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.

  • 1
    $\begingroup$ Suppose you have $N$ independent, identically distributed Bernoulli random variables each with probability of success $p$. How would you determine the chance that at most two of the $N$ random variables is a success? If you assume volcanoes erupt independently and following some common distribution, you can use the calculation you've done above as the probability of "success". $\endgroup$ Oct 11, 2017 at 18:43
  • $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$ Oct 11, 2017 at 18:49
  • $\begingroup$ @Max, how do you find $N$ for the binomial distribution? I think use Poisson approach is a good one. $\endgroup$
    – Deep North
    Oct 11, 2017 at 22:53

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


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

  • $\begingroup$ Thank you for your answer. This actually agrees with the way one treats the one volcano eruption in 50 years if we view it as P(X>0)=1-P(X=0) which is what I got out of the integral. Sweet! Thanks again. $\endgroup$ Oct 11, 2017 at 21:57

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