Mutually exclusive and exhaustive events and addition laws

• Apply the general addition law for non-mutually exclusive events:

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

Exhaustive means nothing is left out

An exhaustive set of events includes every possible outcome. ‘Even’ and ‘odd’ are exhaustive for a six-sided die because every possible number belongs to one of them. The six individual events ‘roll a 1’, ‘roll a 2’, and so on up to ‘roll a 6’ also form an exhaustive set.

Mutually exclusive and exhaustive describe different properties. ‘Roll a 1’ and ‘roll a 2’ are mutually exclusive, but not exhaustive: 3, 4, 5 and 6 are left out. Exhaustive events need not be mutually exclusive; covering all outcomes does not itself rule out overlap.

When events are both mutually exclusive and exhaustive, their probabilities add to 1. Every possible outcome is included exactly once.

In particular, ‘$A$ happens’ and ‘$A$ does not happen’ are mutually exclusive and exhaustive. Therefore,

For example, if a spinner has probability 0.2 of landing on yellow, its probability of not landing on yellow is $1-0.2=0.8$.

Overlapping events and the general addition law

Higher tier: the general addition law handles events that can occur together.

For a die roll, let $A$ be ‘even’ and $B$ be ‘a multiple of 3’. These events overlap because 6 belongs to both. In probability, ‘$A$ or $B$’ means $A$, $B$, or both. ‘$A$ and $B$’ means the overlap.

The Venn diagram places each outcome in exactly one region. To find $P(A)$, include the $A$-only region and the overlap; to find $P(B)$, include the $B$-only region and the overlap. Adding these probabilities counts the overlap twice. Subtracting it once leaves each wanted outcome counted once:

For a fair six-sided die, $P(A)=\frac36$, $P(B)=\frac26$ and $P(A \text{ and } B)=\frac16$. Hence

This matches the four wanted outcomes: 2, 3, 4 and 6. The events are not exhaustive because 1 and 5 are outside both circles.

The general law works for any two events. If they are mutually exclusive, the overlap probability is zero, so it reduces to the simpler addition law.

Finding a missing probability

The general addition law can also be rearranged. Suppose $P(A)=0.2$, $P(A \text{ or } B)=0.7$ and $P(A \text{ and } B)=0.1$. Substitution gives

Therefore $P(B)=0.7-0.2+0.1=0.6$. This is the probability of all of $B$, including its overlap with $A$.

For an event and its complement: