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)

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


Let A = Drawing red ball
$\therefore P\left(A\right)=P\left(B_1\right).P\left(A|B_1\right)+P\left(B_2\right).P\left(A|B_2\right)=\frac{1}{2}\left(\frac{n_1}{n_1+n_2}\right)+\frac{1}{2}\left(\frac{n_3}{n_3+n_4}\right)$
Given, $P\left(B_2|A\right)=\frac{1}{3}$
$\Rightarrow \frac{P(B_2\cap A)}{P\left(A\right)}=\frac{1}{3}\Rightarrow \frac{P\left(B_2\right).P\left(A|B_2\right)}{P\left(A\right)}=\frac{1}{3}\Rightarrow \frac{\frac{1}{2}\left(\frac{n_3}{n_3+n_4}\right)}{\frac{1}{2}\left(\frac{n_1}{n_1+n_2}\right)+\frac{1}{2}\left(\frac{n_3}{n_3+n_4}\right)}=\frac{1}{3}$
options A and B satisfy this condition.