CGPET 2022 Solved Paper

150. (c) Let E1 : Event that bag contains 0 white ball.
E2 : Event that bag contains 1 white ball.
E3 : Event that bag contains 2 white ball.
E4 : Event that bag contains 3 white ball.
E5 : Event that bag contains 4 white ball.
E6 : Event that bag contains 5 white ball.
A : Event that drawn ball is white ball.
Now, P(E1)P(E2)=P(E3)=P(E4)=P(E5)=P(E6)=

1

6

P(

A

E1

)=0
P(

A

E2

)=

‌1C1

‌5C1

=

1

5

P(

A

E3

)=

‌2C1

‌5C1

=

2

5

P(

A

E4

)=

‌3C5

‌5C1

=

3

5

P(

A

E5

)=

‌4C1

‌5C1

=

4

5

P(

A

E5

)=

‌5C1

‌5C1

=

5

5

=1
Required probability
=

=P(E2∕A)P(E2)P(A∕E2)

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

+P(E4)P(A∕E4)+P(E5)P(A∕E5)+P(E6)P(A∕E6)
=

1 6 × 1 5

1 6 ×0+ 1 6 × 1 5 + 1 6 × 2 5 + 1 6 × 3 5 + 1 6 × 4 5 + 1 6 ×1

=

1

1+2+3+4+5

=

1

15