Let `n_(1)and n_(2)` be the number of red and black balls, respectively, in box I. Let `n_(3) and n_(4)` be the numbers of red and black balls, respectively, in the box II.
A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the possible values of `n_(1) and n_(2)` is (are)

Options

A. `n_(1)=4andn_(2)=6`
B. `n_(1)=2andn_(2)=3`
C. `n_(1)=10andn_(2)=20`
D. `n_(1)=3andn_(2)=6`

Answer

C, D

Explanation

Given `(n_(1))/(n_(1)+n_(2)).(n_(1)-1)/(n_(1)+n_(2)-1)+(n_(2))/(n_(1)+n_(2)).(n_(1))/(n_(1)+n_(2)-1)=1/3`
`implies3(n_(1)^(2)-n_(1)+n_(1)n_(2))=(n_(1)+n_(2))(n_(1)+n_(2)-1)`
`implies3n_(1)(n_(1)+n_(2)-1)=(n_(1)+n_(2))(n_(1)+n_(2)-1)`
`implies2n_(1)=n_(2)`