Calculating the Probability that an event happens more than once with an exponential distribution

Question

I am told that the waiting time between now and the moment when the next volcano erupts follows an exponential distribution $$f(t) = \lambda e^{-\lambda t}$$ with parameter $\lambda = 0.1$ events per year. Let $X$ be the random variable representing the waiting time between now and the next eruption. My understanding is that in order to calculate the probability of the next volcano happening within the next 50 years is given by $$P(X<50) = \int_{0}^{50}e^{-\lambda x}dx = -(e^{-50\lambda}-1)\approx 0.99. $$ Now, I am asked to compute the probability that at most 2 volcanoes will erupt within the next 50 years. How do I accomplish this?

What I have tried is doubling the rate $\lambda$ to $2\lambda = 0.2$ for that would be the rate of 2 events per unit time. I obtained $P(X<50) = 0.99995$ which to me seems unrealistic. I don't know if this is a justifiable move. I don't have the answer to the problem, I just thought about it as I am studying for a midterm exam.

Answer 1

Here is how I would attempt this problem, using a Poisson process.

Since the waiting time between volcano eruptions is a random variable with an exponential distribution with rate $$\lambda $$ then if X is the number of volcano eruptions within unit time t, we have

$$ P[X=k] = \frac{(\lambda t)^k e^{-\lambda t}}{k!} $$

In other words, $$ \lambda t $$ can be viewed as the expected count of some event per unit length. In this case, $ \lambda $ is 0.10 volcano eruptions/year, and $t$ is one year.

So, every ten years we would expect a count of one volcano eruption. Now, set t = 50 years. Then we obtain $\lambda t = 0.10 * 50 = 5.$ In other words, in fifty years we would expect five volcano eruptions.

The desired probability is $$ P [X \le 2] = P[X = 0] + P[X=1] + P[X=2]$$.
$$P[X=0] = e^{-5} $$ $$P[X=1] = 5e^{-5}$$ $$P[X=2] = \frac{5^2 e^{-5}}{2!} = 12.5e^{-5} $$

So,

$$ P[X \le 2] = 18.5e^{-5} \approx 0.124652 $$