An urn contains three 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 from 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 _________.

There are n white balls in the turn.

$\Rightarrow$ Probability of Mr. A to draw two balls of same color is

$\frac{{ }^3{\mathrm{C}}_2{+}^{\mathrm{n}}{\mathrm{C}}_2}{{ }^{\mathrm{n}+3}{\mathrm{C}}_2}=\frac{1}{2}\text{ (given) }$

$\begin{array}{l}\text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{6+\mathrm{n}(\mathrm{n}-1)}{(\mathrm{n}+3)(\mathrm{n}+2)}=\frac{1}{2}\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{n}}^2-7\mathrm{n}+6=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{n}=1\text{ or }6 \left(1\right)\end{array}$

Also required probability for Mr. B according to the question is

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

Solving, we get ${\mathrm{n}}^2-10\mathrm{n}+9=0;\mathrm{n}=1\text{ or }9$            (2)

From (1) and (2), n = 1