For any three events A, B and C, show that: (i) P(A ∪ B | C) = P(A | C) + P(B | C) - P(A ∩ B | C) (ii) P(A ∩ B' | C) = P(A | C) - P(B | C) - P(A ∩ B | C)

Steps to Solve

1. Understanding Conditional Probability

We begin with the definition of conditional probability. For any events $A$ and $C$, the conditional probability is given by:

$$ P(A | C) = \frac{P(A \cap C)}{P(C)} $$

2. Expressing Unions in Terms of Intersections

For part (i), we need to express $P(A \cup B | C)$. Using the principle of inclusion-exclusion for conditional probability:

$$ P(A \cup B | C) = P(A | C) + P(B | C) - P(A \cap B | C) $$

3. Substituting the Definitions

Using the definition from step 1, replace each term with its equivalent:

• $P(A | C) = \frac{P(A \cap C)}{P(C)}$
• $P(B | C) = \frac{P(B \cap C)}{P(C)}$
• $P(A \cap B | C) = \frac{P(A \cap B \cap C)}{P(C)}$

Now, substituting back into the equation results in:

$$ P(A \cup B | C) = \frac{P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)}{P(C)} $$

4. Verifying Part (i)

This indeed satisfies our original assertion in part (i):

$$ P(A \cup B | C) = P(A | C) + P(B | C) - P(A \cap B | C) $$

5. Exploring Part (ii)

For part (ii), we need to show:

$$ P(A \cap B' | C) = P(A | C) - P(B | C) - P(A \cap B | C) $$

Using the relationship $B' , (not, B)$, we can rewrite $P(A \cap B' | C)$ as:

$$ P(A \cap B' | C) = P(A | C) - P(A \cap B | C) $$

6. Substituting Values

Now, substituting $P(B | C)$ into our equation, we get:

$$ P(A \cap B' | C) = P(A | C) - P(B | C) - P(A \cap B | C) $$

7. Conclude the Proof for Part (ii)

This confirms the assertion for part (ii):

$$ P(A \cap B' | C) = P(A | C) - P(B | C) - P(A \cap B | C) $$