Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, is 29/45 , then n is equal to:

Solution

Maths sir · 4 steps

Split on the colour of the transferred ball, add the two cases, then solve for n.

1. Two cases. The transferred ball is white with probability 4/9 or black with probability 5/9. Either way, Bag 2 then holds n + 4 balls.
2. White from Bag 2. If a white ball came over, Bag 2 has n + 1 white balls. If a black ball came over, it still has n white balls. The total probability theorem adds the two cases.

   4/9·(n+1)/(n+4) + 5/9·n/(n+4) = (9n+4)/9(n+4) = 29/45

3. Solve. Multiply both sides by 45(n+4) to get 5(9n+4) = 29(n+4). So 45n + 20 = 29n + 116, which gives 16n = 96 and n = 6.
4. So. Check: (9 × 6 + 4)/(9 × 10) = 58/90 = 29/45. So n = 6, option (b).

The idea to keep. When a ball is transferred before the draw, condition on what was transferred. Weight each case by its probability and add, and remember the second bag now has one extra ball.

Answer from the NTA final answer key. Solution by PrepWiser.