Question

L and M make an appointment to meet on 20th Nov. 2005 at the CAT examination centre, but without fixing anything other than that the appointment is between 10 a.m. to 11 a.m. They decide to wait no longer than 10 minutes for each other. Assuming that each is independently likely to arrive at any time during the hour, find the probability that they will meet.

• A.
  
• B.
  
• C.
  
• D. None of these

Answer 1

The correct option is C

There can be 2 approaches to this problem:
Approach 1
We can consider the entire time to be represented by the x-y plot: where on the x-axis we represent the time of one of them coming and on the y-axis the other person coming. L&M will meet provided |x-y|<=10 (x, y) lies on or inside the square with vertices 0(0, 0) A(60, 0) B(60, 60) and C (0, 60).
The region lying inside the square OABC and satisfying the inequality |x - y| > 0 consists two triangles DAE and GFC.
Sum of the areas of two triangles = $\frac{1}{2}$ (50)${}^2$ + $\frac{1}{2}$ (50)${}^2$ = (50)${}^2$
Probability = 1 - $\frac{{50}^2}{{60}^2}$ = 1 - $\frac{25}{36}$ = $\frac{11}{36}$

Alternate approach:

The probability that M will not meet L is $\left(\frac{50}{60}\right)$ (as he cannot meet him any way outside the 10 min he is present at the venue)
Similarly the probability that L will not meet M is also $\left(\frac{50}{60}\right)$
Hence the combined probability that they will not meet is $\frac{{50}^2}{{60}^2}$
Hence the probability that they will meet will be = 1 - $\frac{{50}^2}{{60}^2}$ = 1 - $\frac{25}{36}$ = $\frac{11}{36}$