UPSC NDA Math Model Paper 3

Let the events B1,B2 and B3 be defined as:
B1: The needle is manufactured by machine A
B2: The needle is manufactured by machine B
B3: The needle is manufactured by machine C
Let the event E be 'The needle is defective'.
Given: P(B1)=0.2,P(B2)=0.3 and P(B3)=0.4
Similarly, P(E|B1)=0.04,P(E|B2)=0.03 and P(E|B3)=0.02 ------(Given)
Here, we have to find the value of P(B2|E)
As we know that according to bayes' theorem: P(Ei|A)=

P(Ei)×P(A|Ei)

n ∑ i=1 P(Ei)×P(A|Ei)

, i=1,2,….,n
⇒P(B2|E) =

0.3×0.03

[0.2×0.04+0.3×0.03+0.4×0.02]

=

9

25