Answer: Option C
:
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 vertices0(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 - $\frac{25}{36}$ = $\frac{11}{36}$