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

Let `n_(1)and n_(2)` be the number of red and black balls, respectivel

1.	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)=4andn_(2)=6`B. `n_(1)=2andn_(2)=3`C. `n_(1)=10andn_(2)=20`D. `n_(1)=3andn_(2)=6`
Answer» Correct Answer - C::D Given `(n_(1))/(n_(1)+n_(2)).(n_(1)-1)/(n_(1)+n_(2)-1)+(n_(2))/(n_(1)+n_(2)).(n_(1))/(n_(1)+n_(2)-1)=1/3` `implies3(n_(1)^(2)-n_(1)+n_(1)n_(2))=(n_(1)+n_(2))(n_(1)+n_(2)-1)` `implies3n_(1)(n_(1)+n_(2)-1)=(n_(1)+n_(2))(n_(1)+n_(2)-1)` `implies2n_(1)=n_(2)`

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