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Let n(1)and n(2) be the number of red an...

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)

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`

Text Solution

Verified by Experts

The correct Answer is:
B

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.
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