Assume that at a bus stop, $n$ different bus lines arrive. They follow no reliable plan and the
arrival time of each bus line $X_i$
is exponentially distributed with rate $λ_i$ > 0. Note that the $X_i$ are
independent.
Answer (and prove the answer for) the following.
$(a)$ Define a RV for the arrival time of the next bus arriving and calculate its distribution.
What is the rate of the next bus arriving?
$(b)$ What is the probability of bus line k arriving first?
$(a)$ So I define a RV as X $\sim$ Exp($\lambda$). The cumulative probability distribution $F(x) = 1-e^{-\lambda x}$. Is this equal to the distribution of X?
And the rate of the next bus arriving should be the minimum of X. The minimum of 2 RVs is $min=\{X_1,X_2\}$ = $1-e^{-x(\lambda_1 + \lambda_2)}$. Is the minimum of all $X_i$ then $1-e^{-x\sum_i \lambda_i}$ ?
If $X\sim \mathrm{Expo}(\lambda)$ and $Y\sim\mathrm{Expo}(\mu)$ are independent, then for $t>0$ we have $$ \{X\wedge Y>t\} = \{X>t\}\cap\{Y>t\}, $$ so \begin{align} \mathbb P(X\wedge Y>t) &= \mathbb P(\{X>t\}\cap\{Y>t\})\\ &= \mathbb P(X>t)\mathbb P(Y>t)\\ &= e^{-\lambda t}e^{-\mu t}\\ &= e^{-(\lambda+\mu)t}, \end{align} from which it follows that $X\wedge Y\sim\mathrm{Expo}(\lambda+\mu)$. By induction, if $X_1,\ldots,X_n$ are independent exponentially distributed random variables with respective parameters $\lambda_1,\ldots,\lambda_n$, we have $$ Z:= \bigwedge_{i=1}^n X_i \sim \mathrm{Expo}\left(\sum_{i=1}^n \lambda_i\right). $$ The distribution function of $Z$ is then $$ F_Z(t) = 1-e^{-\left(\sum_{i=1}^n \lambda_i\right)t }. $$ The rate of the next bus arriving is $\sum_{i=1}^n \lambda_i$.
For part (b), if $X\sim \mathrm{Expo}(\lambda)$ and $Y\sim\mathrm{Expo}(\mu)$ are independent, then \begin{align} \mathbb P(X<Y) &= \int_{\{(x,y):0\leqslant x<y<\infty\}} f_{X,Y}\ \mathsf d(x\times y)\\ &= \int_0^\infty\int_0^y \lambda e^{-\lambda x}\mu e^{-\mu y}\ \mathsf dx\ \mathsf dy\\ &= \frac\lambda{\lambda+\mu}. \end{align} The event that bus $k$ arrives first is $\{X_k<Z_k^-\}$, where $$ Z_k^- := \bigwedge_{i=1,i\ne k}^n X_i, $$ and so the probability is $$ \mathbb P( X_k<Z_k^-) = \frac{\lambda_k}{\sum_{i=1}^n \lambda_i}. $$