COMEDK Exam 2014 Solved Paper

Let events T1,T2,T3 be the following
T1: The item is manufactured by factory F1.
T2: The item is manufactured by factory F2.
T3: The item is manufactured by factory F3.
Clearly, T1,T2,T3 are mutually exclusive and exhausted events.
P(T1)=30%=0.3,
P(T2)=20%=0.2
P(T3)=50%=0.5
Let E be the events that items is defective.
Now, P(E|T1)=2%=0.02
P(E|T2)=3%=0.03
P(E|T3)=4%=0.04
Hence, by Baye's theorem, we have
P(T1|E)=

P(T1)P(E|T1)

P(T1)P(E|T1)+P(T2)P(E|T2)+P(T3)P(E|T3)

=

0.3×0.02

0.3×0.02+0.2×0.03+0.5×0.04

=

0.006

0.006+0.006+0.20

=

0.006

0.032

=

6

32

=

3

16