BITSAT 2023 Shift 1 Paper

Let E1,E2 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

2

When E2 has occurred i.e., a blue ball has already been transferred from bag P 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