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02. Uniform Sample Space

$\gdef \N{\mathbb{N}} \gdef \Z{\mathbb{Z}} \gdef \Q{\mathbb{Q}} \gdef \R{\mathbb{R}} \gdef \C{\mathbb{C}} \gdef \setcomp#1{\overline{#1}} \gdef \sseq{\subseteq} \gdef \pset#1{\mathcal{P}(#1)} \gdef \covariant{\operatorname{Cov}} \gdef \of{\circ} \gdef \p{^{\prime}} \gdef \pp{^{\prime\prime}} \gdef \ppp{^{\prime\prime\prime}} \gdef \pn#1{^{\prime\times{#1}}} $

\(\newcommand{\u}{\cup}\newcommand{\n}{\cap}\newcommand{\comp}[1]{\overline{**1**{: #1 .hash}}}\newcommand{\trig}[1]{\overline{#1}\,\theta}\)
\(\trig{cos}\)

Probability Space

Definition:

A probability space is a sample space \Omega together with a probability function (mapping events [subsets of \(\Omega\)] to the interval [0,1]) which satisfies the following two axioms:

Axioms of a Probability Space
  1. \(P(\Omega)=1\)
  2. If \(A\) and \(B\) are disjoint events, then:
\[P(A\cup B)=P(A)+P(B)\]

Example 1

\(A=\) The high temp for today \(\in [28^\of, 33^\of)\)
\(B=\) The high temp for today \(\in [33^\of, 38^\of]\)

\(A,B\) are disjoint, because there is no outcome that can be in both sets.
Suppose \(P(A)=0.7\), \(P(B)=0.15\).
Then, \(P(A\cup B)=0.7 + 0.15=0.85\)

Example 2

\(A=\) Pass the math exam
\(B=\) Pass the computers exam

\(A,B\) are not disjoint, because an outcome can be in both \(A\) and \(B\). (Passing both exams)
\(P(A\cup B)\) is not \(P(A)+P(B)\).

Example 3

A 6-sided die is rolled.
\(A=\) roll ≥ 2
\(B=\) roll ≥ 4

\(A\subseteq B\) because every outcome in \(B\) is also in \(A\).

Properties of a Probability Space
  1. \(P(\overline A)=1-P(A)\)
    • Proof: \(A, \overline A\) are disjoint events (\(A\cap\overline A=\varnothing\)), and \(A\cup \overline A=\Omega\). Therefore, \(P(A) + P(\overline A)=P(\Omega)=1\Longrightarrow P(\overline A)=1-P(A)\).
  2. \(P(\varnothing)=0\)
    • Proof: \(\varnothing=\overline\Omega\), so by property 1: \(P(\varnothing)=1-P(\Omega)=1-1=0\)
  3. If \(B\subseteq A\), then \(P(B)\leq P(A)\)
    • Proof Omitted
  4. If \(B\subseteq A\), then \(P(A)-P(B)=P(A\cap\overline B)\)
    • Proof Omitted
      Excalidraw Excalidraw
  5. For any events \(A,B\):
    1. \(P(A\cap B)\leq P(A)\leq P(A\cup B)\)
    2. \(P(A\cap B)\leq P(B)\leq P(A\cup B)\)
      Excalidraw Excalidraw
      - Proof Omitted
  6. For any events A,B: \(P(A\u B)=P(A)+P(B)-P(A\n B)\)
    • Proof Omitted

Example of property 6

\(A=\) I pass math exam
\(B=\) I pass computers exam
\(P(A)=0.7\)
\(P(B)=0.6\)
\(P(A\u B)=0.4\)
What is the probability of failing both exams?

The event of failing both exams is \(\comp A\n\comp B=\comp{A\u B}\) (De Morgan’s Law)

\(P(\comp{A\u B})=1-P(A\u B)\)

\(P(A\u B)=P(A)+P(B)-P(A\n B)=0.7+0.6-0.4=0.9\)

\[P(\comp{A\u B})=1-0.9=0.1\]