Axioms for Probabilities

Given a sample space $S$ , we say that $P (A)$ for some event $A$ is the probability of the event $A$ . These must satisfy the following axioms:

For any event $A$ , $P (A) \geq 0$
$P (S) = 1$
If $A_{1}, A_{2}, . . .$ is an infinite collection of disjoint events (so no two events have the same outcomes in common) then:

P (A_{1} \cup A_{2} \cup \dots) = \sum_{i = 1}^{\infty} P (A_{i})

For context:

Axiom 1 says all events must be non-negative
The sample space is all events, so the probability any event occurs is 1.
The probability of one event in a collection is the sum of disjoint events (no two can occur simultaneously) then you just add up the probabilities.

Axiom 3 Finite Case

The case where we have $A_{1}, . . ., A_{n}$ events that are disjoint is similarly:

P (A_{1} \cup \dots \cup A_{n}) = \sum_{i = 1}^{n} P (A_{i})

But this is derived from Axiom 3, so it itself isn't an axiom.

P (\emptyset) = 0

$P (\emptyset) = 0$ where $\emptyset$ is the null event.

Using this proposition allows us to derive the finite case of axiom 3.

Proof
First consider the infite collection $A_{1} = \emptyset, A_{2} = \emptyset, . . .$ . Since $\emptyset \cap \emptyset = \emptyset$ then the events in this collection are disjoint and just $⋃ A_{i} = \emptyset$ . Axiom 3 tells us:

P (\emptyset) = P (⋃ A_{i}) = \sum P (A_{i}) = \sum P (\emptyset)

This can only happen when $P (\emptyset) = 0$ .

Now suppose that $A_{1}, . . ., A_{k}$ are disjoint events, and append to these the infinite collection $A_{k + 1} = \emptyset, . . .$ . Then the events $A_{1}, . . ., A_{k}, A_{k + 1}, . . .$ are disjoint, since $A \cap \emptyset = \emptyset$ for all events $A$ . Again using Axiom 3 then:

\begin{aligned} P (⋃_{i = 1}^{k} A_{i}) & = P (⋃_{i = 1}^{\infty} A_{i}) \\ = \sum_{i = 1}^{\infty} P (A_{i}) \\ = \sum_{i = 1}^{k} P (A_{i}) + \sum_{i = k + 1}^{\infty} P (A_{i}) \\ = \sum_{i = 1}^{k} P (A_{i}) + \sum_{i = k + 1}^{\infty} 0 \\ = \sum_{i = 1}^{k} P (A_{i}) \end{aligned}

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Thumbtack Example

If you toss a thumbtack in the air, either it points up or down. Thus $S = {U, D}$ as it's sample space. Since $U, D$ are disjoint then since $P (S) = 1$ :

1 = P (S) = P (U) + P (D) \Rightarrow P (D) = 1 - P (U)

So if you know either $P (U)$ or $P (D)$ then you know the other.

References

[[Matthew A. Carlton, Jay L. Devore - Probability with STEM Applications-Wiley (2020).pdf#page=43]]