Find probability with the help of Poisson distribution

Question

There are $3$ cyclone on average in a year in russia .

What is the probability that there will be cyclone in the next TWO years ?

I just want to know the value of $\lambda$ in Poisson distribution! As we know $\lambda$ actually works on the same time interval. My calculations: as it's Two years, $\lambda = 3\cdot 2 = 6$.

But my confusion is should we calculate the present year too? I mean if we don't calculate this year we can calculate as time = $2$ years otherwise $3$ ...

Is my calculation ok?

Answer 1

$\lambda=6$ thus your probability is

$$P[X>0]=1-e^{-6}\approx 99.75\%$$

Answer 2

I suppose that your question means that there is some misunderstanding in the question about taking $2 \lambda$: otherwise there would be no questions.

Let $\xi$ be the number of cyclones during the first year and $\eta$ is the number of cyclones during the second year. Thus $\xi \sim Pois(\lambda)$, $\eta \sim Pois(\lambda)$. We implicitly suppose that $\xi$ is independent of $\eta$ - otherwise this approach will not work.

$E \xi = \lambda =3$.

We use the next fact: the sum of independent r.v. with distributions $Pois(\lambda_1)$ and $Pois(\lambda_2)$ has distribution $Pois(\lambda_1 + \lambda_2)$. Hence $\xi+ \eta \sim Pois(2 \lambda)$.

Hence $P(\xi + \eta > 0) = 1 - e^{-6}$ is the probability of at least one cyclone and $P(\xi + \eta = 1) = 3e^{-6}$ is the probability that there will be exactly one cyclone.