# Probability question with multiple random variables

Suppose Frederika arrives at bus stop at time, S, uniformly distributed on the interval [ − 5, 5]. Suppose that the bus departs from the bus stop at a time, T, which is normally distributed with mean 0 and standard deviation 4 (S and T are measured in minutes). Suppose S and T are independent. Find the probability that Frederika catches the bus. This is the probability of the event (S < T). Hint: Integrate by parts to do an integral of the form $\int F_T (s)ds$, where F_T is the cumulative distribution of T. Also note that if the bus departs after t=5, then Frederika will catch the bus.

I'm not sure how to get the cumulative distribution of T and then be able to integrate it with respect to s.

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Sketch the region of the plane where the point $(S,T)$ can be. Then figure out the joint density of $S$ and $T$ (hint: they are independent and $T$ is normal so you should know its pdf and CDF) and integrate the joint density over the region $\{(s,t) : s < t\}$. Try not to set up the integrals without drawing a sketch first. –  Dilip Sarwate Nov 30 '11 at 3:18
@DilipSarwate What would I graph for T? I know I need to find the area under T in [-5,5], but I'm not sure what I need to graph for the normal distribution. –  thezboe Nov 30 '11 at 3:31
The joint density $f_{S,T}(s,t)$ is a function of two variables and is a surface above the plane with coordinate axes $s$ and $t$. Mark on the plane all points $(s,t)$ where $f_{S,T}(s,t) > 0$. Hint: This set of points is not the entire plane. Mark all points $(s,t)$ where $s < t$. You need to integrate (double integral!) the joint density over this entire second region but the integrand is zero over parts of the region and so you can simplify the limits a little bit if you have the two regions described in front of you. Doing it by sheer brainpower is not easy for beginners. –  Dilip Sarwate Nov 30 '11 at 3:55
I calculated the CDF to be $\rho_{S,T}(s,t)=\frac{1}{40\sqrt{2\pi}}e^{-t^2/32}$. Is this correct? –  thezboe Nov 30 '11 at 4:45