# Math Help - Probability Meeting Problem

1. ## Probability Meeting Problem

A difficult homework problem I received:
Two people meet between 1:00 and 2:00. The person that arrives first will wait for 15 minutes and then leave. Find the probability that the two actually meet, if arrival times are random.
Thanks.

2. Originally Posted by C^2
A difficult homework problem I received:
Two people meet between 1:00 and 2:00. The person that arrives first will wait for 15 minutes and then leave. Find the probability that the two actually meet, if arrival times are random.
Thanks.
Here is an outline of one possible solution:

$X_1$ ~ $U(0, 60)$ and $X_2$ ~ $U(0, 60)$.

Get the pdf for $U= X_1 - X_2$.

Calculate $\Pr(-15 < U < 15)$.

3. The question only asked to find the probability of this event.
So, I wouldn't find the density of $U_1-U_2$.
Instead I would just integrate the region in $R^2$, where
$-15. Naturally, one must assume that these rvs are independent.
But we must becareful that if one person arrives at 1:05, then the other person must arrive between 1 and 1:20.
So drawing this region in $R^2$ is quite important.

4. And you need not integrate this. Just draw [0,60]x[0,60] and figure out your region. The third dimension, f(u1,u2)= 1/(60)^2. But it's easier to get the the complement, which consists of two triangles, that are equal in probability.

5. Originally Posted by mr fantastic
Another approach (which boils down to the same thing as meymathis did):
Originally Posted by mr fantastic

Let X be the random variable number of minutes after noon that person X arrives. X ~ U(0, 60).
Let Y be the random variable number of minutes after noon that person Y arrives. Y ~ U(0, 60).
1. When $0 \leq X \leq 15$ you require $0 \leq Y \leq X + 15$ for the two people to meet.
2. When $15 \leq X \leq 45$ you require $X - 15 \leq Y \leq X + 15$ for the two people to meet.
3. When $45 \leq X \leq 60$ you require $X - 15 \leq Y \leq 60$ for the two people to meet.
On the XY-plane draw the square bounded by the lines X = 0, Y = 0, X = 60 and Y = 60. The area of this square is 3600 square minutes.
Cases 1-3 above define areas inside this square (unit is square minutes) for which the two people will meet:
1. 675/2
2. 900
3. 675/2
Total area = 1350 square minutes.
Therefore the probability of the two people meeting is 1350/3600 = 7/16.
Note: This geometric approach only works because X and Y follow uniform distributions.

Mr. Fantastic, can you please explain how you arrived at the individual areas?