Question

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 colour is $1/2$.
Mr. B draws one ball from the urn, notes its colour and replaces it. He then draws a second ball from the urn and finds that the probability that both balls have the same colour is. $5/8$. The possible value of $n$ is

• A. $9$
• B. $6$
• C. $5$
• D. $1$

Answer 1

The correct option is D $1$

P (drawing two balls of same colour) = $\frac{{}^3{C}_2}{{}^{n+3}{C}_2}\frac{{}^n{c}_2}{{}^{n+3}{C}_2}=\frac{3+n(n-1)}{(n+3)(n+2)}$

$\Rightarrow 12+2n^2-2n=n^2+5n+6\Rightarrow n^2-7n+6=0\Rightarrow n=6_1$__ (1)

From case 2

P (drawing two balls of same colour ) = 5/8

$\Rightarrow \frac{\frac{{}^3{C}_1{\times }^3{C}_1}{(n{+}^3{C}_1{)}^2}+\frac{{}^n{C}_1{\times }^n{C}_1}{(n{+}^3{C}_1{)}^2}}{\frac{{(}^3{C}_1{)}^2+{(}^n{C}_1{)}^2+3n+3n}{{(}^{n+3}{C}_1{)}^2}}=\frac{9+n^2}{9+n^2+6n}=\frac{5}{8}$

$\Rightarrow 72+8n^2=45+5n^2+30n$

$\Rightarrow 3n^2-30n-27=0\Rightarrow n^2-10n-9=0$

$\Rightarrow n=9,1$ __ (2)

From (1) and (2), we have

$n=1$