Probability Rules

Probability Rules are the small set of identities, derived from the Axioms of Probability, used to combine and decompose event probabilities: the complement, addition, multiplication, and total-probability rules.

P(A^c) = 1 - P(A)

P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A \cap B) = P(A \mid B)\,P(B)

P(A) = \sum_i P(A \mid B_i)\,P(B_i)

• The addition rule subtracts the overlap so it is not counted twice; for Mutually Exclusive Events the intersection is empty and the rule reduces to P(A) + P(B). Three or more events need the Inclusion-Exclusion Principle.
• The multiplication rule simplifies to P(A)\,P(B) exactly when A and B are independent.
• The last identity is The Law of Total Probability, valid when \{B_i\} partitions the Sample Space; reversing the conditioning gives Bayes Theorem.
• Mutually exclusive is not independent. Disjoint events with positive probability are strongly dependent: knowing A occurred guarantees B did not, so P(A \mid B) = 0 \neq P(A).
• The complement rule is the workhorse shortcut — "at least one" is almost always faster as 1 - P(\text{none}).

Policyholders Holding Auto or Home Coverage 60% of policyholders hold auto coverage, 45% hold home coverage, and 25% hold both. Find the probability a policyholder holds at least one, and the probability they hold neither.

Answer By the addition rule:

\begin{align*}
P(A \cup H) &= P(A) + P(H) - P(A \cap H) \\
            &= 0.60 + 0.45 - 0.25 \\
            &= 0.80
\end{align*}

By the complement rule, P(\text{neither}) = 1 - 0.80 = 0.20.

"At Least One" via the Complement Each of 5 independent policies files a claim with probability 0.10. Find the probability that at least one files a claim.

Answer Enumerating one, two, three... claims is slow. Complement instead:

\begin{align*}
P(\text{at least one}) &= 1 - P(\text{no claims}) \\
                       &= 1 - (0.90)^5 \\
                       &= 1 - 0.59049 \\
                       &= 0.40951 \approx 0.41
\end{align*}

Chaining the Multiplication Rule An urn holds 4 defective and 6 sound parts. Three are drawn without replacement. Find the probability all three are sound.

Answer Without replacement the draws are dependent, so chain conditionals rather than raising 0.6 to the third power:

\begin{align*}
P(S_1 \cap S_2 \cap S_3) &= P(S_1)\,P(S_2 \mid S_1)\,P(S_3 \mid S_1 \cap S_2) \\
                         &= \frac{6}{10} \cdot \frac{5}{9} \cdot \frac{4}{8} \\
                         &= \frac{120}{720} \\
                         &= \frac{1}{6}
\end{align*}