JEE Main 2026 — 28 January, Morning Shift — Question 9

Let $E$ be the event that all three drawn balls are black. We need $P(k=1 \mid E)$. By Bayes' theorem, $P(k=1 \mid E) = \frac{P(E \mid k=1) \cdot P(k=1)}{\sum_{k=0}^{10} P(E \mid k) \cdot P(k)}$. Assuming each $k$ from 0 to 10 is equally likely, $P(k) = \frac{1}{11}$. For a given $k$, $P(E \mid k) = \frac{\binom{10-k}{3}}{\binom{10}{3}}$ if $10-k \ge 3$, else 0. Thus, $P(k=1 \mid E) = \frac{\binom{9}{3}}{\sum_{k=0}^{7} \binom{10-k}{3}}$. Compute numerator: $\binom{9}{3} = 84$. Denominator: $\sum_{j=3}^{10} \binom{j}{3} = \binom{11}{4} = 330$ (using hockey-stick identity). Hence, $P = \frac{84}{330} = \frac{14}{55}$.