Mathematics · Probability

JEE Advanced 2020 — Paper 2 — Question 48

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

Numerical answer — enter this value.

Step-by-step solution

Possible sums from two dice range from 2 to 12. Sums that are perfect squares: 4 and 9. Outcomes for sum 4: $(1,3), (2,2), (3,1)$ — 3 outcomes. Outcomes for sum 9: $(3,6), (4,5), (5,4), (6,3)$ — 4 outcomes. Total outcomes for perfect square: $3+4=7$ . Among these, only sum 9 is odd, giving 4 favorable outcomes. Since we condition on the sum being a perfect square (and no overlap with prime sums), the required probability is $p = \frac{4}{7}$ . Thus, $14p = 14 \times \frac{4}{7} = 8$ .

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Mathematics
Chapter: Probability
Topic: Introduction to Probability

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