Q45 / 75 · JEE Main 2026 (23 January Shift 2) · Mathematics
Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is $\frac{\mathrm{p}}{\mathrm{q}}, \operatorname{gcd}(\mathrm{p}, \mathrm{q})=1$, then $\mathrm{p}+\mathrm{q}$ is equal to
- Correct answerA23
- B24
- C21
- D22
Solution
Bag A: 9W, 8B. Bag B: 6W, 4B.
Case 1: White transferred from B (prob $= \frac{6}{10}$). A becomes 10W, 8B. $P(\text{white from A}) = \frac{10}{18} = \frac{5}{9}$.
Case 2: Black transferred from B (prob $= \frac{4}{10}$). A becomes 9W, 9B. $P(\text{white from A}) = \frac{9}{18} = \frac{1}{2}$.
Total probability $= \frac{6}{10} \cdot \frac{5}{9} + \frac{4}{10} \cdot \frac{1}{2} = \frac{1}{3} + \frac{1}{5} = \frac{8}{15}$.
$p = 8, q = 15, \gcd(8,15) = 1$. So $p + q = 23$.
Case 1: White transferred from B (prob $= \frac{6}{10}$). A becomes 10W, 8B. $P(\text{white from A}) = \frac{10}{18} = \frac{5}{9}$.
Case 2: Black transferred from B (prob $= \frac{4}{10}$). A becomes 9W, 9B. $P(\text{white from A}) = \frac{9}{18} = \frac{1}{2}$.
Total probability $= \frac{6}{10} \cdot \frac{5}{9} + \frac{4}{10} \cdot \frac{1}{2} = \frac{1}{3} + \frac{1}{5} = \frac{8}{15}$.
$p = 8, q = 15, \gcd(8,15) = 1$. So $p + q = 23$.
Hence, the correct answer is option: 'A'.