I'm trying to solve the problem:

Given that an corporation gets usually 360 e-mails in each 6 hours of work what is the probability that in 10 minutes the corporation will get at least 2 e-mails?

The problem asks to model that question using Poisson probability distribution ($x\in \mathbb{N}$)

$$P(x) = e^{-\lambda}\frac{\lambda^x}{x!}$$

Then I tried to find lambda. I think lambda should be the mean value of e-mails that the corporation receives so $\lambda = 1$; this is the number of e-mails per minute of work. Now we get that the probability that at least two e mails are get it is the sum of probabilities up to $x=2$:

$$P(0)+P(1)+P(2) = e^{-1}+e^{-1}+e^{-1}/2 = 5e^{-1}/2 = 0,91969$$

This makes sense because the probability must be very high. But it is not the correct answer given by the professor: $0,9995$.

Where I'm wrong?

Was my approach to the problem correct?


The probability of at least two is the probability of receiving two, or three, or four, etc. You computed the probability of at most two.

Now, you can do it as $$P(X\geq2)=P(X=2)+P(X=3)+\dots=\sum_{x=2}^\infty P(X=x),$$

and compute the infinite sum, or do it by finding the complementary event of "at least 2", which is "at most one":


You found the correct rate, $\lambda=1$ messages/minute. But you are asked about a 10 minute period. For any Poisson process, the number of arrivals in a period $[0,t]$ has the following probability: $$P(X=x)=\frac{e^{-\lambda t}(\lambda t)^x}{x!}.$$

So in your case, with $t=10$ and $\lambda=1$, $$P(X=0)=\frac{e^{-10}(10)^0}{0!}=e^{-10}=0.00004539992,$$ $$P(X=1)=\frac{e^{-10}(10)^1}{1!}=10e^{-10}=0.00045399929.$$

So $$P(X\geq2)=1-0.00004539992-0.00045399929=0.99950060077.$$


You have two errors.

Your $\lambda$ is wrong. The expected rate of one per minute is correct, but we have a run of ten minutes, so the expected number is ten.

Your computation adding the values for $0,1,2$ computes the chance that the corporation will get (in your model) at most $2$ e-mails, not at least $2$ e-mails. You should add the probabilities for $0$ and $1$, then subtract the sum from $1$ to get the chance of at least two.


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