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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.

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

### 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)


Bayes’ rules, Conditional probability, Chain rule