Multiple choice

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 $i^{th}$ 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 -

  1. $P(G) = \dfrac{3}{7}$
  2. $P(R) = \dfrac{6}{7}$
  3. $P(R) > P(G)$
  4. $P(R) = P(G)$
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C Correct answer
Explanation

a1, a2, a3 are consecutive primes with minimum sum as odd prime. Primes are 2, 3, 5. a1=2, a2=3, a3=5. Urn probabilities are proportional to i^2 (1:4:9). Total weight = 14. P(R) = (1/14 * 2/7) + (4/14 * 3/6) + (9/14 * 5/7) = 2/98 + 2/14 + 45/98 = (2+14+45)/98 = 61/98. P(G) = 1 - 61/98 = 37/98. P(R) > P(G).

AI explanation

Using the minimum odd prime sum condition for a prime increasing arithmetic progression, the terms are 3, 5, and 7. Using the proportionality P(i) is proportional to i squared, the probabilities for urns 1, 2, and 3 are 1 divided by 14, 4 divided by 14, and 9 divided by 14, respectively. The total probability of drawing a red ball, calculated by the law of total probability, is (1/14)(3/10) plus (4/14)(1/2) plus (9/14)*(7/10), which equals 0.65. The total probability of drawing a green ball is 1 minus 0.65, which equals 0.35. Since 0.65 is greater than 0.35, the probability of drawing a red ball is greater than the probability of drawing a green ball.

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