Tricky Poisson distribution question Am new to probability and have been struggling to solve the following question.

Assume $X_1, \ldots,X_4$ are IID with $X_i ∼ \text{Po}(λ)$. Let $Y = \frac{1}{4} (X_1 + \ldots + X_4)$.
  Find $\text{Pr}(Y < 2)$ in terms of $\lambda$.

Can anyone explain the relevance of the random variables being IID? this seems to be putting me off.
 A: "IID" stands for "independent, identically distributed", but the "identically distributed" part isn't really that important here (outside of convenience), at least in the sense that one could ask a similar question with different $\lambda$'s.
However, the independent part is critical, because it allows you to apply the result that a sum of independent Poisson random variables has a Poisson distribution, with mean equal to the sum of the component means. (When you don't have independence you don't generally have that a sum of Poissons is Poisson; an obvious exception is the case where all the $X_i$'s are equal.)
So rather than throwing you off, it makes the problem much easier than if the variables were dependent.
You can see that independence makes a difference by also considering the case where $X_1=X_2=X_3=X_4$.
[There's another neat trick that makes answering easier still. You're asked for $P(Y<2)$, but $Y=\overline{X}$, so $Y$ is not Poisson. The trick is to convert the event into an equivalent event that is easy to work with, given the abovementioned result.]
A: The sum of independent Poisson variables is itself Poisson distributed. In particular, if $X_1 \sim Poisson(\lambda_1)$ is independent of $X_2 \sim Poisson(\lambda_2)$, then the sum $X_1 + X_2 \sim Poisson(\lambda_1 + \lambda_2)$. Thus, the sample sum $S_n$ of $n$ independent and identical $Poisson(\lambda)$ variables:
$$S_n  = \sum _{i=1}^n X_i \quad \sim \quad Poisson(n \lambda) $$
with pmf, say, $f(s)$:

(source: tri.org.au)
We can then find a general solution to $P(\frac{S}{n} < 2)$ not just for $n = 4$, but for any value of $n$, and for any desired probability (whether < 2 or a general parameter):

(source: tri.org.au)
where I am using the Prob function from the mathStatica package for Mathematica to compute, and Gamma[a,z] is the incomplete gamma function.
This is the same as the solution presented by David Stork, but more compact, and more general (solved for general $n$). For comparative purposes, when $\lambda = 1$ and $n=4$, the above probability is:

0.948866 ...

A: Probability[x1 + x2 + x3 + x4 < 8,
 {x1 \[Distributed] PoissonDistribution[\[Lambda]],
  x2 \[Distributed] PoissonDistribution[\[Lambda]],
  x3 \[Distributed] PoissonDistribution[\[Lambda]],
  x4 \[Distributed] PoissonDistribution[\[Lambda]]}]

$
\frac{1}{315} e^{-4 \lambda } \left(1024 \lambda ^7+1792 \lambda ^6+2688 \lambda ^5+3360
   \lambda ^4+3360 \lambda ^3+2520 \lambda ^2+1260 \lambda +315\right)
$
