NDA 1 2025 Maths, question 107

Solution.

Answer: (c) 1/4

1. $9\binom{6}{4}k^{4}(1 - k)^{2} = \binom{6}{2}k^{2}(1 - k)^{4}$, and $\binom{6}{4} = \binom{6}{2} = 15$.
2. Cancel the common factors (with $0 < k < 1$): $9k^{2} = (1 - k)^{2}$, so $3k = 1 - k$.
3. $k = 1/4$.

Trap: $k = 0$ and $k = 1$ also make both sides zero, but a binomial probability is taken strictly between 0 and 1.