Mutually exclusive events — addition rule

WHAT are we talking about?

WHY this matters: "Roll a 2" and "roll a 5" on one die cannot both happen — mutually exclusive. But "roll an even number" and "roll a number $>3$" can both happen (a 4 or 6) — not mutually exclusive.

HOW to derive the addition rule from first principles

We start from the general rule and earn the special case.

Step 1 — Count outcomes. For equally likely outcomes, $P(E) = \dfrac{n(E)}{n(S)}$ where $n(\cdot)$ counts outcomes.

Why this step? Probability of an event = (favourable outcomes) ÷ (total outcomes). Everything flows from honest counting.

Step 2 — Count $A \cup B$. How many outcomes are in "$A$ or $B$"? If we naively add $n(A) + n(B)$, any outcome sitting in both $A$ and $B$ gets counted twice. To fix the double count, subtract the overlap once: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Why this step? This is the inclusion–exclusion idea: add the pieces, then remove what you double-added.

Step 3 — Divide by $n(S)$. $\frac{n(A\cup B)}{n(S)} = \frac{n(A)}{n(S)} + \frac{n(B)}{n(S)} - \frac{n(A\cap B)}{n(S)}$ which is the general addition rule: $\boxed{P(A \cup B) = P(A) + P(B) - P(A \cap B)}$

Step 4 — Apply mutual exclusivity. If $A$ and $B$ are mutually exclusive, then $A\cap B = \varnothing$, so $P(A\cap B) = 0$. The correction term vanishes:

Handy corollary (complement): Since $A$ and its complement $A'$ are mutually exclusive and together fill $S$: $P(A) + P(A') = 1 \quad\Rightarrow\quad P(A') = 1 - P(A)$

Worked examples

Forecast-then-Verify

Common mistakes (Steel-man them)

80/20 — the one thing to keep

Active-recall flashcards

Connections

• Probability — basic definitions & sample space
• General addition rule & inclusion–exclusion
• Independent events — multiplication rule
• Complementary events
• Conditional probability
• Venn diagrams in probability

Concept Map

Hinglish (regional understanding)