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. A ball is drawn at random from box 1 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 option(s) with the possible values of `n_1 and n_2` , is(are)A. `n_(1)=4, n_(2)=6`B. `n_(1)=2, n_(2)=3`C. `n_(1)=10, n_(2)=20`D. `n_(1)=3, n_(2)=6`

Let `n_1, and n_2`, be the number of red and black balls, respectively

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 number one red and b of red and black balls, respectively, in box II. A ball is drawn at random from box 1 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 option(s) with the possible values of `n_1 and n_2` , is(are)A. `n_(1)=4, n_(2)=6`B. `n_(1)=2, n_(2)=3`C. `n_(1)=10, n_(2)=20`D. `n_(1)=3, n_(2)=6`
Answer» Correct Answer - C::D Consider the following events: `E_(1)`=Transferring a red ball from box-I to box-II `E_(2)`=Transferring a black ball from box-I to box-II A=Drawing a red ball from box-I, after transferring a ball from box-I to box-II By, addition theorem of probability `P(A)=(n_(1))/(n_(1)+n_(2))xx(n_(1)-1)/(n_(1)+n_(2)-1)+(n_(2))/(n_(1)+n_(2))xx(n_(2))/(n_(1)+n_(2))xx(n_(1))/(n_(1)+n_(2)-1)` `implies (1)/(3)=(n_(1))/(n_(1)+n_(2))xx(n_(1)-1)/(n_(1)+n_(2)-1)+(n_(2))/(n_(1)+n_(2))xx(n_(1))/(n_(1)+n_(2)-1)` We observe that the values in options (c ) and (d) satisfy the above relation. Hence, options (c ) and (d) are correct.

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