability of defective toy

= P(B)P(D|B) + P(A)P(D|A) + P(C)P(D|C)

= 0.40 × 0.03 + 0.30 × 0.02 + 0.30 × 0.01

= (12 + 6 + 3) × 10-3

= 21 × 10-3

Now, \({\rm{P}}\left( {{\rm{B|D}}} \right) = \frac{{{\rm{P}}\left( {{\rm{D|B}}} \right){\rm{P}}\left( {\rm{B}} \right)}}{{{\rm{P}}\left( {\rm{D}} \right)}}\)

\(= \frac{{0.03\; \times \;0.40}}{{21\; \times \;{{10}^{ - 3}}}}\)

\(= \frac{{12\; \times \;{{10}^{ - 3}}}}{{21\; \times \;{{10}^{ - 3}}}}\)

= 4/7