Law of Total Probability

Law

• $\forall$ Partition $A_1,A_2\dots$
• For arbitrary event $B$
• $P(B)=P(B\cap S)={\sum }_{i=1}^{n}P(B\cap A_i)$

Generalization

$P(A)=P(A\cap B)+P(A\cap B^c)$

Proof

1. Note that $P(B\cap S)={\sum }_{i=1}^{n}P(B\cap A_i)$
2. Then, the sample space is also $S={\cup }_i{A}_i$ by defn Disjoint
3. Then, $P(B\cap S)=P(B\cap ({\cup }_i{A}_i))=P({\cup }_i(B\cap A_i))$
4. Note that $B\cap A_i$ are also Disjoint
5. Then, using the third Probability Axiom, $P({\cup }_i(B\cap A_i))={\sum }_{i=1}^{n}P(B\cap A_i)$

Intuition

The probability of a event intersecting the sample space is the sum of probability of the intersection of each disjoint.