An urn contains 3 red balls and n white balls. Mr. A draws two balls together from the urn. The probability that they have the same color is 1/2. Mr. B draws one balls form the urn, notes its color and replaces it. He then draws a second ball from the urn and finds that both balls have the same color is 5/8. The possible value of n is

In the first case, the urn contains 3 red and n white balls. The probability that color of both the balls matches is

$\frac{{ }^3{\mathrm{C}}_2{+}^{\mathrm{n}}{\mathrm{C}}_2}{{ }^{\mathrm{n}+3}{\mathrm{C}}_2}=\frac{1}{2}$

or   $\frac{6+\mathrm{n}(\mathrm{n}-1)}{(\mathrm{n}+3)(\mathrm{n}+2)}=\frac{1}{2}$

$\begin{array}{l}\mathrm{or} 2\left({\mathrm{n}}^2-\mathrm{n}+6\right)={\mathrm{n}}^2+5\mathrm{n}+6\\ \mathrm{or} {\mathrm{n}}^2-7\mathrm{n}+6=0\\ \Rightarrow \mathrm{n}=1\text{ or }6 \left(1\right)\end{array}$

In the second case,

$\frac{3}{\mathrm{n}+3}\frac{3}{\mathrm{n}+3}+\frac{\mathrm{n}}{\mathrm{n}+3}\frac{\mathrm{n}}{\mathrm{n}+3}=\frac{5}{8}$

Solving, we get

${\mathrm{n}}^2-10\mathrm{n}+9=0$

$\Rightarrow \mathrm{n}=9\text{ or }1$.                        (2)

From Eqs. (1) and (2), we have n = 1.