Prove that a given probability forms a probability measure with binomial distribution I currently have some troubles in proving the given probability forms a probability measure for the following problem:
Assume we have $m$ distinct balls distributed to $n$ urns, and each of the $n^m$ distributions have equal probability. Given the probability of an urn $A$ containing exactly $k$ balls $(k=0,1,...,m)$, is $p_k = \frac{{m\choose k}(n-1)^{(m-k)}}{n^m}$. Show  that $p_k$ forms a probability measure binomial$(p, m)$ where $p = \frac{1}{n}$.
My attempt: We need to show that the $3$ following properties for $p_k$ is satisfied:
(1) $\ p_k = p(A_k)\geq 0$ where $A_k =$ an event when urn $A$ has exactly $k$ balls for $k=0,1,...,m$. This is trivially true given definition of $p_k$.
(2)  Need to show that $\ p(A) = 1$ where $A = \bigcup_{k=0}^{m} A_k$ is the sample space. But isn't this ALWAYS true?? If not, is all I need to do is to verify $\sum_{k=0}^{m} p_k = 1$? But to do this, I need to prove $p$ satisfies the 3rd property below.
(3) Need to show that for a countable union of disjoint sets ${A_k}$, $p(\bigcup_{k=0}^{m} A_k)\  = \ \sum_{k=0}^{m} p_k$? But isn't this also true by using Principle of Inclusion-Exclusion, since $m$ is fixed and {$A_{k}$}$_{k=0}^{m}$ are disjoint sets due to its definition?
Finally, from the formula of $p_k$, we could rewrite it as ${m\choose k}(1-\frac{1}{n})^{m-k}(\frac{1}{n})^k$, thus $p_k$ follows binomial distribution with $p = \frac{1}{n}$ and $n=r$.
My question: Could anyone help on proving step $(2)$ and $(3)$ above? I am quite confused since they all seem obviously true in this setting.
 A: Note: In order to properly answer this problem, we have to clearly distinguish and precisely state three parts: assumption, claim and proof.

  
*
  
*Assumption:
  
  
*
  
*We have $m$ distinct balls, distributed to $n$ urns giving $n^m$ distributions.
  
*We have marked a specific urn and identied by the name $A$.
  
*The probability $p_k$ that urn A contains $k$ balls is as you already noted
  \begin{align*}
 p_k= \binom{m}{k}\left(1-\frac{1}{n}\right)^{m-k}\left(\frac{1}{n}\right)^k\qquad\qquad 0\leq k\leq m
  \end{align*}
  
  
*Claim:
  
  
*
  
*Show that $p_k$ forms a probability measure $\text{binomial}(p,m)$ with $p=\frac{1}{n}$
  
  
*Proof:
In order to show that $p_k$ is a probability measure we have to specify the sample space, let's say $S$. Note the urn $A$ is not the sample space. 
The sample space $S$ is the set of all possible outcomes, i.e. the set of number of balls in $A$ when distributing the $m$ balls in $n$ urns. We encode this by
  \begin{align*}
S=\{0,1,2,\ldots,m\}
\end{align*}

Regarding OPs remark in (2): isn't this ALWAYS true?
This is not the appropriate question. A sample space is not explicitly given. So, we have to define it and since it is a part of our proof we have to argue that our definition is correct. But, here the correctness of the definition is obvious by its construction.

In   order to show that $p_k$ is a probability measure, we have to show the following
(1) $\quad p_k\in[0,1]\qquad\qquad 0\leq k\leq  m$
(2) $\quad P(S)=1$

Ad (1):

Since for $0\leq k\leq m$
\begin{align*}
0&\leq\binom{m}{k}\left(1-\frac{1}{n}\right)^{m-k}\left(\frac{1}{n}\right)^k\\
&=p_k\\ 
&\leq \sum_{k=0}^m\binom{m}{k}\left(1-\frac{1}{n}\right)^{m-k}\left(\frac{1}{n}\right)^k=1
\end{align*}
  (1) is valid.

Ad (2):

The outcomes $k_1,k_2\in S$ with different $k_1,k_2$ are pairwise disjoint. So,    we  obtain
  \begin{align*}
P(S)&=P\left(\bigcup_{k=0}^m(X=k )\right)
=\sum_{k=0}^mP(X=k)
=\sum_{k=0}^m p_k\\
&= \sum_{k=0}^m\binom{m}{k}\left(1-\frac{1}{n}\right)^{m-k}\left(\frac{1}{n}\right)^k=1\tag{1}
\end{align*}
  and (2) is fulfilled.

Finally:

We have to show that $p_k$ is a $\text{binomial}(\frac{1}{n},m)$ distribution.
A binomial distribution of type $(\frac{1}{n},m)$ is per definitionem
  \begin{align*}
P(X=k)=\binom{m}{k}\left(\frac{1}{n}\right)^k\left(1-\frac{1}{n}\right)^{m-k}
\end{align*}
  This is precisely  the  formula in (1), so  we are done.

Conclusion: The proof itself is rather easy, since we have to do only few calculations.  Despite that the challenge is to precisely specify what is given, what is the sample space, what is the probability measure and to get a certain familiarity how these objects interact.
