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Geometric Probability

Hardmathematics

Two friends agree to meet between five and six o'clock, each arriving at a uniformly random time and waiting fifteen minutes for the other; what is the probability that the two friends actually meet?

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About This Question

Subject
mathematics
Chapter
statistics and probability
Topic
geometric probability
Difficulty
Hard
Year
2025
Tags
advanced-calculus-drillgeometric-probabilitycontinuous-sample-spacearea-ratiomeeting-problem

Solution

Correct Answer:

Geometric probability replaces counting with measuring areas when outcomes are continuous and uniformly distributed, so probability equals favourable area divided by total area. This area-based modelling is a hallmark of harder JEE Advanced probability. Let the two arrival times x and y, in minutes after five o'clock, range over the 60 × 60 square of area 3600. The friends meet when their arrival times differ by at most 15 minutes, that is |x − y| ≤ 15. The non-meeting region consists of two right triangles where |x − y| > 15, each with legs of length 60 − 15 = 45, so each triangle has area (1/2)(45)(45) = 1012.5, totalling 2025. The favourable meeting area is 3600 − 2025 = 1575, giving probability 1575/3600 = 7/16. Option 1/4 wrongly squares the single ratio 15/60. Option 9/16 reports the complement, the non-meeting probability. Option 1/2 ignores the geometry entirely. The method uses the uniform-area definition of probability over the square sample space. Plausibility check: 7/16 ≈ 0.44 is below one-half, sensible because a 15-minute window within a 60-minute interval gives less than even odds of overlap.

This hard difficulty mathematics question is from the chapter statistics and probability, covering the topic of geometric probability. It appeared in the 2025 exam.

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