Equally Likely Outcomes

Usually for $N$ outcomes, we assume that each one is equally likely (ex: fair coin, dice, etc.). With $p = P (E_{i})$ for each $i \in {1, . . ., N}$ :

1 = \sum_{i = 1}^{N} P (E_{i}) = \sum_{i = 1}^{N} p = p \cdot N \Rightarrow p = \frac{1}{N}

Now instead if we have an event $A$ with $N (A)$ denoting the number of outcomes contained in $A$ , then:

P (A) = \sum_{E \in A} P (E) = \sum_{E \in A} \frac{1}{N} = \frac{N (A)}{N}

References

[[Matthew A. Carlton, Jay L. Devore - Probability with STEM Applications-Wiley (2020).pdf#50]]