Discussion Forum

Let n₁  and n₂  be the number of red and black balls. respectively in box I. Let  n₃ and n₄  be the number of red and black balls respectively in box II.

One of the two boxes, box I and box II a 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 n₁,n₂,n₃  and n₄ is/are

Answer:

Explanation:

Let A= Drawing red ball

  $\therefore$        $ P(A)=P(B_{1}).P(A/B_{1})+P(B_{2}).P(A/B_{2})$

$=\frac{1}{2}\left(\frac{n_{1}}{n_{1}+n_{2}}\right)+\frac{1}{2}\left(\frac{n_{3}}{n_{3}+n_{4}}\right)$

 Given      $ P(B_{2}/A)$=$\frac{1}{3}$

$\Rightarrow$             $\frac{P(B_{2}).P(B_{2}\cap A)}{P(A)}=\frac{1}{2}$

$\Rightarrow$         $ \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}$

$\Rightarrow$           $\frac{n_{3}(n_{1}+n_{2})}{n_{1}(n_{3}+n_{4})+n_{3}(n_{1}+n_{2})}=\frac{1}{3}$

 Now, check options, then clearly options (a) and (b) satisfy.