Two people plan to meet to go to the bar. Each of them arrives at a time uniformly distributed between midnight and 1am and independently of the other. Denote $\displaystyle X$ (respectively $\displaystyle Y$) the random variable representing the arrival time of the first person (respectively, the second). The joint probability distribution is given by:

$\displaystyle f_{(X,Y)}{(x,y)}=\left\{\begin{array}{cc}2,&\mbox{ if}\ 0\leq x\leq y\\0, &\mbox{otherwise}\end{array}\right.$

(a)Find the probability that the first person is waiting for his friend for more than 10 minutes.

(b)Determine the marginal probability density functions of $\displaystyle X$ and $\displaystyle Y$. Check that thery are indeed probability density functions.

What I did so far is $\displaystyle P(x>10)=1-P(x\leq10)$

Then I use formula $\displaystyle F_{X}{(x)}=\int_{0}^{x} f(y) dy$

And the formula $\displaystyle f_{Y}{(y)}=\int_{-\infty}^{\infty}f_{(X,Y)}{(x,y)}dx$

But I end up $\displaystyle F_{X}{(x)}=x^2$, there for, the probability is no longer less than 1.

I really don't know what to do now, I thought I got the right formulae, can anyone help me please. It is same for part (b), I kind of get something, use formulae, but as question required, when I check my answer, I realised the integration of density functions don't give me 1, which means I must be wrong.

Please help me. Thanks a lot.