JEE Main 2024 — 27 January, Shift 2 — Question 13

Total number of ways to draw 4 balls from 15: $\binom{15}{4}$. Number of ways to draw 4 white balls from 6: $\binom{6}{4}$. Probability of first draw all white: $P_1 = \frac{\binom{6}{4}}{\binom{15}{4}} = \frac{15}{1365} = \frac{1}{91}$. After first draw, balls left: 2 white, 9 black, total 11. Number of ways to draw 4 black from 9: $\binom{9}{4}$. Number of ways to draw any 4 from remaining 11: $\binom{11}{4}$. Probability of second draw all black given first was all white: $P_2 = \frac{\binom{9}{4}}{\binom{11}{4}} = \frac{126}{330} = \frac{21}{55}$. Required probability: $P = P_1 \times P_2 = \frac{1}{91} \times \frac{21}{55} = \frac{21}{5005} = \frac{3}{715}$.