Probability of two people meeting Problem:

Two people are to arrive at the same location some time between 3pm
and 4pm. They both arrive at random times within the hour and only
stay for ten minutes. Since they only go one time, what is the
probability that they will meet during that hour?

I have found the correct answer to this problem here (the second formula P2):
http://www.mathpages.com/home/kmath124/kmath124.htm

$$ P2 = 1 - \frac{(1-w_1)^2 + (1-w_2)^2}{2} = w_1 + w_2 - \frac{w_1^2}{2} - \frac{w_2^2}{2} $$
Plugging in ($1/6$) for both $w_1$ and $w_2$ is $11/36$ which is the correct answer.
But I don't understand how they came to that conclusion and what formula they derived it from. Any help in explaining this would be appreciated!
 A: The side of the square is one, so its surface $= 1 * 1 = 1$.
The side of the upper white square-cornered triangle is $1 - w_1$, so its surface is $(1-w_1)*(1-w_1)/2$.
Similarly with the bottom triangle.
So: the surface of the gray area = square - two white triangles = formula $P2$.
Finally note that each 'point' in the square is equally likely to occur, so that the area of the 'valid' points (the grey area) divided by the area of 'all' points (the white square) is the probability of a 'valid' event (i.e. the two persons meeting). Since the surface of the square is 1, the result is still formula $P2$.
A: First of all this is a very good example of application of geometric probability. It's related to continuous distribution.
The idea behind taking the 1×1 square is as below. Imagine a square with both axes as time.let the time be 1 hour each.this means the area is now 1hour ×1 hour.Let the 2 people be A and B  . If A arrives at time t=0second  he waits for 10 minutes.If he arrives at t=0.001second ,he waits till 10mins,0.001 seconds. similarly if he comes at t=1min he waits till 11 mins and so on...lets associate A with horizontal time axis and B with vertical time axis.now imagine a vertical slot of 10minutes moving along x axis as time passes.To be clear, the slot is a vertical bar,moving along x axis as time.so, now we are clear that A arrives at any instant in the big square frame of 1hour ,waits for 10 minutes.
Similar is the case with B also.This time consider a horizontal bar of 10 minutes moving upwards as time passes. This means B also arrives at some instant within 1 hour, waits for 10 minutes.
Now the task is to see where is possibility of the two people meeting in this frame of 1 hour × 1 hour.
When we move the horizontal and vertical bar together we get an area that is common to both the bars .this represents the time at which they are together. The common region we obtain by moving the 2 bars is diagonal as u can see in one of the above links.
Finding probability is now easy as we know the favourable area and the total area.
Probability= favourable area/total area
A: Beside the geometric interpretation, the probability that the two people meet in the hour is given by the probability that the second person arrives in a suitable interval ($\pm 10$ minutes) conditioned to the time of arrival of the first person:
$$p(meet) = p(y \in \Delta T_x | x) \cdot p(x) =  p(y \in \Delta T_x) \cdot p(x),$$
where we can assume the independence between the arrival times $x$ and $y$ and we define $\Delta T_x $ as the time window suitable for meeting:
$$\Delta T_x = [max(0, x-10), min(x+10, 60)].$$
As depicted in the figure, $\Delta T$ is symmetric and it is constant except in the first and the last 10 minutes within the hour, where it exhibits a linear behaviour.

Putting all together, and considering three separate time intervals for people to meet (0-10 min, 10-50, 50-60 min), the overall probability is given by:
$$p(meet) = p(y \in \Delta T_x)\cdot p(x) = p(y \in \Delta T_x)\cdot p(x \in [0,10]) + \\
+ p(y \in \Delta T_x)\cdot p(x \in [10,50]) + p(y \in \Delta T_x)\cdot p(x \in [50,60])= \\
= \frac{1}{60}\cdot\int_0^{10}\frac{10+x}{60}dx + \frac{2}{3}\cdot\frac{20}{60} +
\frac{1}{60}\cdot\int_{50}^{60} \frac{10+x}{60}dx = \frac{2}{60}\cdot\int_0^{10} \frac{10+x}{60}dx + \frac{2}{3}\cdot\frac{20}{60} = \\
= \frac{1}{30}\cdot\frac{150}{60} + \frac{2}{9} = \frac{11}{36}.$$
A: First of all, we can see how many different ways the first person can arrive
i.e., 
          if he comes at 3pm then he goes again at 3 10 pm, similarly 
  3 01 pm to 3 11 pm
  3 02 pm to 3 12 pm   ...................and so on
  ...
  ...
  ...
  3 50 pm to 4 00 pm

Therefore, he can arrive in 51 ways if we calculate the above combinations, and similarly the second person can also arrive in 51 ways.
So, the combinations of arrival of both the persons is 51*51 and in so much above ways there will be 51 ways in which both will arrive the same time
Therefore, probability of arriving at the same time:  $\frac{51}{51*51}$, hence probability = $\frac{1}{51}$
