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 p is

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answer is 8.

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Detailed Solution

Prime :2,3,5,7,11
1 2 4 6 2

$p (prime) = 15 / 36$

$Perfect square = 4, 9$

$P (perfect square) = 7 / 36$

required probability

$\begin{aligned} = \frac{\frac{4}{36} + \frac{14}{36} \times \frac{4}{36} + {(\frac{14}{36})}^{2} \frac{4}{36} + \dots}{\frac{7}{36} + \frac{14}{36} \times \frac{7}{36} + {(\frac{14}{36})}^{2} \frac{7}{36} + \dots} \\ P = 4 / 7 ∴ 14 P = 14 \cdot \frac{4}{7} = 8 \end{aligned}$

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