I am trying to do the following question:
Number 24 and number 42 buses arrive independently at the corner of
Mayeld Road at a random rate of 3 and 4 per hour respectively. You
arrive at the bus stop at 10 a.m. What is the probability that it
takes at least 20 minutes for a bus (either the 24 and 42 bus) to
arrive?
I am kinda of stuck on how to start it. So I've learned about Uniform distribution, normal distribution and exponential distribution. I actually not really sure which one to use here. Could you also explain what sort of problems to apply each distribution to?
If I set up a random variable, what should it denote? The number of minutes pass 10 for a bus to arrive? Or that the 42 and 24 bus arrive?
Best Answer
You're looking for the waiting time distribution for either bus, which is exponential. This is because you know the bus has equal probability of arriving at any given time in an hour, and that the distribution is memoryless. So the waiting time for a bus has density $f(t)=\lambda e^{-\lambda t}$, where $\lambda$ is the rate. To understand the rate, you know that $f(t)dt$ is a probability, so $\lambda$ has units of $1/[t]$. Thus if your bus arrives $r$ times per hour, the rate would be $\lambda=r$. Since the expectation of an exponential distribution is $1/\lambda$, the higher your rate, the quicker you'll see a bus, which makes intuitive sense.
So define $X=\min(B_1,B_2)$, where $B_1$ is exponential with rate $3$ and $B_2$ has rate $4$. It's easy to show the minimum of two independent exponentials is another exponential with rate $\lambda_1+\lambda_2$. Thus you want:
$$P(X>\mbox{20 minutes}) = P(X> 1/3)=1-F(1/3),$$
where $F(t)=1-e^{-t(\lambda_1+\lambda_2)}$.