Poisson Process - earthquakes question

Question

The earthquakes occur in a region in accordance with a Poisson Process, at a rate of seven per year. Assume that all required conditions for the Poisson process hold.

What is the probability that in exactly three of the next eight years no earthquakes will occur?

My answer: We look for parameter $\lambda$ for 3 years, so $\lambda_3 = 7*3 = 21$

Let X be no. of earthquakes that occur in 3 years, P(X=0) = e^(-$\lambda_3)$ = e^(-21)

Let Y be no. of years without an earthquake in 8 years, then Y is a binomial random variable, Y~Binom(N=8, p=e^(-21)).

Then, P(Y=3) = (8C3)((e^(-21))^3)(1-e^(-21))^5 = 2.4414*10^(-26)

May I know if this is correct?

Answer 1

You have the right idea, but the wrong application.

To understand where you went wrong, note that if $Y$ is the random number of calendar years (i.e. non-overlapping time intervals from January 1 to December 31) in which no earthquakes occur, then for $8$ such years, $Y$ has a binomial distribution with $n = 8$ and some probability $p$ that represents the probability that no earthquakes occur in one calendar year.

Think of it this way: suppose at the beginning of each year on January 1, you decide to count how many earthquakes happen during that year, and at the very end of December 31 of that year, you tally up how many occurred. If no earthquakes occurred, you consider that year a "success." Each year's outcome is an independent Bernoulli trial, and the total number of years in which no earthquakes occurred is a binomial random variable.

So, while you are correct that $Y$ is binomially distributed, you incorrectly determined the success parameter $p$. The correct value should be the probability that, in a single year, no earthquakes occurred. Since their occurrences follow a Poisson process with rate $\lambda = 7$ per year, the random number of earthquakes in a year is $$X \sim \operatorname{Poisson}(\lambda = 7).$$ Hence $$p = \Pr[X = 0] = e^{-7}.$$

Then the probability that we observed exactly $Y = 3$ years without earthquakes in an $8$-year period is simply $$\Pr[Y = 3] = \binom{8}{3} p^3 (1-p)^{8-3} = 56 e^{-21} (1 - e^{-7})^5.$$ This is a very small number--approximately $1$ in $23.66$ million, but nowhere near as small as the one you obtained.

Your solution is calculating something different, which is the probability that there are exactly three three-year periods in which no earthquakes occur over a total observation time frame of $24$ years.