Two people wish to explore Asia, where both tourists have a lifetime that is exponentially distributed with mean $\theta$. $T$ is the length of time that Asia is being explored by at least one of the tourists.
How do I find the cdf of the distribution given the following: 1) Both tourists leave the hotel at time $t = 0$, where their lifetimes are independently distributed. 2) The second tourist leaves the hotel on the event of the first tourist becoming sick.
The more difficult part of the problem is translating the "words" into technical language.
For 1), let $X$ be the lifetime of Tourist A, and $Y$ the lifetime of Tourist B. Exploration continues until both die. If $U$ is the length of time that exploration continues, then $U=\max(X,Y)$.
The probability that $U\le u$ is the probability both are dead by time $U$. The density function of $X$ is $\frac{1}{\theta}e^{-x/\theta}$ (for $\theta\gt 0$). It follows that the probability A is dead by time $u$ is $\int \frac{1}{\theta}e^{-x/\theta}\,dx$. This is $1-e^{-u/theta}$ (for $u\gt 0$). We get the same expression for $\Pr(Y\le u)$. Thus by independence $$F_U(u)=(1-e^{-u/\theta})^2$$ for $u\gt 0$, and $F_U(u)=0$ elsewhere.
2) suppose that Tourist A leaves first. Then exploration time $V$ is given by $V=X+Y$. To find $\Pr(V\le v)$ for $v\gt 0$, we integrate the joint density function over the part of the first quadrant that is below the line $x+y=v$. Thus for positive $v$ we have $$F_V(v)=\int_{x=0}^v \left(\int_{y=0}^{v-x}\frac{1}{\theta^2}e^{-x/\theta}e^{-y/\theta}\,dy\right)\,dx.$$
The integration is straightforward. For generalizations, please see the gamma distribution, or more specifically the Erlang distribution.