Classical Probability
Two fair dice are rolled together once, and the recorded sum of the two top faces turns out to be a prime number; what is the probability of this event?
Select the correct option:
Solution
5/12
Classical probability assigns equal weight to each outcome in a finite, symmetric sample space, so the probability of an event is the count of favourable outcomes divided by the total count. Rolling two distinguishable fair dice produces 6 × 6 = 36 equally likely ordered pairs, which is the standard JEE setup for sum-based events. The possible sums range from 2 to 12, and the prime sums in that range are 2, 3, 5, 7, and 11. Counting ordered pairs giving each: sum 2 occurs 1 way, sum 3 occurs 2 ways, sum 5 occurs 4 ways, sum 7 occurs 6 ways, and sum 11 occurs 2 ways, totalling 1 + 2 + 4 + 6 + 2 = 15 favourable outcomes. Hence the probability is 15/36 = 5/12. Option 1/2 wrongly includes sum 9 as prime. Option 7/18 = 14/36 omits one favourable pair, typically the sum-2 outcome. Option 1/3 = 12/36 forgets the sum-11 contributions. The governing principle is the equally-likely-outcomes definition of probability over the 36-element sample space. Plausibility check: 15 lies between 0 and 36 and 5/12 ≈ 0.417 is a sensible probability strictly between 0 and 1, confirming the count.
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About This Question
- Subject
- mathematics
- Chapter
- statistics and probability
- Topic
- classical probability
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
5/12
Classical probability assigns equal weight to each outcome in a finite, symmetric sample space, so the probability of an event is the count of favourable outcomes divided by the total count. Rolling two distinguishable fair dice produces 6 × 6 = 36 equally likely ordered pairs, which is the standard JEE setup for sum-based events. The possible sums range from 2 to 12, and the prime sums in that range are 2, 3, 5, 7, and 11. Counting ordered pairs giving each: sum 2 occurs 1 way, sum 3 occurs 2 ways, sum 5 occurs 4 ways, sum 7 occurs 6 ways, and sum 11 occurs 2 ways, totalling 1 + 2 + 4 + 6 + 2 = 15 favourable outcomes. Hence the probability is 15/36 = 5/12. Option 1/2 wrongly includes sum 9 as prime. Option 7/18 = 14/36 omits one favourable pair, typically the sum-2 outcome. Option 1/3 = 12/36 forgets the sum-11 contributions. The governing principle is the equally-likely-outcomes definition of probability over the 36-element sample space. Plausibility check: 15 lies between 0 and 36 and 5/12 ≈ 0.417 is a sensible probability strictly between 0 and 1, confirming the count.
This easy difficulty mathematics question is from the chapter statistics and probability, covering the topic of classical probability. It appeared in the 2025 exam.
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