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. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probablity that this red ball was drawn from box II is `1/3` then the correct option(s) with the possible values of `n_1, n_2, n_3, and n_4`, 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

B

Explanation

Let A = Drawing red ball
`therefore " " (A)=P(B_(1))*P(A//B_(1))+P(B_(2))*P(A//B_(2))`
`=(1)/(2)((n_(1))/(n_(1)+n_(2)))+(1)/(2)((n_(3))/(n_(3)+n_(4)))`
Given, `P(B_(2)//A)=(1)/(3)`
`rArr" " (P(B_(2))*P(B_(2)nnA))/(P(A))=(1)/(3)`
`rArr " " ((1)/(2)((n_(3))/(n_(3)+n_(4))))/((1)/(2)((n_(1))/(n_(1)+n_(2)))+(1)/(2)((n_(3))/(n_(3)+n_(4))))=(1)/(3)`
`rArr " " (n_(3)(n_(1)+n_(2)))/(n_(1)(n_(3)+n_(4))+n_(3)(n_(1)+n_(2)))=(1)/(3)`
Now, check options,then clearly options (a) and (b) satisfy.