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Probability basics: axioms, independence, total probability and Bayes

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Sample spaces, events and the axioms

  • The sample space SS is the set of all outcomes. An event is a subset of SS.
  • Rolling two dice gives ∣S∣=36\lvert S\rvert = 36 equally likely ordered pairs.

Kolmogorov's axioms.

  1. P(A)≥0P(A) \ge 0.
  2. P(S)=1P(S) = 1.
  3. For mutually exclusive (disjoint) events, P(A∪B∪… )=P(A)+P(B)+…P(A \cup B \cup \dots) = P(A) + P(B) + \dots

Consequences.

  • P(Ac)=1−P(A)P(A^c) = 1 - P(A) and P(∅)=0P(\varnothing) = 0.
  • If A⊆BA \subseteq B, then P(A)≤P(B)P(A) \le P(B).
  • P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B).
  • P(A∪B∪C)=∑P(A)−∑P(A∩B)+P(A∩B∩C)P(A \cup B \cup C) = \sum P(A) - \sum P(A \cap B) + P(A \cap B \cap C).
  • Union bound: P(A∪B)≤P(A)+P(B)P(A \cup B) \le P(A) + P(B).

Example 1. P(A)=0.5P(A) = 0.5, P(B)=0.4P(B) = 0.4 and P(A∩B)=0.2P(A \cap B) = 0.2.

  • P(A∪B)=0.7P(A \cup B) = 0.7.
  • P(neither)=0.3P(\text{neither}) = 0.3.
  • P(exactly one)=0.7−0.2=0.5P(\text{exactly one}) = 0.7 - 0.2 = 0.5.
  • P(A∩Bc)=0.5−0.2=0.3P(A \cap B^c) = 0.5 - 0.2 = 0.3.

Example 2: two dice.

  • P(sum=7)=636=16P(\text{sum} = 7) = \frac{6}{36} = \frac16.
  • P(at least one six)=1−(56)2=1136P(\text{at least one six}) = 1 - \left(\frac56\right)^2 = \frac{11}{36}.
  • P(doubles)=16P(\text{doubles}) = \frac16.

Example 3: the complement trick. The probability of at least one six in 4 rolls is 1−(5/6)4≈0.5181 - (5/6)^4 \approx 0.518. "At least one" questions are almost always easier through the complement.

Conditional probability

P(A∣B)=P(A∩B)P(B),P(B)>0.P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) > 0.

Conditioning shrinks the sample space to BB. Multiplication rule: P(A∩B)=P(B) P(A∣B)=P(A) P(B∣A)P(A \cap B) = P(B)\,P(A \mid B) = P(A)\,P(B \mid A). For more events, P(A∩B∩C)=P(A)P(B∣A)P(C∣A∩B)P(A \cap B \cap C) = P(A)P(B \mid A)P(C \mid A \cap B).

Example 4. Two cards are drawn without replacement. P(both red)=2652⋅2551=25102≈0.245P(\text{both red}) = \frac{26}{52}\cdot\frac{25}{51} = \frac{25}{102} \approx 0.245.

Example 5: conditioning on partial information. A family has two children, and at least one is a boy. What is the probability that both are boys?

  • The equally likely cases are BB, BG and GB (GG is ruled out).
  • So the answer is 13\frac13, not 12\frac12.

If instead we are told the elder child is a boy, the cases are BB and BG, so the answer is 12\frac12. The exact information matters.

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