Probability

Published

Sunday Jun 8, 2025

1 Probability functions

For a given experiement and associated sample space \(S\), a probability function \(P\) is a real-valued function whose domain is the power set of the sample space, \(S\), and satisfies the following:

  1. \(P(A) \geq 0\) for all \(A \subset S\) (probability can’t be negative)
  2. \(P(S) = 1\) (Something must happen)
  3. Suppose \(A_1, A_2, \ldots\) is an infinite sequence of disjoint events. Then \(P(A_1 \cup A_2 \cup \ldots) = P(A_1) + P(A_2) + \ldots\)

These three conditions are known as the three axioms of probabiliyt. They do not completely specify \(P\), but merely ensure that \(P\) is ‘sensible’. It remains for \(P\) to be precisely defined in any given situation. Typically, \(P\) is defined by assigning ‘reasonable’ probabilities to each of the same points (or simple events) in \(S\).

If the die is fair, then all of the possible outcomes 1, 2, 3, 4, 5, 6 are equally likely.

So it is reasonable to assign probability function \(P\) in case by \[ P(\{1\}) = P(\{2\}) = \cdots = P(\{6\}) = 1 / 6 \]

Equivalently, we may write \(P(\{k\}) = 1/6\), \(k=1, \ldots, 6\) or \(P(\{k\}) = 1/6 \; \forall k = S\).

Theorem 1 \(P(\emptyset) = 0\)

Proof. Apply Axiom 3 with \(A_i = \emptyset\) for all \(i\).

\(\emptyset = \emptyset \cup \emptyset \cup \ldots\) Also \(\emptyset \cap \emptyset = \emptyset\) (i.e. \(\emptyset\) and \(\emptyset\) are disjoint). It follows that \(P(\emptyset) = P(\emptyset \cup \emptyset \cup \ldots) = P(\emptyset) + P(\emptyset) + \cdots\). We now subtract \(P(\emptyset)\) from both sides. Hence \(0 = P(\emptyset) + P(\emptyset) + \cdots\). Therefore, \(P(\emptyset) = 0\).

Theorem 2 Axiom 3 also holds for finite sequences. Thus if \(A_1, A_2, \ldots, A_n\) are disjoint events, then

\[ P(A_1 \cup A_2 \cup \ldots \cup A_n) = P(A_1) + P(A_2) + \cdots + P(A_n) \]

Proof. Apply Axiom 3 and Theorem 1, with \(A_i = \emptyset\) for all \(i = n + 1, n + 2, \ldots\).

Theorem 3 \[ P(\bar{A}) = 1 - P(A) \]

Proof. \[\begin{align*} 1 & = P(S) \tag{by Axiom 2} \\ & = P(A \cup \bar{A}) \tag{by the definition of complementation} \\ & = P(A) + P(\bar{A}) \tag{by Theorem 2 with $n = 2$, since $A$ and $\bar{A}$ are disjoint.} \end{align*}\]