How do I calculate probability of a number of events happening in a time period? Would appreciate some help with a question. Before anyone asks, it's not homework and I'm not a professional maths person, only someone who is somewhat inquisitive!
I have a series of probabilities related to "events" or "incidents" that may occur on a particular day.
$P(n)$ is the probability of "n" incidents occurring on a particular day.
$P(0)=0.55, P(1)=0.28, P(2)=0.15, P(3)=0.01, P(4)=0.01$
I would like to calculate the probability of more than a total of "x" incidents occurring over a period of "y" days.
For instance, what is the probability of more than 20 incidents happening over a period of 30 days?
I have no clue how I would go about doing this. Can anyone suggest a method? 
 A: Assuming that the number incidents $N_i$ occuring on particular days are independent, the probability in question is
$$P(\sum_{i=1}^y N_i >x)=\sum_{k=x+1}^{4y} P(N_i =k). $$
For a given $k>x$ we have to find partitions of $k$ as $k= \sum_{j=0}^4 ja_j$ where $a_j$ denotes the number of days when $j$ incidents occured. For a given partition $(a_0,...,a_4)$ the probability is $\frac{y!}{a_0!a_1!...a_4!}p_0^{a_0}p_1^{a_1} ... p_4^{a_4}$. So
$$P(\sum_{i=1}^y N_i >x)= \sum_{k=x+1}^{4y} \sum_{\sum_{j=0}^4 ja_j=k}\frac{y!}{a_0!a_1!...a_4!}p_0^{a_0}p_1^{a_1} ... p_4^{a_4} $$
For general $x$ and $y$ this is not an easy task. 
A: Let's say $X_i$ is the number of incidents on day $i$, and $Y$ is the total number of incidents in $30$ days, so $Y=\sum_{i=1}^{30} X_i$.  We assume that the $X_i$'s are independent and all have the stated distribution.
By calculation from the probabilities provided, the mean of $X_i$ is $\mu_X=0.65$, and the variance is $\sigma^2_X =0.7075$. So the mean of $Y$ is $\mu_Y = 30 \mu_X = 19.5$, and the variance is $\sigma^2_Y = 30 \sigma^2_X = 21.225$. 
It seems reasonable to approximate $Y$ with a Normal distribution with mean $\mu_Y$ and variance $\sigma^2_Y$.  Then $Z= (Y-\mu_y)/\sigma_Y$ has a Normal(0,1) distribution, and we can find using tables of the normal distribution or by use of software that $P(Y < 20.5) \approx 0.586$, so $P(Y > 20.5) = 1 - P(Y < 20.5) \approx \boxed{0.414}$.
It is also possible to find an exact probability that $Y > 20$ using a probability generating function and a computer algebra system, such as Mathematica.  It turns out that the exact probability is $0.400276$ (to six  digits), so the Normal approximation is pretty good.  For those interested in such things, the probability generating function of $Y$ is
$$f(x) = \left(0.55\, +0.28 x+0.15 x^2+0.01 x^3+0.01 x^4\right)^{30}$$
The probability that $Y<=20$ is the sum of the coefficients of $x^n$ for $0 \le n \le 20$ when $f(x)$ is expanded.
If you are interested in learning about generating functions in general (including probability generating functions), a number of resources can be found in the answers to this question: How Can I Learn About Generating Functions?
