different probability spaces for $P(X=k)=\binom{n}{k}p^k\big(1-p\big)^{ n-k}$? Consider the following formula, which is from the Binomial distribution:
$$P(X=k)=\binom{n}{k}p^k\big(1-p\big)^{ n-k}$$
On the left hand side, it is the probability of one event: $\{\omega\in\Omega:X(\omega)=k\}$. One the right hand side, one can consider $n$ independent variables, and using the addition and multiplication formula. It seems that the underlying sample spaces can be different for the two sides of the equality. For the left hand side, as I understand, the sample space(up to set isomorphism) $\Omega_1=\{(a_i)_{i=1}^{n}:a_i=0,1\}$. And the random variable $X:\Omega_1\to{\mathbb R}$,
$$X(\omega)=\sum_{i=1}^{n}a_i$$ 
While the right hand side, it can be $\Omega_2=\{0,1\}$. And for the random variable $Y:\Omega_2\to{\mathbb R}$, $Y(0)=0,Y(1)=1$.
Here is my question:
How can one reach the equality by working with different sample spaces(and hence different probability spaces)?
Edit:
According to the comments and the answers to the question, I would like to modify the question as the following:
Is it possible to interpret the formula $P(X=k)=\binom{n}{k}p^k\big(1-p\big)^{ n-k}$ in terms of the following probability space $(\Omega,\Sigma,P)$ where


*

*$\Omega=\{0,1\}$;

*$\Sigma=2^{\Omega}$;

*$P(0)=p,P(1)=1-p\quad 0\leq p\leq 1$.


On the right-hand side, all I can see is a number.  That number could be found in several ways.
There is no random variable mentioned on the right-hand side of the equation, and no sample space. 
And in any case, one can (and often does) show that two probabilities are equal when the underlying sample spaces are entirely different. Think of tossing a coin, event "Head", and tossing a die, event "even number." 
