Task about probability (frequency of event) I have the following task:
The phone rings $12$ times a week. For each ringing there's an equal probability to happen in each of the week days. What is the probability for the phone to ring at least $1$ time per day?
The answer is $p=0.2285$, but I don't know how we get it. Is this a Poisson distribution or not? Can anybody tell me how do we get this answer?
 A: The probability that a given ring will occur on a given day is $\frac{1}{7}$. Let $R_i$ be the number of rings on day $i\in [1...7]$.
The event in question is that each day has at least 1 ring. We can formalize this event as the complement of the event where at least one day has zero calls:
$$A:=\left(\bigcup_{i=1}^7 \{R_i=0\}\right)^c$$
Thus:
$$P(A)=1-P\left(\bigcup_{i=1}^7 \{R_i=0\}\right)$$
Using the Inclusion-Exclusion Formula for unions of events, we get:
$$P\left(\bigcup_{i=1}^7 \{R_i=0\}\right)=\sum_i P(R_i=0)-\sum_{i<j} P(R_i=0\cap R_j=0)+...+P\left(\bigcap_{i=1}^7 \{R_i=0\}\right)$$
For brevity, lets redefine each term on the RHS as:
$$S_1=\sum_i P(R_i=0),S_2=\sum_{i<j} P(R_i=0\cap R_j=0),...,S_7=P\left(\bigcap_{i=1}^7 \{R_i=0\}\right)$$
So our RHS becomes a simpler formula:
$$P\left(\bigcup_{i=1}^7 \{R_i=0\}\right)=\sum_{i=1}^7 (-1)^{i+1}S_i$$
This looks somewhat forbidding, but your underlying probabilities are simple, since each call's arrival is independent of the others and identically distributed, so the intersections (joint probabilities) are pretty simple to write out. Let $K$ be a subset of the numbers $[1...7]$, where $|K|$ denotes the number of elements in $K$, then the probability that each of the days in $K$ has no calls is:
$$P\left(\bigcap_{i\in K}\{R_i=0\}\right)=\left(\frac{7-|K|}{7}\right)^{12}$$
Now, for any given value for $|K|$ there are $7 \choose |K|$ ways to pick which $|K|$ days will not get any calls. We now can make a nice formula for $S_n$:
$$S_n={7\choose n}\left(\frac{7-n}{7}\right)^{12}$$
And our overall formula becomes:
$$P\left(\bigcup_{i=1}^7 \{R_i=0\}\right)=\sum_{i=1}^7 (-1)^{i+1}{7\choose i}\left(\frac{7-n}{7}\right)^{12}$$
Using brute calculations, I get:
$$P\left(\bigcup_{i=1}^7 \{R_i=0\}\right) \approx 0.77154 \implies P(A)=1-P\left(\bigcup_{i=1}^7 \{R_i=0\}\right) \approx 0.2285$$
This matches the answer.
A: I had originally read "an equal probability to happen in each of the week days" as meaning a five day week, which would make the probability higher.  But apparently it means a seven day week.
The solution is $$\dfrac{S_2(12,7)\times 7!}{7^{12}}= \dfrac{627396\times 5040}{13841287201} \approx 0.22845$$ where $S_2(n,k)$ is a Stirling number of the second kind.  
Alternatively let $p(n,k)$ be the probability that $n$ calls in the week are distributed among exactly $k$ of the $7$ days; then $$p(n,k)=\tfrac{(8-k)}{7}\,p(n-1,k-1)+ \tfrac{k}{7} \,p(n-1,k)$$ starting with $p(0,0)=1$ and you will get the same result for $p(12,7)$.
