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

`Red ton_(1)`
` "Box" IltBlack ton_(2)`
`Red to n_(3)`
`"Box II"lt"Black" to n_(4)`
`P(R)=1/2.(n_(1))/(n_(1)+n_(2))+1/2.(n_(3))/(n_(3)+n_(4))`
`P (II//R) =(1/2(n_(3))/(n_(3)+n_(4)))/(1/2(n_(1))/(n_(1)+n_(2))+1/2.(n_(3))/(n_(3)+n_(4)))`
`((n_(3))/(n_(3)+n_(4)))/((n_(1))/(n_(1)+n_(2))+(n_(3))/(n_(3)+n_(4)))`
by option `n_(1)=3,n_(2)=3,n_(3)=5,n_(4)=15`
`P(II//R)=(5/20)/(3/6+5/20)=(1/4)/(1/2+1/4)=1/4xx(4)/(2+1)=1/3`
Also, when `n_(1)=3, n_(2),=6,n_(3)=10, n_(4)=50,P(II//R)=1//3`