Let $n_{1} a n d n_{2}$ be the number of red and black balls respectively, in box 𝐼. Let $n_{3}$ and n₄ 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 1/3, then the incorrect option(s) with the possible values of $n_{1}, n_{2}, n_{3}$ and $n_{4}$ is (are)

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$n_{1} = 3, n_{2} = 3, n_{3} = 5, n_{4} = 15$

$n_{1} = 8, n_{2} = 6, n_{3} = 5, n_{4} = 20$

$n_{1} = 3, n_{2} = 6, n_{3} = 10, n_{4} = 50$

$n_{1} = 6, n_{2} = 12, n_{3} = 5, n_{4} = 20$

answer is A, B.

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Detailed Solution

Box I : red balls $\to n_{1}$ black balls $\to n_{2}$
Box I : red balls $\to n_{3}$ black balls $\to n_{4}$
P(R) = $\frac{1}{2} . \frac{n_{1}}{n_{1} + n_{2}} + \frac{1}{2} . \frac{n_{3}}{n_{3} + n_{4}}$
P(BOX II/R) = $\frac{\frac{1}{2} . \frac{n_{3}}{n_{3} + n_{4}}}{\frac{1}{2} . \frac{n_{1}}{n_{1} + n_{2}} + \frac{1}{2} . \frac{n_{3}}{n_{3} + n_{4}}} = \frac{1}{1 + (\frac{\frac{n_{1}}{n_{1} + n_{2}}}{\frac{n_{3}}{n_{3} + n_{4}}})} = \frac{1}{3}$
Thus, $\frac{n_{1}}{n_{1} + n_{2}} = 2 \frac{n_{3}}{n_{3} + n_{4}}$