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 number one red and b of red and black balls, respectively, in box II. A ball is drawn at random from box 1 and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is `1/3` then the correct option(s) with the possible values of `n_1 and n_2` , is(are)

Options

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

Answer

D

Explanation

`therefore P("drawing red ball from" B_(1))=(1)/(2)`
`rArr ((n_(1)-1)/(n_(1)+n_(2)-1))((n_(1))/(n_(1)+n_2))+((n_(2))/(n_(1)+n_(2)))((n_(1))/(n_(1)+n_(2)-1))=(1)/(3)`
`rArr " " (n_(1)^(2)+n_(1)n_(2)-n_(1))/((n_(1)+n_(2))(n_(1)+n_(2)-1))=(1)/(3)`
Clearly, options (c) and (d) satisfy.