Problem

# Show that for any three events A, B, and C with P(C) > 0, P(A ∪B|C) = P(A|C) + P(B|C)...

Show that for any three events A, B, and C with P(C) > 0, P(A ∪B|C) = P(A|C) + P(B|C) – P(A∩B|C).

#### Step-by-Step Solution

Solution 1

For any three events $$A, B$$ and $$C$$ with $$P(C)>0$$, then

\begin{aligned} P(A \cup B \mid C) &=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C) \\ \begin{aligned} P(A \cup B \mid C) &=\frac{P[C \cap(A \cup B)]}{P(C)} \\ &=\frac{P[(C \cap A) \cup(C \cap B)]}{P(C)} \end{aligned} \end{aligned}

Expand the Numerator using $$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$

\begin{aligned} P(A \cup B \mid C) &=\frac{P(C \cap A)+P(C \cap B)-P[(C \cap A) \cap(C \cap B)]}{P(C)} \\ &=\frac{P(C \cap A)+P(C \cap B)-P(A \cap B \cap C)}{P(C)} \\ &=\frac{P(C \cap A)}{P(C)}+\frac{P(C \cap B)}{P(C)}-\frac{P(A \cap B \cap C)}{P(C)} \\ &=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C)\left(\text { using } P(A / B)=\frac{P(B \cap A)}{P(B)}\right) \end{aligned}