Question

An urn contains three red balls and $n$ white balls. Mr. $\text{Roushan}$ draws two balls together from the urn. The probability that he has the same colour is $\frac{1}{2}$. Mr. $\text{Rajesh}$ draws one ball from the urn, notes its colour and replaces it. He then draws a second ball from the urn and finds the probability that both balls have the same colour is $\frac{5}{8}.$ The possible value of $n$ is

Answer 1

There are $n$ white balls in the urn.
Probability of Mr. $\text{Roushan}$ to draw two balls of same color is $\frac{{}^3{C}_2{+}^n{C}_2}{{}^{n+3}{C}_2}$
$\Rightarrow \frac{6+n(n-1)}{(n+3)(n+2)}=\frac{1}{2}\text{(given)}$
$\Rightarrow n^2-7n+6=0$
$\Rightarrow n=1\text{or}6\cdots \left(i\right)$
Also required probability for Mr. $\text{Rajesh}$according to the question is
$\frac{3}{n+3}\frac{3}{n+3}+\frac{n}{n+3}\frac{n}{n+3}=\frac{5}{8}\text{(given)}$
Solving, we get
$n^2-10n+9=0\Rightarrow n=1\text{or}9\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$ we get
$n=1$