Why $\mathbb{P}\left(X_{1}+\cdots+X_{n}=k\right)=\sum_{s_1+\cdots+s_n=k}\mathbb{P}\left(X_{1}=s_{1},\ldots,X_{n}=s_{n}\right)$ in Bernouli experiment? Problem ( Bernouli experiment ) :
We repeat some experiment ( for example, tossing a coin ) $ n $ times, where each experiment can end in success or failure. It is known that the probability for the success of each experiment separately is $ p \in (0,1) $. And that all the experiments are independent. What is the probability that there will be exactly $k$ successes out of $ n $ experiments?
Solution ( from lecture notes ):
The probability space that corresponds to a single experiment is $ \left(\{0,1\}, \mathbb{P}_{1}\right) $, where $ \mathbb{P(1)} = p $ ( success ) and $ \mathbb{P(0)} = 1-p $ ( failure ). The corresponding space for $ n $ independent experiments is the product space
$ (\Omega, \mathbb{P})=\bigotimes_{i=1}^{n}\left(\{0,1\}, \mathbb{P}_{1}\right) $
Denote as $ A_k $ the event in which the $k$-th experiment succeeded. From the given, it holds that $ \mathbb{P}(A_k) = p $ and $ A_1,...,A_n $ are independent. Denote the indicators of these events as $ X_1,...,X_n  $, correspondingly.
From the definition of independence-ship of the events it occurs that these variables are independent. In addition $\mathbb{P}\left(X_{i}=0\right)=1-p, \mathbb{P}\left(X_{i}=1\right)=p$
For the sake of further calculations, we'll denote it shortly as:
$\mathbb{P}\left(X_{i}=s_{i}\right)=p^{s_{i}}(1-p)^{1-s_{i}}, \forall s_{i} \in\{0,1\}$.
Define the random variable $ N = X_1 + \cdots + X_n $.
Notice that $\{N=k\}=\left\{X_{1}+\cdots+X_{n}=k\right\}=\{$ exactly $k$ experiments out of $n$ succeeded   $\}$ $\color{red}{\text{And it occurs that  } }   $
$\mathbb{P}(N=k)=\mathbb{P}\left(X_{1}+\cdots+X_{n}=k\right)=\sum_{s_{1}+\cdots+s_{n}=k} \mathbb{P}\left(X_{1}=s_{1}, \ldots, X_{n}=s_{n}\right)$
Where the sum is on all $ s_1,...,s_n \in \{ 0,1 \} $ such that their sum is $k $. From independence-ship we'll get that for all $ s_1,...,s_n \in \{ 0,1 \} $ it occurs that
$\begin{aligned} \mathbb{P}\left(X_{1}=s_{1}, \ldots, X_{n}=s_{n}\right) &=\mathbb{P}\left(X_{1}=s_{1}\right) \ldots \mathbb{P}\left(X_{n}=s_{n}\right) \\ &=\prod_{i=1}^{n} p^{s_{i}}(1-p)^{1-s_{i}}=p^{\sum_{i=1}^{n} s_{i}}(1-p)^{n-\sum_{i=1}^{n} s_{i}} \end{aligned}$
Meaning if $\sum_{i=1}^{n} s_{i} = k $ ( meaning there'll be exactly $ k $ successes out of $ n $ ) then $\mathbb{P}\left(X_{1}=s_{1}, \ldots, X_{n}=s_{n}\right)=p^{k}(1-p)^{n-k}$
Hence,
$\begin{aligned} \mathbb{P}(N=k) &=\sum_{s_{1}+\cdots+s_{n}=k} \mathbb{P}\left(X_{1}=s_{1}, \ldots, X_{n}=s_{n}\right) \\ &=\sum_{s_{1}+\cdots+s_{n}=k} p^{k}(1-p)^{n-k}=\left(\begin{array}{l}n \\ k\end{array}\right) p^{k}(1-p)^{n-k} \end{aligned}$
And notice that from Newton's binomial formula, it occurs that
$\mathbb{P}(0 \leq N \leq n)=\sum_{k=0}^{n} \mathbb{P}(N=k)=\sum_{k=0}^{n}\left(\begin{array}{l}n \\ k\end{array}\right) p^{k}(1-p)^{n-k}=(p+(1-p))^{n}=1$
$ \square $
My Questions:

*

*Why $\mathbb{P}(A_k) = p$ and not $ \mathbb{P}(A_k) = p^k $?

