A compound event is made of two or more simple events, like "heads and a 6." With a sample space you can find its probability by counting, and multiplying fractions gives a shortcut.
Key idea
For equally likely outcomes, $$P(\text{event}) = \frac{\text{favorable outcomes}}{\text{all outcomes in the sample space}}$$ When the events do not affect each other (independent events), $P(A \text{ and } B) = P(A) \times P(B)$.
Why it works
Half of all coin flips are heads, and $\frac{1}{6}$ of those also come with a 6 on the number cube. A fraction of a fraction is a product: $\frac{1}{6}$ of $\frac{1}{2}$ is $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$. The tree shows the same thing: multiply along a path.
Worked examples
Worked example: Coin and number cube
A coin is flipped and a number cube is rolled. Find the probability of heads and a 6.
- Sample space: $2 \times 6 = 12$ equally likely outcomes. Only H6 fits.
- $P = \frac{1}{12}$. Shortcut: $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ ✓.
Answer: $\frac{1}{12}$
Worked example: At least one head
Three coins are flipped. Find the probability of at least one head.
- Of the 8 outcomes, only TTT has no head.
- So 7 outcomes fit: $P = \frac{7}{8}$.
Answer: $\frac{7}{8}$
Worked example: Two number cubes
Find the probability that two number cubes have a sum of 8.
- Sum 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2). That is 5 outcomes.
- $P = \frac{5}{36}$.
Answer: $\frac{5}{36}$
Common mistake
Adding instead of multiplying for "and": $\frac{1}{2} + \frac{1}{6} = \frac{2}{3}$ for heads and a 6. That is bigger than either chance alone, but needing both things to happen must be less likely than needing one. Check: is your answer smaller than each single probability?