Calculating probability of a poisson process over a different interval

Question

If I can model an event with a Poisson distribution such as:

"A large asteroid hits the earth once every 100,000,000 years"

how do I calculate the probability of a single successful event with a different interval?

For example:

"what is the probability of a large asteroid hitting the earth in the next 100 years?"

Answer 1

"A large asteroid hits the earth once every 100,000,000 years"

Standing to this sentence, you must define the hitting rate $\lambda$ (number of hit in a year) as:

$$\lambda = \frac{1}{100000000} = 10^{-8}.$$

If the process is Poisson, then:

$$P(n, t) = \frac{e^{-\lambda t}(\lambda t)^n}{n!}$$

is the probability to have exactly $n$ hits in $t$ years.

"what is the probability of a large asteroid hitting the earth in the next 100 years?"

This probability is:

$$p = \sum_{n=1}^{+\infty} P(n,100),$$

since you want the earth to be hit, no matter how many times (it can be $1$, $2, \ldots, +\infty$ times).

For the properties of the probability density distribution, you have that:

$$ \sum_{n=0}^{+\infty} P(n,100) = 1,$$

and hence

$$P(0,t) + \sum_{n=1}^{+\infty} P(n,t) = 1\Rightarrow P(0,100) + p = 1 \Rightarrow p = 1- P(0,100),$$

which is much easier than evaluating $\sum_{n=1}^{+\infty} P(n,100).$

To conclude:

$$P(0,100) = \frac{e^{-10^{-8} \cdot 100}(10^{-8} \cdot 100)^0}{0!} = \frac{e^{-10^{-6}} \cdot 1}{ 1} = e^{-10^{-6}},$$

and

$$p = 1- e^{-10^{-6}}.$$

Answer 2

Since we want to model it with a Poisson process, let $N_t$ be the number of asteoroids hitting Earth between time $0$ and time $t$. Then its intensity should be $\lambda = 10^{-8} \, \mathrm{years}^{-1}$, so that $\mathbb E N_{10^8} = 1$ - the average number of asteroids in $10^8$ years is equal one.

The probability you want to find is $$ \mathbb P(N_{100} \ge 1) = 1 - \exp(-100 \, \mathrm{years} \cdot \lambda) = 1 - e^{-10^{-6}} \approx 10^{-6}. $$