JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Probability XII

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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 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

14 p

is ____

Answer: 8

Solution

The possible sums are 2 to 12. Prime sums: 2, 3, 5, 7, 11 (15 outcomes out of 36). Perfect square sums: 4 (3 outcomes), 9 (4 outcomes) — total 7 outcomes. The probability of getting a prime is

\frac{15}{36}

, perfect square is

\frac{7}{36}

, and neither is

\frac{14}{36}

. Given that a perfect square occurs before a prime, the probability that it's 9 (odd) is:

p = \frac{\frac{4}{36} + \frac{14}{36} \cdot \frac{4}{36} + ( \frac{14}{36} ) ^{2} \cdot \frac{4}{36} + \dots}{\frac{7}{36} + \frac{14}{36} \cdot \frac{7}{36} + ( \frac{14}{36} ) ^{2} \cdot \frac{7}{36} + \dots} = \frac{4/36}{1 - 14/36} \div \frac{7/36}{1 - 14/36} = \frac{4}{7}

Thus

14 p = 14 \cdot \frac{4}{7} = 8

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Probability XII

Solution

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