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)

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$n_{1} = 12, n_{2} = 12, n_{3} = 10, n_{4} = 30$

$n_{1} = 10, n_{2} = 12, n_{3} = 10, n_{4} = 12$

$n_{1} = 7, n_{2} = 14, n_{3} = 11, n_{4} = 55$

$n_{1} = 14, n_{2} = 16, n_{3} = 10, n_{4} = 20$

answer is A, C.

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Detailed Solution

$P (Re d B a l l) = P (I) . P (R | I) + P (I I) . P (R | I I) P (I I | R) = \frac{1}{3} = \frac{P (I I) . P (R | I I)}{P (I) . P (R | I) + P (I I) . P (R | I I)} \frac{1}{3} = \frac{\frac{n_{3}}{n_{3} + n_{4}}}{\frac{n_{1}}{n_{1} + n_{2}} + \frac{n_{3}}{n_{3} + n_{4}}} \frac{n_{1}}{n_{1} + n_{2}} = \frac{2 n_{3}}{n_{3} + n_{4}}$