Question

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 of red and black balls, respectively in 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 $\frac{1}{3}$, then the correct option(s) with the possible values of $n_1$ and $n_2$ is/are

• A. $n_1=4and{n}_2=6$
• B. $n_1=2and{n}_2=3$
• C. $n_1=10and{n}_2=20$
• D. $n_1=3and{n}_2=6$

Answer 1

Answer: (C), (D)

$n_1=3and{n}_2=6$


$\therefore$ P(drawing red ball from $B_1)=\frac{1}{3}$
$\Rightarrow \left(\frac{n_1-1}{n_1+n_2-1}\right)\left(\frac{n_1}{n_1+n_2}\right)+\left(\frac{n_2}{n_1+n_2}\right)\left(\frac{n_1}{n_1+n_2-1}\right)=\frac{1}{3}\Rightarrow \frac{n_1^2+n_1{n}_2-n_1}{(n_1+n_2)(n_1+n_2-1)}=\frac{1}{3}$
Clearly, options C. and D. satisfy.