Probability involved in information theory.

Question

I was reading Information, Entropy, and the Motivation for Source Codes chapter 2 MIT 6.02 DRAFT Lecture Notes(https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-02-introduction-to-eecs-ii-digital-communication-systems-fall-2012/readings/MIT6_02F12_chap02.pdf, 2.1.2 Examples), trying to understand the mathematics behind information gain when I came across this:

Now suppose there are initially N equally probable and mutually exclusive choices, and I tell you something that narrows the possibilities down to one of M choices from this set of N. How much information have I given you about the choice? Because the probability of the associated event is M/N, the information you have received is log2(1/(M/N)) = log2(N/M) bits. (Note that when M = 1, we get the expected answer of log2 N bits.)

I could not understand how the probability of the associated event is M/N. Please explain in detail.

Answer 1

It is quite immediate :

Suppose that $C$ is a random variable with values in a set $\mathcal{C}$ having a cardinality $N$. Suppose that all possible values of $C$ have the same probability.

Consider a subset $\mathcal{C}'$ of $\mathcal{C}$ that contains $M$ elements.

Then, if you consider the event $E$ : $C \in \mathcal{C}'$ then $P(E)=\frac{M}{N}$