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Set Theory

In this blog, we donated probability of event \(A\) by \(P(A)\).

  1. The set of all possible outcomes or sample space of random experiment donates by \(\Omega\). Our probability \(P(A)\) map from subsets of \(\Omega\) to \(\mathbb{R}\), where it valued with \([0,1] \in \mathbb{R}\). Each outcome of experiment \(\omega \in \Omega\) encapsulates all information at the end of specific experiment.

  2. Event space \(\Sigma : A\) is the set of all outcomes of experiment takes. \(\Sigma\) is subset of \(\Omega\).

  3. Properties of \(P\), \(\Sigma\) has to stisfy three of these properties:

    • \(P(\Omega) = 1\)
    • If \(A\subset\Omega\) then \(P(A) \ge 0\)
    • If \(A_i\) and \(A_j\) are disjoint events, then \(A_i \cap A_j = \emptyset\) with \(i\neq j\). Disjoint means \(A_i\) and \(A_j\) are separate.

Those three of properties are known as Axioms of Probability. Then,

\[ P : \Sigma \rightarrow [0, 1] \]

Other Properties :

  • If \(A_i\) and \(A_j\) are separate, \(P(A_i\cup A_j) = P(A_i) + P(A_j)\)
  • If \(A_i\) and \(A_j\) are not separate and independent, \(P(A_i \cap A_j) = P(A_i) \times P(A_j)\)
  • \(P(A_i \cup A_j) = P(A_i) + P(A_j) - P(A_i \cap A_j)\)
  • \(A^c = 1-P(A)\)
  • If \(A\) is empty set, then \(P(A) = P(\emptyset)=0\)
  • If \(A \subseteq B\), then \(P(A) \leq P(B)\)

Example : If we throw 3 coin flips. The sample space we have is :

\[ \Omega = \{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \]

There are 8 possible outcomes, with H is head and T is tail.

We take 2 events where \(A\) and \(B\) are independent, the set of outcomes where first flip is a head \(\{HHH, HHT, HTH, HTT\}\), donated by \(A\).The set of outcomes where second flip is a tail \(\{HTH, HTT, TTH, TTT\}\), donated by \(B\). So,

\[ P(A) = \frac{4}{8} = \frac{1}{2}\\ P(B) = \frac{4}{8} = \frac{1}{2} \]
  • Every outcomes in both \(A\) and \(B\), \(P(A \cap B) = P(A) \times P(B)\)
\[ P(A \cap B) = \{HHH, HTT\} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]
  • Every outcomes in \(A\) or \(B\), \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\[ P(A \cup B) = \{HHH, HHT, HTH, HTT, TTH, TTT\} = \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = \frac{3}{4} \]
  • Every outcomes are not in \(A\), \(A^c = 1-P(A)\)
\[ A^c = \{THH, THT, TTH, TTT\} = 1 - \frac{1}{2} = \frac{1}{2} \]

Info

Most of this section refered to Review of Probability Theory by Maleki and Do, A First Course in Probability by Sheldon Ross, and Probability Bootcamp by Steve Bruton