What's $P(\{k\})$ in this exercise? The question is based on the following problem in the book Probability Essentials by Jacod:
 
Here is my question:

What does $P(\{k\})$ mean in the second problem?

The probability on a finite or countable measurable space $(\Omega,\Sigma)$ is determined by $P(\{\omega\})$ where $\omega\in\Omega$. As I understand, for the binomial distribution $B(p,n)$ the sample space is $\Omega=\{(a_1, a_2):a_1,a_2=0,1\}^n$. How does $P(\{k\})$ come out here?
 A: $k$ is the number of success which is an event $\in\Sigma$. Namely trying $n$ times getting the success of $k$ times. Therefore this event has some probability $P(\{k\})$. I also think that it is defined in $4.1$ and there is an abuse of notation.
A: $P(\{k\})$ in Exercise 4.2 is $P(k\ \text{successes})$ in Exercise 4.1.
A: The binomial random variable is the number of success in n trials.  It can take on any integer value from 0 to n  and the probability that a binomial with success proportion p and n trials equals k (i.e. there are exactly k successes in the n trials) is C$_k$$^n$ p$^k$ (1-p)$^n$$^-$$^k$ where C$_k$$^n$ is the number of ways to pick k objects out of a total of n objects. So P({k}) is just this probability of getting exactly k successes in n trials.
A: It is an abuse of notation.  If we denote the probability space as $(\Omega,\Sigma,P)$, and let $X:\Omega\to{\Bbb R}$ be a random variable with Binomial distribution $B(p,n)$. 
$P$ in $P(\{k\})$ should be understood as the distribution measure of $X$ and $P(\{k\})$ means $P(X=k)$, which is defined as $P(\{\omega\in\Omega:X(\omega)=k\})$. 
