Complements and Mutually Exclusive Events

Many probability questions involve “or” and “not”: the chance of drawing a king or a heart, or of not rolling a six. Two simple rules handle these, as long as you watch for outcomes that get counted twice. Venn diagrams make that easy to see.

Key ideas

Complements

The complement of event $A$, written $A'$ (or $\sim\!A$), is everything in the sample space that is not in $A$. Since either $A$ happens or it doesn’t:

$$P(A') = 1 - P(A)$$

The complement is often the easy way to find “at least one” probabilities.

Mutually exclusive events

Events are mutually exclusive if they can’t happen at the same time: they share no outcomes. Rolling a $2$ and rolling an odd number are mutually exclusive. Drawing a king and drawing a heart are not (the king of hearts is both).

For mutually exclusive events:

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

The additive principle

When events can overlap, adding $P(A) + P(B)$ counts the overlap twice, so subtract it once:

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

This works for all events. For mutually exclusive events, $P(A \text{ and } B) = 0$, which gives the simpler rule.

Venn diagrams

A Venn diagram shows events as overlapping circles inside a rectangle (the sample space). The overlap is ”$A$ and $B$”; both circles together are ”$A$ or $B$”; outside the circles is “neither”.

A class of $30$: $12$ in band, $9$ in choir, $4$ in both.

Worked examples

Example 1: Complements

Find $P(\text{not rolling a } 6)$ on one die, and $P(\text{at least one head})$ when three coins are flipped.

Solution.

$$P(\text{not } 6) = 1 - \frac{1}{6} = \frac{5}{6}$$

“At least one head” is the complement of “no heads” (TTT), which has probability $\tfrac{1}{8}$:

$$P(\text{at least one head}) = 1 - \frac{1}{8} = \frac{7}{8}$$

Example 2: Mutually exclusive events

One card is drawn from a standard deck. Find $P(\text{king or queen})$.

Solution. A card can’t be both, so the events are mutually exclusive:

$$P(\text{king or queen}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$$

Example 3: Overlapping events

One card is drawn. Find $P(\text{king or heart})$.

Solution. The king of hearts is in both events, so subtract it once:

$$P(\text{king or heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$$

Example 4: Using a Venn diagram

In the class above, a student is chosen at random. Find $P(\text{band or choir})$, $P(\text{neither})$, and $P(\text{choir only})$.

Solution.

$$P(\text{band or choir}) = \frac{12 + 9 - 4}{30} = \frac{17}{30}$$

(Or add the regions: $8 + 4 + 5 = 17$.)

$$P(\text{neither}) = \frac{13}{30}, \qquad P(\text{choir only}) = \frac{5}{30} = \frac{1}{6}$$

Common mistakes

Adding probabilities of overlapping events. $P(\text{king or heart})$ is not $\tfrac{4}{52} + \tfrac{13}{52}$: that counts the king of hearts twice.

Confusing “mutually exclusive” with “independent”. Mutually exclusive events can’t happen together. Independent events don’t affect each other (see independent events). They’re different ideas.

Putting the total in the “only” region. In a Venn diagram, if $12$ are in band and $4$ of them are also in choir, the “band only” region holds $12 - 4 = 8$.

Forgetting the “neither” region. It’s part of the sample space too.

Practice

1. (Warm-up) If $P(A) = 0.35$, find $P(A')$.

Solution

$1 - 0.35 = 0.65$

2. (Warm-up) Are the events mutually exclusive?

• (a) rolling a $2$; rolling an odd number
• (b) drawing a red card; drawing a king
• (c) a student being $15$ years old; the same student being $16$ years old

Solution

(a) Yes. (b) No: the king of hearts and king of diamonds are both. (c) Yes.

3. (Warm-up) $A$ and $B$ are mutually exclusive, with $P(A) = 0.2$ and $P(B) = 0.45$. Find $P(A \text{ or } B)$.

Solution

$0.2 + 0.45 = 0.65$

4. (Core) A die is rolled. Find $P(\text{even or greater than } 3)$.

Solution

Even: $\{2, 4, 6\}$. Greater than $3$: $\{4, 5, 6\}$. Both: $\{4, 6\}$.

$$\frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3}$$

5. (Core) $P(A) = 0.5$, $P(B) = 0.4$, and $P(A \text{ or } B) = 0.7$. Find $P(A \text{ and } B)$. Are $A$ and $B$ mutually exclusive?

Solution

$0.7 = 0.5 + 0.4 - P(A \text{ and } B)$, so $P(A \text{ and } B) = 0.2$. Since that isn’t $0$, they’re not mutually exclusive.

6. (Core) In a survey of $50$ students, $28$ play a sport, $20$ play an instrument, and $8$ do both. Draw a Venn diagram, then find the probability that a randomly chosen student plays a sport or an instrument, plays neither, and plays a sport only.

Solution

Regions: sport only $20$, both $8$, instrument only $12$, neither $50 - 40 = 10$.

$P(\text{sport or instrument}) = \tfrac{40}{50} = 0.8$, $P(\text{neither}) = \tfrac{10}{50} = 0.2$, $P(\text{sport only}) = \tfrac{20}{50} = 0.4$.

7. (Core) Two dice are rolled. Find $P(\text{sum is } 7 \text{ or doubles})$.

Solution

No double adds to $7$ (it would need $3.5 + 3.5$), so the events are mutually exclusive:

$$\frac{6}{36} + \frac{6}{36} = \frac{12}{36} = \frac{1}{3}$$

8. (Challenge) Two dice are rolled. Find $P(\text{sum is } 8 \text{ or doubles})$.

Solution

Sum $8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$, so $5$ outcomes. Doubles: $6$ outcomes. Both: $(4,4)$.

$$\frac{5}{36} + \frac{6}{36} - \frac{1}{36} = \frac{10}{36} = \frac{5}{18}$$

9. (Challenge) Explain why $P(A \text{ or } B)$ can never be more than $P(A) + P(B)$, and when the two are equal.

Solution

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$, and $P(A \text{ and } B) \ge 0$, so subtracting it can only make the total smaller or leave it the same. They’re equal exactly when $P(A \text{ and } B) = 0$, which means the events are mutually exclusive.