Q.7 A and B are friends. They decide to meet between 1 PM and 2 PM on a given day. There is a
condition that whoever arrives first will not wait for the other for more than 15 minutes. The
probability that they will meet on that day is
(A) 1/4 (B) 1/16 (C) 7/16 (D) 9/16
Probability That Two Friends Will Meet (15-Minute Waiting Rule)
Problems based on random arrival times and limited waiting conditions
are classic examples of geometric probability. This question analyzes
the probability that two friends meet when neither waits for more than 15 minutes.
Assumptions
- Arrival times are independent and uniformly distributed
- Time is measured in minutes after 1 PM
Variable Definition
Let:
- x = arrival time of A (0 ≤ x ≤ 60)
- y = arrival time of B (0 ≤ y ≤ 60)
Condition for Meeting
A and B will meet if the difference between their arrival times is not more than
15 minutes:
|x − y| ≤ 15
Sample Space
The sample space is a square of side 60 minutes.
Total area = 60 × 60 = 3600
Unfavorable Region
The unfavorable region consists of two right-angled triangles where
the arrival time difference exceeds 15 minutes.
Area of one triangle:
(1/2) × 45 × 45 = 1012.5
Total unfavorable area:
2 × 1012.5 = 2025
Favorable Region
Favorable area = Total area − Unfavorable area
= 3600 − 2025 = 1575
Probability Calculation
Probability that A and B meet:
P = 1575 / 3600
P = 7/16
Summary Table
| Region | Area |
|---|---|
| Total sample space | 3600 |
| Unfavorable region | 2025 |
| Favorable region | 1575 |
| Probability | 7/16 |
Final Answer
The probability that the two friends meet is 7/16.
