Compound Probability: Two Events at Once

Listing every outcome

A compound event involves more than one thing happening. The method from single events still works, as long as every possible outcome is listed.

Flip two coins. The sample space is:

$$HH, \quad HT, \quad TH, \quad TT$$

Four equally likely outcomes, so:

$$P(\text{two heads}) = \frac{1}{4} \qquad P(\text{one of each}) = \frac{2}{4} = \frac{1}{2}$$

$HT$ and $TH$ are different outcomes. The coins land in an order, and treating “one head one tail” as a single outcome would give $\tfrac{1}{3}$, which is wrong — it happens twice as often as two heads do.

A table for two events

When each event has several outcomes, a table catches them all. Here are the sums from rolling two dice:

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

Thirty-six equally likely outcomes. Counting them:

$$P(\text{sum of } 7) = \frac{6}{36} = \frac{1}{6} \qquad P(\text{sum of } 2) = \frac{1}{36}$$

The table also settles a question people get wrong by instinct. The sums run from $2$ to $12$, and they are not equally likely: seven appears six times and two appears once.

Tree diagrams

A tree shows each stage as a branching, which suits three or more events or unequal probabilities.

Flipping a coin twice:

        ┌── H ── HH
   ┌ H ─┤
   │    └── T ── HT
───┤
   │    ┌── H ── TH
   └ T ─┤
        └── T ── TT

Each complete path is one outcome, and there are four of them. A third flip would double the branches again to eight.

Multiplying along a branch

For independent events — where the first does not affect the second — multiply:

$$P(A \text{ and } B) = P(A) \times P(B)$$ $$P(\text{two heads}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ $$P(\text{six then six}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$$

Both match what listing gives, which is the point: multiplying is a shortcut for counting the paths, and it stays usable when the list would be far too long to write out.

Why dependent events need care

Multiplying only works when the second probability is unchanged by the first. Often it is not.

A bag holds $3$ red and $2$ blue marbles. Draw two, keeping the first.

$$P(\text{first red}) = \frac{3}{5}$$

After a red is taken out, there are $4$ marbles left and only $2$ are red:

$$P(\text{second red}) = \frac{2}{4} = \frac{1}{2}$$ $$P(\text{both red}) = \frac{3}{5} \times \frac{1}{2} = \frac{3}{10}$$

Both numbers changed — the total fell from $5$ to $4$, and the reds fell from $3$ to $2$. Using $\tfrac{3}{5}$ twice would give $\tfrac{9}{25}$, an answer to a different question.

	Independent	Dependent
example	flipping twice	drawing without replacing
second probability	unchanged	recalculated
how to tell	the first leaves no trace	the first removes or alters something

Putting the marble back makes the draws independent again, and then $\tfrac{3}{5} \times \tfrac{3}{5}$ is correct. Whether the item is replaced is the whole question, which is why exam questions always say.

Simulation

Some compound events are too tangled to list. When that happens, run the process many times and count — a simulation.

To find how often a family of three children is all girls, you could list the eight outcomes. Or you could flip three coins repeatedly and record how often all three come up heads. Both give $\tfrac{1}{8}$, and only the second still works when the situation gets complicated enough that nobody can write down the sample space.

Worked examples

Common mistakes

Practice problems