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 color is$\frac{1}{2}$ . 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 colour is $\frac{5}{8}$. The possible value of n is

• A. 9
• B. 6
• C. 5
• D. 1

Answer 1

The correct option is D 1

In the first case, the urn contains 3 red and n white balls. The probability that colour of both the balls matches is $\frac{{}^3{C}_2{+}^n{C}_2}{(n+3)C_2}=\frac{1}{2}or\frac{6+n(n-1)}{(n+3)(n+2)}=\frac{1}{2}or2(n^2-n+6)n^2+5n+6or{n}^2-7n+6=0\Rightarrow n=1or\mathrm{6...................}\left(1\right)$ In the second case, $\frac{3}{n+3}\frac{3}{n+3}+\frac{3}{n+3}\frac{3}{n+3}=\frac{5}{8}$ Solving, we get $n^2-10n+9=0$ $\Rightarrow n=9or1$ From Eqs. (1) and (2), we have n =1.............(2)