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

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Detailed Solution

Prime/ Perfect square Prime →2,3,5,7,11(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2)(6,1),(5,2),(4,3),(1,6),(2,5),(3,4),(5,6),(6,5) Square =4,9→(1,3),(3,1),(2,2),(6,3),(3,6),(5,4),(4,5) (Perfect square before prime )=736+1536736+15362736+…… P (Perfect square is odd) =436+1536436+15362436+…….=4/361−15367/361−1536=47⇒14p=47×14=8

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