1. ## [SOLVED] jointly distribution

A man and a woman decide to meet at a certain location. If each of them independently arrives at a time uniformly distributed between 12 noon and 1 P.M., find the probability that the first to arrive has to wait longer than 10 minutes.

Let X and Y denote, respectively, the time past 12 that the man and the woman arrive, then X and Y are independently random variables, each of which is uniformly distributed over (0, 60). The desired probability,
P[X + 10 < Y] + P[Y + 10 < X]= 2P[X + 10 < Y]
$\displaystyle =2 \int_{x+10<y} \int f(x, y) dx dy$
$\displaystyle =2 \int_{x+10<y} \int f_X(x) f_Y(y) dx dy$
$\displaystyle =2 \int_{10}^{60} \int_{0}^{y-10} (\frac{1}{60})^{2} dx dy$
But why $\displaystyle \int_{10}^{60}$ and not $\displaystyle \int_{0}^{60}$ in the first part of the integral??

2. Hello,
Originally Posted by noob mathematician
A man and a woman decide to meet at a certain location. If each of them independently arrives at a time uniformly distributed between 12 noon and 1 P.M., find the probability that the first to arrive has to wait longer than 10 minutes.

Let X and Y denote, respectively, the time past 12 that the man and the woman arrive, then X and Y are independently random variables, each of which is uniformly distributed over (0, 60). The desired probability,
P[X + 10 < Y] + P[Y + 10 < X]= 2P[X + 10 < Y]
$\displaystyle =2 \int_{x+10<y} \int f(x, y) dx dy$
$\displaystyle =2 \int_{x+10<y} \int f_X(x) f_Y(y) dx dy$
$\displaystyle =2 \int_{10}^{60} \int_{0}^{y-10} (\frac{1}{60})^{2} dx dy$
But why $\displaystyle \int_{10}^{60}$ and not $\displaystyle \int_{0}^{60}$ in the first part of the integral??

more formally :

We know that X+10<Y
X can have values over 0 to 60
So we have 0<X<60
Then 10<X+10<70

Since Y>X+10>10, we have Y>10. This explains the lower bound.
The upper bound is 60 because Y has to be <60.

Looks clear ?