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 correct options (s) with the possible values of `n_(1) and n_(2)` is (are)

Options

A. `n_(1)=3,n_(2)=3, n_(3)=5, n_(4)=15`
B. `n_(1)=3, n_(2)=6, n_(3)=10, n_(4)=50`
C. `n_(1)=8, n_(2)=6, n_(3)=5, n_(4)=20`
D. `n_(1)=6, n_(2)=12, n_(3)=5, n_(4)=20`

Answer

A, B

Explanation

Consider the following events :

`E_(1)`= Selecting Box-I, `E_(2)`=Selecting Box-II
A=Getting a red ball.
Clearly,
`P(E_(1))=P(E_(2))=(1)/(2),P(A//E_(1))=(n_(1))/(n_(1)+n_(2)),P(A//E_(2))=(n_(3))/(n_(3)+n_(4))`
It is given that `P(E_(2)//A)=(1)/(3)`
`implies ((1)/(2)xx(n_(3))/(n_(3)+n_(4)))/((1)/(2)xx(n_(1))/(n_(1)+n_(2))+(1)/(2)xx(n_(3))/(n_(3)+n_(4)))=(1)/(3)`
`implies ((n_(3))/(n_(3)+n_(4)))/((n_(1))/(n_(1)+n_(2))+(n_(3))/(n_(3)+n_(4)))=(1)/(3)`
We observe that values in options (a) and (b) satify the above relation. Hence, options (a) and (b) are correct.