Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If p is the probability that this perfect square is an odd number, then the value of 14p is ______.

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answer is 8.

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Detailed Solution

Prime (2,3,5,7,11)=

$\{(1, 1), (1, 2), (2, 1), (1, 4), (2, 3) (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)\}$

n(odd prime) = 14

$∴ P (o d d p r i m e) = \frac{14}{36}$

Perfect square $= (4, 9) = \{(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)\}$

n(perfect square) = 7

$∴ P (p e r f e c t s q u a r e) = \frac{7}{36}$ and P(odd perfect square) $= \frac{4}{36}$ Required probability

$= \frac{\frac{4}{36} + \frac{14}{36} \times \frac{4}{36} + {(\frac{14}{36})}^{2} \frac{4}{36} + ...}{\frac{7}{36} + \frac{14}{36} \times \frac{7}{36} + {(\frac{14}{36})}^{2} \frac{7}{36} + ...} = P = \frac{4}{7}$

$∴ 14 P = 14. \frac{4}{7} = 8$

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