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Research Scientist @ Foundation AI
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Probability Series

Basic Probability Concepts

Conditional Probability and Bayes’ Rule

Discrete Random Variables

Continuous Random Variables

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Conditional Probability

The probability of event X given that Y has already occurred is denoted by \(P(X|Y)\)

  • If X and Y are independent: \(P(X|Y) = P(X)\) because event X is not dependent on event Y.

  • If X and Y are mutually exclusive: \(P(X|Y) = 0\) because X and Y are disjoint events.

Product Rule

From \eqref{1}, following can be concluded,

  • \(X \subseteq Y\) implies \(P(X|Y) = P(X)/P(Y)\) because \(X \cap Y = X\)
  • \(Y \subseteq X\) implies \(P(X|Y) = 1\) because \(X \cap Y = Y\)

The distributive, associative and De Morgan’s laws are valid for conditional probability.

\[P(X \cup Y|Z) = P(X|Z) + P(Y|Z) - P(X \cap Y|Z)\] \[P(X^{c}|Z) = 1-P(X|Z)\]

Chain Rule

Bayes’ Theorem

Where \(P(X) = P(X \cap Y) + P(X \cap Y^{c})\) from the sum rule.

Derivation of Bayes’ Theorem

From \eqref{1},

Using the commutative law,

From \eqref{2}, \eqref{3} and, \eqref{4},

Hence,

EXAMPLE

  • \(p_ot\) is probability of reaching on time when no car trouble.
  • \(p_ct\) is probability of car trouble.
  • Commute by train if car trouble occurs.
  • N is the number of trains available.
  • Only 2 of the N trains would reach on time.
  • What is the probability of reaching on time.

  • Explaination:
    • \(O\): reach on time
    • \(C^c\): car not working
    • \(P(O) = P(O \cap C) + P(O \cap C^c)\)
    • \(P(O \cap C) = P(O|C) * P(C) \text{ where } P(C) = 1 - P(C^c)\)
    • \(P(O \cap C^c) = P(O|C^c) * P(C^c) \text{ where } P(O|C^c) = 2/N\)
p_ct = float(input()) # P(Car Trouble)
p_ot = float(input()) # P(On Time | No Car Trouble)
N = float(input()) # Number of trains
p_rt = 2.0/N # P(Correct Train)
p_o = p_ct * p_rt + (1-p_ct) * p_ot # P(On Time)
print("%.6f" % p_o)

REFERENCES:

Bayes’ rules, Conditional probability, Chain rule

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