$a_{1}, a_{2}, a_{3}$ are three consecutive terms of an increasing A.P., where $a_{1} a n d a_{2}$ are prime numbers such that their sum is minimum possible odd prime number.
Urn-1: Contains $a_{1}$ red and $a_{3}$ green balls
Urn-2: Contains $a_{2}$ red and $a_{2}$ green balls
Urn-3: Contains $a_{3}$ red and $a_{1}$ green balls
$p (i)$ represents the probability of choosing $i^{t h}$ urn and $P (R)$ represents probability of choosing red ball and similarly $P (G)$ represents the probability of choosing green ball.

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If $P (i) \propto i^{2}$ and one ball is drawn from one of these urns then $P (R) = 6 / 7$

If $P (i) = 1 / 3$ for all $i = 1, 2, 3 a n d 2$ balls are drawn randomly from one of these urns then the chance of drawing balls of different colours is $\frac{5}{9}$

If $P (i) \propto i^{2}$ and one ball is drawn from one of these urns then $P (R) > P (G)$

If $P (i) \propto i^{2}$ and one ball is drawn from one of these urns then $P (G) = 3 / 7$

answer is A, D.

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Detailed Solution

A,B,C) $a_{1} = 2, a_{2} = 3, a_{3} = 4$
If $P (i) \propto i^{2}$
$P (1) = \frac{1}{14}, P (2) = \frac{4}{14}, P (3) = \frac{9}{14} P (R) = \frac{1}{14} . \frac{2}{6} + \frac{4}{14} . \frac{3}{6} + \frac{9}{14} . \frac{4}{6} = \frac{25}{42} P (G) = \frac{1}{14} . \frac{4}{6} + \frac{4}{14} . \frac{3}{6} + \frac{9}{14} . \frac{2}{6} = \frac{17}{42}$