No of goals in a game of soccer using Poisson Suppose on average the number of goals in a soccer game follows a Poisson process with $\lambda = 4$ goals per game. In particular the distribution of the numbers of goals $X$ in one random game has density function
$$f_X(x) = P(X = x) = e^{-4}\frac{4^x}{x!}, \ \ x = 0, 1, 2,...$$ What is the probability of no goals in two games?
If it was the probability of no goals in one game, I would just sub in $0$ for the $x$'s for the answer. But how do you take account of the fact that it is over 2 games.
 A: We are expected to assume that if $X$ is the number of goals in Game A and $Y$ is the number of goals in Game B, then $X$ and $Y$ are independent.
Thus the probability that $X=0$ and $Y=0$ is the product of $\Pr(X=0)$ and $\Pr(Y=0)$. 
A: Andre's solution is great and I do not mean to step on any toes by extending this discussion further however I am a fan of the Poisson distribution and I think the following could be useful if you are going to be solving a lot of Poisson problems.
Another way of solving this problem (and others like it involving the Poisson distribution) is to write $Y = X_1 + X_2$ for the sum of the number of goals in games $1$ and $2$.  It turns out that the distribution of $Y$ is also Poisson with parameter $ \lambda_1 + \lambda_2$ where $\lambda_i$ is the parameter of $X_i's$ distribution.  In this case we see that the answer is $e^{-8}$ which agrees with Andre's solution.  This might not seem like a quicker way to solve the problem, but in general for: $$Y_n = \sum_{i=1}^{n}X_i \quad \ \quad X_i \sim Poisson(\lambda_i) \ , X_i \perp  X_j \ ;i \neq j$$ We get that: $$Y_n \sim Poisson(\sum_{i=1}^{n} \lambda_i)$$
You may immediately see the usefulness to this if you were instead asked: What is the probability that the same player above scores 15 goals in their first 10 games of the season?  Using the above technique you would need to calculate: $$\sum_{g_1 +\cdots +g_{10}=15}Pr(X_1=g_1)\cdots Pr(X_{10}=g_{10})$$
However with the knowledge of $Y_{10}'s$ distribution we can rattle off the answer quickly as: $$\frac{40^{15}e^{-40}}{15!} \approx 3.48\times 10^{-6}$$
We can also find all moments, cumulative probabilities, etc. very easily now as well.
