Mutually Exclusive Events

Roll a single die. Can it come up 2 and 5 on the same roll? Of course not — it shows one face, so it is either a 2, or a 5, but never both. Two events that cannot both happen at once like this are called mutually exclusive. They rule each other out.

This "can't-both-happen" fact is the whole reason the next idea works. When two events can't overlap, the chance that one or the other occurs is simply their probabilities added together — the OR rule:

P(A \text{ or } B) = P(A) + P(B)

There is no double-counting to worry about, because no single outcome belongs to both events at the same time. That is exactly what "mutually exclusive" buys us.

Three worked examples

1. A 2 or a 5 on one die. Each face has probability \tfrac{1}{6}, and one roll can't be both, so they're mutually exclusive. Add:

P(2 \text{ or } 5) = \tfrac{1}{6} + \tfrac{1}{6} = \tfrac{2}{6} = \tfrac{1}{3}.

2. Not a six — the complement. The chance of rolling a six is \tfrac{1}{6}. Rather than add up the chances of 1, 2, 3, 4 and 5, flip it round with the complement rule:

P(\text{not a six}) = 1 - P(\text{six}) = 1 - \tfrac{1}{6} = \tfrac{5}{6}.

3. A spinner. A spinner has red, blue and green sectors with P(\text{red}) = 0.4 and P(\text{blue}) = 0.25. Red and blue can't both win, so P(\text{red or blue}) = 0.4 + 0.25 = 0.65. And since red, blue and green form a complete set that must sum to 1, green is the leftover: P(\text{green}) = 1 - 0.65 = 0.35.

See it on a spinner

Each sector is equally likely. Landing on red and landing on blue are mutually exclusive, so add their chances. Step through the spinner.

Exhaustive events: it all adds up to 1

On that spinner, red, blue and green between them cover every possible result. A set of outcomes that leaves nothing out like this is called exhaustive, and when the pieces are also mutually exclusive their probabilities must add up to exactly 1 — because on every spin something has to happen:

\tfrac{3}{6} + \tfrac{2}{6} + \tfrac{1}{6} = \tfrac{6}{6} = 1.

This is why the complement rule works. "A six" and "not a six" are mutually exclusive and exhaustive — together they cover everything — so their chances add to 1, which rearranges to P(\text{not a six}) = 1 - P(\text{six}).

Where this shows up

Mutually exclusive events are everywhere once you look. A traffic light is red, amber or green — never two at once — and those three chances must add to 1. A weather forecast of "70% chance of rain" instantly tells you the chance of a dry day: 1 - 0.7 = 0.3, straight from the complement rule. A single exam grade is one letter; a football match is a home win, an away win, or a draw. In each case the outcomes can't overlap, so you add to find "one or the other" and subtract from 1 to find "not this one".

The habit worth building: when a problem lists outcomes that can't coincide, adding and the complement rule will almost always be the quickest route to the answer.

The tidy rule P(A) + P(B) works only for mutually exclusive events. The moment two events can happen together, plain adding double-counts the overlap and gives an answer that is too big.

Take a normal pack of cards. There are 13 hearts, so P(\text{heart}) = \tfrac{13}{52}, and 4 kings, so P(\text{king}) = \tfrac{4}{52}. Add them and you get \tfrac{17}{52} — but that's wrong, because the king of hearts is a heart and a king, and you just counted it twice. The real answer is \tfrac{16}{52}.

Overlapping events like this need the general addition rule, which subtracts the double-counted bit — you'll meet it with Venn diagrams. Until then, always ask the safety question first: "can both happen at once?" If no, add away. If yes, adding will overshoot.

The complement rule P(\text{not } A) = 1 - P(A) looks almost too simple to be useful. It is one of the most-used shortcuts in all of probability.

Suppose you roll a die four times and want the chance of at least one six. Counting that head-on is a nightmare: exactly one six, exactly two sixes, three, four… so many cases. But "at least one six" is just the opposite of "no sixes at all", and no sixes is easy — each roll misses with chance \tfrac{5}{6}:

P(\text{at least one six}) = 1 - \left(\tfrac{5}{6}\right)^4 \approx 0.52.

Whenever a question says "at least one", reach for the complement — it turns a horrible pile-up of cases into a single easy subtraction.