VITEEE 2010 Paper

Let E1,E2‌and‌E3 denote the events of selecting boxes A, B, C respectively and A be the event that a screw selected at random is defective. Then,
P(E1)=

1

3

,P(E2)=

1

3

,P(E3)=

1

3

,
P(

A

E1

)=

1

5

,P(

A

E2

)=

1

6

,P(

A

E3

)=

1

7

,
By Baye's rule, the required probability
P(

E1

A

)=

P(E1)P( A E1 )

P(E1)P( A E1 )+P(E2)P( A E2 )+P(E3)P( A E3 )

⇒ P(

E1

A

)=

1 3 . 1 5

1 3 . 1 5 + 1 3 . 1 6 + 1 3 . 1 7

=

1 5

1 5 + 1 6 + 1 7

=

42

107