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

Options

A. 9
B. 6
C. 5
D. 1

Answer

D

Explanation

In the first case, the urn contains 3 red and n white balls. The probability that color of both the balls matches is
`(""^(3)C_(2)+""^(n)C_(2))/(""^(n+3)C_(2))=1/2`
`or(6+n(n-1))/((n+3)(n+2))=1/2`
`or2(n^(2)-n+6)=n^(2)+5n+6`
`orn^(2)-7n+6=0`
`impliesn=1 or 6" "(1)`
In the second case,
`(3)/(n+3)(3)/(n+3)+(n)/(n+3)(n)/(n+3)=5/8`
Solving, we get
`n^(2)-10n +9=0" "(2)`
`implies n=9 or 1.
From Eqs. (1) and (2), we have n=1.