I have the following problem and I having trouble in finding it solution. I need a hint.
The problem:
Two people arranged to meet between 12:00 and 13:00. The arriving time of each one is i.i.d. and follows an uniform distribution. Given that someone has already arrived, find the probability that the waiting time is at least 30 minutes.
My attempt
I was trying to calculate that by doing that but it's wrong
P(X$\ge$0.5) = $\int_.5^\infty$ $\frac{1}{0.5} dx$
Can someone help me in solving that question?
Best Answer
Let $\ X,Y \tilde{} \mathcal{U} (\lbrack 0,1\rbrack) $ be uniformly distributed on $\lbrack0,1\rbrack$. 0 for 12.00, 1 for 13.00. Now we want to compute the probability that $X-Y \geq 0.5$. For this we will need to find the distribution function of the $Z=X-Y$. We can do this by convolution. For further details please check, http://www.math.wm.edu/~leemis/chart/UDR/PDFs/StandarduniformStandardtriangular.pdf
Then we will get, $ f_Z(z) = \left\lbrace \begin{array}{c} z+1 \quad -1\leq z\leq0 \\ 1-z \quad 0<z\leq1 \end{array} \right. $
Now you just have to compute, $ \mathbb{P}(Z\geq 0.5) = \int_{0.5}^1 1-z \ \mathrm{d}z = \frac{1}{8} $.
Hope this helps.