BITSAT Mathematics Question (2023) — Solution

Step-by-step solution

Let E_1, E_2 and A be the events defined as follows: E _1= red ball is transferred from bag P to bag Q E _2= 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 (E_1 )= 6 10 = 3 5 and P (E_2 )= 4 10 = 2 5 Note that E_1 and E_2 are mutually exclusive and exhaustive events. When E_1 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 E_1 )= 6 12 = 1 2 When E_2 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 E_2 )= 7 12 By using law of total probability, we get P(A)=P (E_1 ) P (A E_1 )+P (E_2 ) P (A E_2 ) = 3 5 1 2 + 2 5 7 12 = 8 15