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 (a)`9` b. `6` c. `5` d. 1

Answer

1

Explanation

There are n white balls in the turn.
`implies` Probability of Mr. A to draw two balls of same color is
`(""^(3)C_(2)+""^(n)C_(2))/(""^(n+3)C_(2))=1/2("given")`
`or (6+n(n-1))/((n+3)(n+2))=1/2`
`or n^(2)-7n+6=0`
`impliesn=1or6" "(1)`
Also required probability for Mr. B according to the question is
`(3)/(n+3)(3)/(n+3)+(n)/(n+3)(n)/(n+3)=5/8"(given)"`
Solving, we get `n^(2)-10n +9=0,n=1or9" "(2)`
From (1) and (2), `n=1`