Q.If A, B, C are three events associated with a random experiment, prove that $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$.

Concept: Set Operations — the inclusion-exclusion principle extended to three events.

Step 1: Start with the union of $A$ and $B \cup C$:

Step 2: Expand $P(B \cup C)$ and the intersection:

Step 3: Substitute everything back:

$P(A \cup B \cup C) = P(A) + \big[P(B) + P(C) - P(B \cap C)\big] - \big[P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)\big].$ …

We need to prove the inclusion-exclusion principle for three events. The core idea is simple: when we count outcomes in $A \cup B \cup C$, we must avoid double-counting outcomes that belong to more than one event.

Think of it like this: if you add $P(A) + P(B) + P(C)$, you've counted every outcome that lies in exactly one event once, but outcomes in two events get counted twice, and outcomes in all three events get counted three times. So we subtract the pairwise intersections to fix the double-counting — but then we've subtracted the triple intersection three times (once for each pair), when it was originally added three times. That leaves it counted zero times, so we must add it back once.

Let's prove it formally.

1. Start with the union of A and (B ∪ C). We know the addition rule for two events:

Let $X = A$ and $Y = B \cup C$. Then:

2. Expand $P(B \cup C)$ using the same rule:

3. Handle the intersection $A \cap (B \cup C)$. By the distributive law of set operations:

So:

4. Apply the two-event rule again to $(A \cap B)$ and $(A \cap C)$:

5. Simplify the triple intersection. Notice that:

So:

6. Now substitute everything back into the original expression. From step 1: …