*In the line where it says " $\mathbb{P}(N=k)=\mathbb{P}\left(X_{1}+\cdots+X_{n}=k\right)=\sum_{s_{1}+\cdots+s_{n}=k} \mathbb{P}\left(X_{1}=s_{1}, \ldots, X_{n}=s_{n}\right)$ " ( near red marking above), how is the transition $ \mathbb{P}\left(X_{1}+\cdots+X_{n}=k\right) = \sum_{s_{1}+\cdots+s_{n}=k} \mathbb{P}\left(X_{1}=s_{1}, \ldots, X_{n}=s_{n}\right)$ justified? how the fact that $X_1,...,X_n $ are independent is used here? I can't see how $ \sigma $-additivity is used here.

Notations:

*

*$ (\Omega, \mathbb{P})=\bigotimes_{i=1}^{n}\left( \Omega_i, \mathbb{P}_i\right) \iff (\Omega, \mathbb{P}) = ( \Omega_1 \times \cdots \times \Omega_n  , \mathbb{P}_1 \cdots \mathbb{P}_n ) $

*$ \mathbb{P}(A,B) = \mathbb{P}(A \cap B ) $
Also, all the discussion above relates to discrete probability spaces ( I don't  learn about continuous probability spaces in the course )
Thanks in advance for any help!
 A: Question 1.
By definition, $A_k$ is the event where the $k^{th}$ experiment is a success, meaning
\begin{align}
A_k &= \{\omega\in \Omega\,:\, \omega_k=1\}\\
&= \underbrace{\{0,1\}\times \cdots \times \{0,1\}}_{\text{$k-1$ times}}\times \{1\}
\times \underbrace{\{0,1\}\times \cdots \times \{0,1\}}_{\text{$n-k$ times}}
\end{align}
What is the measure of this set? Well, by definition $\Bbb{P}$ is the $n$-fold product of the measure $\Bbb{P}_1$, so
\begin{align}
\Bbb{P}(A_k)&=\Bbb{P}_1(\{0,1\})^{k-1}\cdot \Bbb{P}_1(\{1\})\cdot \Bbb{P}_1(\{0,1\})^{n-k}\\
&=1^{k-1}\cdot p\cdot 1^{n-k}\\
&=p.
\end{align}

Question 2.
For each $s=(s_1,\dots, s_n)\in \{0,1\}^n$, let us consider the event
$B_s=\{X_1=s_1,\dots, X_n=s_n\}$, or more formally, $B_s= \{\omega\in\Omega\,:\,X_1(\omega)=s_1,\dots, X_n(\omega)=s_n\}$, and consider the set $\mathcal{S}_k$ of all $s\in\{0,1\}^n$ such that $s_1+\cdots + s_n=k$. Then, we can write
\begin{align}
\{N=k\}&=\bigcup_{s\in \mathcal{S}_k}B_s,
\end{align}
and the latter is a disjoint collection of sets (so you have to show the two set inclusions $\subset$ and $\supset$, and also that the sets on the right are pairwise disjoint). Thus, by additivity of measures (a trivial consequence of $\sigma$-additivity), we have
\begin{align}
\Bbb{P}(\{N=k\})&=\sum_{s\in \mathcal{S}_k}\Bbb{P}(B_s)\\
&=\sum_{s\in\mathcal{S}_k}\Bbb{P}(\{X_1=s_1\})\cdots \Bbb{P}(\{X_n=s_n\}),
\end{align}
where the last line used independence.
