I'm waiting at a bus stop, I can take the bus 34, which I need to wait for, on average, 10 min, and the bus 45, which I need to wait for, on average 4 minutes. The times of arrival of the buses are assumed to be independent.
What is the probability that I would wait more than 5min.
My Attempt
I'm not sure how having two buses affect the probability, but I know I could represent them separately as exponential distributions.
Bus 34: Our exponential distribution parameter is $\lambda=\frac{1}{10}$. Our p.d.f. is:
\begin{equation} f(x)= \begin{cases} \frac{1}{10}e^{-\frac{x}{10}} &\text{if } x\geq0\\ 0 & \text{otherwise} \end{cases} \end{equation} Integrating from $0$ to $t$ gives us our c.d.f. $$F(t)=1-e^{-\frac{t}{10}}$$
Bus 45: Our exponential distribution parameter is $\lambda=\frac{1}{4}$. Our p.d.f. is:
\begin{equation} f(x)= \begin{cases} \frac{1}{4}e^{-\frac{x}{4}} &\text{if } x\geq0\\ 0 & \text{otherwise} \end{cases} \end{equation} Integrating from $0$ to $t$ gives us our c.d.f. $$F(t)=1-e^{-\frac{t}{4}}$$
Then the probability of waiting over five minutes for bus 34 is $$\mathbb{P}(X>5)=1-\mathbb{P}(x\leq 5)=1-F(5)=1-(1-e^{-\frac{5}{10}})=0.6065$$ and the probability of waiting over five minutes for bus 45 is $$\mathbb{P}(X>5)=1-\mathbb{P}(x\leq 5)=1-F(5)=1-(1-e^{-\frac{5}{4}})=0.2865$$
My thinking is that the probability that you wait more than $5$ minutes is the average of these two probabilities, but I was told that this is wrong. Can I get some pointers on how to approach this problem?