91Ó°ÊÓ

Let X =Time of arrival of 1^st person.

Y = Time of arrival of 2^nd person.

Now, we can say that X and Y are independent and uniformly distributed on the interval (5pm, 6pm).

Thus, the joint pdf of X and Y is \({f_{\left( {x,y} \right)}}\left( {x,y} \right) = \left\{ \begin{array}{l}1{\rm{ if 5 pm < x < 6 pm}}\\0{\rm{ elsewhere}}{\rm{.}}\end{array} \right.\)

Defining\(W = \left| {X - Y} \right|\)is the time one person will have to wait for the other one.

The W takes the values, w, between 0 and 1 hour.

The probability that a person have to wait more than 10 minutes is

\({\rm P}\left( {W > \frac{1}{6}} \right)\).

Thus, the probability that two persons will meet is

\(\begin{array}{c}1 - {\rm P}\left( {W > \frac{1}{6}} \right) = {\rm P}\left( {W \le \frac{1}{6}} \right)\\ = {F_W}\left( {\frac{1}{6}} \right)\end{array}\)

Now, to find the cdf of W

A is the event that is

\(A = \left\{ {\left( {x,y} \right) \in {\mathbb{R}^2}|5pm < x < 6pm,5pm < y,6pm,\left| {x - y} \right| \le w} \right\}\)

In the above figure above We can see the event A.

Area of A can be found by subtracting from 1 the area of the two triangles

\(\begin{array}{c}{\rm P}\left( {W \le w} \right) = 1 - {\left( {1 - w} \right)^2}\\ = 2w - {w^2}\end{array}\)

Hence,\({F_W}\left( w \right) = 2w - {w^2}for0 < w < 1.\)

We can conclude that the probability that two persons will meet is

\(\begin{array}{c}{F_W}\left( {\frac{1}{6}} \right) = 2 \times \frac{1}{6} - {\left( {\frac{1}{6}} \right)^2}\\ = \frac{{11}}{{36}}\end{array}\)