Question

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. $n_1=4$ and $n_2=6$
• B. $n_1=2$ and $n_2=3$
• C. $n_1=10$ and $n_2=20$
• D. $n_1=3$ and $n_2=6$

Answer 1

Answer: (C), (D)

The correct options are
C $n_1=3$ and $n_2=6$
D $n_1=10$ and $n_2=20$

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 $II$.
Let $E_3$ be the event that transferred ball is red and $E_4$ be the event that transferred ball is black.
$R$ be the event of selecting a Red ball.
$P\left(R\right)=P\left(E_3\right).P\left(R/E_3\right)+P\left(E_4\right).P\left(R/E_4\right)$
$\Rightarrow P\left(R\right)=\frac{1}{3}=\frac{n_1}{n_1+n_2}.\left(\frac{n_1-1}{n_1+n_2-1}\right)+\frac{n_2}{n_1+n_2}.\left(\frac{n_1}{n_1+n_2-1}\right)=\frac{n_1}{n_1+n_2}$
only options $C$ and $D$ satisfies the above relation.