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Experiment and Outcomes

Let a1, a2,...

Question

Let $a_{1}, a_{2}, a_{3}$ are three consecutive terms of an increasing $A . P .$ , where $a_{1}$ and $a_{2}$ 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 & $P (R)$ represents probability of choosing red ball & similarly $P (G)$ represents the probability of choosing green ball.
On the basis of above information answer the following:
If $P (i) \propto i^{2}$ and one ball is drawn from one of these urns then -

P (G) = \frac{3}{7}

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P (R) = \frac{6}{7}

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P (R) > P (G)

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P (R) = P (G)

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Solution

The correct option is B $P (R) > P (G)$
$a_{1}, a_{2}, a_{3}$ are in A.P.
$a_{1}, a_{2}$ are prime numbers such that their sum is minimum possible odd prime number
Urn $1 : a_{1}$ Red, $a_{3}$ Green
Urn $2 : a_{2}$ Red, $a_{2}$ Green
Urn $3 : a_{3}$ Red, $a_{1}$ Green
$P (i) \propto i^{2}$
$P (i) = k i^{2}$
$P (1) + P (2) + P (3) = 1$
$k + 4 k + 9 k = 1$
$k = 1 / 14$
$P (U r n 1) = 1 / 14, P (U r n 2) = 4 / 14, P (U r n 3) = \frac{9}{14}$
$P (G) = \frac{1}{14} \times \frac{a_{3}}{a_{1} + a_{3}} + \frac{4}{14} \times \frac{a_{2}}{2 a_{2}} + \frac{9}{14} \times \frac{a_{1}}{a_{1} + a_{3}}$
$= \frac{1}{14} [\frac{a_{3} + 9 a_{1}}{a_{1} + a_{3}}] + \frac{1}{7}$
$= \frac{1}{14} [\frac{a_{3} + a_{1} + 8 a_{1}}{a_{1} + a_{3}}] + \frac{1}{7}$
$= \frac{1}{14} [1 + \frac{8 a_{1}}{2 a_{2}}] + \frac{1}{7} = \frac{1}{14} + \frac{2}{14} + 4 \frac{a_{1}}{a_{2}}$
$= \frac{3}{14} + 4 \frac{a_{1}}{a_{2}} [\frac{a_{1}}{a_{2}} < 1]$
$P (R) = \frac{1}{14} \times \frac{a_{1}}{a_{1} + a_{3}} + \frac{4}{14} \times \frac{a_{2}}{2 a_{2}} + \frac{9}{14} \times \frac{a_{3}}{a_{1} + a_{3}} - \frac{2}{14} + \frac{1}{14} [\frac{a_{1} + 9 a_{3}}{a_{1} + a_{3}}]$
$P (R) = \frac{2}{14} + \frac{1}{14} + \frac{8 a_{3}}{2 a_{2}} = \frac{3}{14} + 4 \frac{a_{3}}{a_{2}} [\frac{a_{3}}{a_{2}} > 1]$
$∴ P (R) > P (G)$

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