BITSAT 2019 Paper

(b) Let E ‌1 , E ‌2 and A be the events defined as follows:
E‌1 = red ball is transferred from bag P to bag Q
E‌2 = blue ball is transferred from bag P to bag Q
A = the ball drawn from bag Q is blue
As the bag P contains 6 red and 4 blue balls,
P(E‌1) =

6

10

=

3

5

and P(E‌2) =

4

10

=

2

5

Note that E‌1 and E‌2 are mutually exclusive and exhaustive events.
When E‌1 has occurred i.e., a red ball has already been transferred from bag P to Q, then bag Q will contain 6 red and 6 blue balls, So, P(A|E‌1) =

6

12

=

1

6

When E‌2 has occurred i.e., a blue ball has already been transferred from bagP to Q, then bag Q will contain 5 red and 7 blue balls, So,P(A|E‌2) =

7

12

By using law of total probability, we get P(A) = P(E‌1) P(A|E‌1) + P(E‌2) P(A|E‌2)
=

3

5

×

1

2

+

2

5

×

7

12

=

8

15