Proof: P(A∪B) = 1 − P(A′)·P(B′) for Independent Events

NCERT Class 12 Mathematics — Probability, Example 14.

Problem

If $A$ and $B$ are two independent events, prove that:

$$P(A \cup B) = 1 – P(A’)\cdot P(B’)$$

Key insight. “Independent” only tells us how to compute $P(A \cap B)$ — as a plain product $P(A)P(B)$. Once that’s substituted into the ordinary union formula, the rest is just algebra: grouping terms to reveal $(1-P(A))(1-P(B))$ hiding inside the expression.

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Approach

Start from the standard union formula $P(A\cup B) = P(A)+P(B)-P(A\cap B)$. Since $A$ and $B$ are independent, replace $P(A\cap B)$ with $P(A)P(B)$. Then rearrange the resulting expression algebraically until it matches the form $1-P(A’)P(B’)$, using the complement rule $P(A’)=1-P(A)$.

Solution

Step 1 — Start from the general union formula

For any two events:

$$P(A \cup B) = P(A) + P(B) – P(A \cap B)$$

Step 2 — Use independence to simplify the intersection

Since $A$ and $B$ are independent, by definition:

$$P(A \cap B) = P(A)\cdot P(B)$$

Substituting:

$$P(A \cup B) = P(A) + P(B) – P(A)P(B)$$

Step 3 — Rewrite in terms of the complements

Recall $P(A’) = 1-P(A)$ and $P(B’)=1-P(B)$, so $P(A)=1-P(A’)$ and $P(B)=1-P(B’)$. It’s easier to work backwards: expand the product $P(A’)P(B’)$ and see what it gives.

$$P(A’)P(B’) = (1-P(A))(1-P(B)) = 1 – P(A) – P(B) + P(A)P(B)$$

Step 4 — Rearrange to isolate the union expression

$$1 – P(A’)P(B’) = P(A)+P(B)-P(A)P(B)$$

But the right-hand side is exactly what Step 2 gave for $P(A\cup B)$.

Step 5 — Conclude

$$P(A \cup B) = 1 – P(A’)P(B’)$$

Answer

Common mistakes

• Using $P(A\cap B)=P(A)P(B)$ without checking independence is actually given. This substitution is only valid for independent events; for general events, $P(A\cap B)$ must be handled separately or given directly.
• Trying to prove the identity by picking specific numeric values for $P(A)$ and $P(B)$. Checking a specific case can build intuition, but it doesn’t constitute a proof — the algebraic derivation above holds for all valid probabilities, which a numeric example alone can’t establish.

Practise next

• If $A$, $B$ and $C$ are mutually independent events, express $P(A\cup B\cup C)$ in terms of $P(A’)$, $P(B’)$ and $P(C’)$, extending the same complement-based argument. Then check whether pairwise independence alone would have been enough.