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 $II$.

One of the two boxes, box $I$ and box $II$, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $II$ is $\frac{1}{3}$, then the correct option(s) with the possible values of $n_1,n_2,n_3$ and $n_4$ 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. $n_1=6,n_2=12,n_3=5,n_4=20$

Answer 1

Answer: (A), (B)

$n_1=3,n_2=6,n_3=10,n_4=50$

Box $I:$
Red balls $-n_1$ and black balls $-n_2$
Box $II:$
Red balls $-n_3$ and black balls $-n_4$
Probability of drawing a red ball $=P\left(R\right)$
$\therefore P\left(R\right)=\frac{1}{2}\cdot \frac{n_1}{n_1+n_2}+\frac{1}{2}\cdot \frac{n_3}{n_3+n_4}$

Now probability that box $II$ was selected,
$P\left(II/R\right)=\frac{P(II\cap R)}{P\left(R\right)}\Rightarrow \frac{\frac{1}{2}\cdot \frac{n_3}{n_3+n_4}}{P\left(R\right)}=\frac{1}{3}\Rightarrow \frac{\frac{n_3}{n_3+n_4}}{\frac{n_1}{n_1+n_2}+\frac{n_3}{n_3+n_4}}=\frac{1}{3}\Rightarrow \frac{\frac{n_1}{n_1+n_2}}{\frac{n_3}{n_3+n_4}}+1=3\Rightarrow \frac{2n_3}{n_3+n_4}=\frac{n_1}{n_1+n_2}$