JEE Advanced 2020 · Paper 2 · Mathematics Q15

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MathematicsNumerical+4Question 51 of 54

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 and worked solution

Official answer: 8

Possible sums.

Primes: 2, 3, 5, 7, 11.
Perfect squares: 4 and 9 (the two sets do not overlap).

Probabilities per roll. $P(4)=\frac3{36}$ and $P(9)=\frac4{36}$ .

Conditioning. Given that the process stops on a square, the stopping roll is a square, and it is 9 with probability
$p=\frac{4/36}{7/36}=\frac47.$
So $14p=8$ .

Answer: 8

Topics:Conditional probability Classical probability

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JEE Advanced 2020 · Paper 2, Mathematics question 15