Combining two exponential distributions onto one graph In a probability question I am working through, the question says the length of time of a football game,  $s$ , is exponentially distributed with a mean of  $\bar s$ . 
It then says to write out the probability distribution for $s$, which I can do. 
The time after which the game is cancelled due to bad weather, $t$, is also exponentially distributed with a mean of $\bar t$, and  $t$ is independent of $s$ .
It then asks ''Draw a pair of axes at right-angles to each other labeling one $s$ and one $t$. Indicate on the diagram the region where the game is not interrupted by bad weather.
I am confused by this, as surely I need a third axis orthogonal to the $st$ plane in order to sketch the probability distribution for both of these distributions. If I have $t$ as the independent variable, and s as the dependent variable, I can sketch $t$ as just a general exponential distribution, but then what does the sketch of $s$ against $t$ look like ?
 A: EDIT: @Did's comment on the main question is the actual solution; I will leave this faulty answer to show how wrong you can go when not correctly understanding a question!
This is a bit of an odd question, because we generally do not work directly with random variables, but rather with their distributions. So I am not sure exactly how to state this problem formally. Nonetheless, I will lend my thoughts.
Let $\lambda = \frac1{\bar s}$ and $\mu=\frac1{\bar t}$, and $E$ denote the event "the game is not interrupted by bad weather". $E$ is simply the set $\{S<T\}$, and it is straightforward to compute this probability (assuming $S$ and $T$ are independent). Let $f_S$ and $f_T$ be the probability density functions of $S$ and $T$ respectively, then the joint density is equal to the product of the marginal densities, i.e. $f_{S,T}=f_Sf_T$. Hence
\begin{align}
\mathbb P(S<T) &= \iint\limits_{\{(s,t)\in\mathbb R^2\ :\ s<t\}} f_{S,T}\ \mathsf d(s\times t)\\
&= \int_0^\infty \int_0^t f_S(s)f_T(t)\ \mathsf ds\ \mathsf dt\\
&= \int_0^\infty \mu e^{-\mu t} \int_0^t \lambda e^{-\lambda s}\ \mathsf ds\ \mathsf dt\\
&= \int_0^\infty \mu e^{-\mu t}(1-e^{-\lambda t})\ \mathsf dt\\
&= \int_0^\infty \mu e^{-\mu t}\mathsf dt - \mu\int_0^\infty e^{-(\lambda+\mu) t}\ \mathsf dt\\
&= 1 - \frac\mu{\lambda+\mu}\\
&= \frac\lambda{\lambda+\mu}.
\end{align}
By symmetry it is clear that $\mathbb P(S<T)=\frac12$ if $\lambda=\mu$, in which case your plot would simply be of the set
$$\{(s,t)\in\mathbb R^2: 0<s<t\}. $$
In general, I believe you are being asked to plot the set
$$\left\{(s,t)\in\mathbb R^2: 0 < s < \frac\mu\lambda t \right\}.$$
My intuition is that for a fixed $\mu$, we have
\begin{align}
\lim_{\lambda\to0}\frac\lambda{\lambda+\mu}&=0\\
\lim_{\lambda\to\infty}\frac\lambda{\lambda+\mu}&=\infty,
\end{align}
and the plot with the slope $\frac\mu\lambda$ is a function that comes to mind that satisfies that property.
