Question

A bag contains $N$ balls, some of which are white, the others are black, white being more in number than black. Two balls are drawn at random from the bag, without replacement. It is found that the probability that the two balls are of the same colour is the same as the probability that they are of different colour. It is given that $180<N<220$. If $K$ denotes the number of white balls, find the exact value of $(K+N)$.

Answer 1

Answer: 301

$\text{ Given }\frac{{}^K{C}_2+{}^{N-K}{C}_2}{{}^N{C}_2}=\frac{{}^K{C}_1{}^{N-K_C}{C}_1}{{}^N{C}_2}$
$K(K-1)+(N-K)(N-K-1)=2K(N-K)$
$K^2-K+N^2+K^2-2NK-N+K=2NK-2K^2$
$4K^2-4NK+\left(N^2-N\right)=0$
$k=\frac{4N\pm \sqrt{16N^2-16\left(N^2-N\right)}}{8}=\frac{4N\pm 4\sqrt{N}}{8}=\frac{N+\sqrt{N}}{2}\text{ or }\frac{N-\sqrt{N}}{2}$
$\text{ Hence }N\text{ must be a perfect square }$
$\Rightarrow N=196$
$\text{ [As }N\in (180,220)]$
$K=\frac{196\pm 14}{2}=105\text{ or }91$
$\text{ as }k>N-K\Rightarrow 2K>N\Rightarrow K>\frac{196}{2}\Rightarrow K>98$
$\text{ Hence }K=105\text{ and }N=196$
$\therefore N+K=301$