Probability of Compound & Independent Events

A compound event is two or more events happening together — flip a coin and roll a die, spin a spinner twice, draw a marble and then draw another. Instead of building a giant sample space for the whole combination, there is a shortcut whenever the events are independent: multiply their probabilities.

Two events are independent when one does not change the other's probability. The coin has no idea what the die rolled; the second spin does not remember the first. When that holds, the rule is $P(A \text{ and } B) = P(A) \cdot P(B)$ — find each probability on its own, then multiply.

What independent means

Independence is about influence, not timing. Rolling a die and flipping a coin are independent because neither result affects the other. Two spins of the same spinner are independent for the same reason — the spinner resets completely between spins.

Drawing objects from a bag is the case to watch. If you draw a marble, put it back, and draw again, the second draw faces the exact same bag, so the draws are independent. If you keep the first marble out, the bag has changed — fewer marbles, different counts — and the events are no longer independent. At this level, problems signal independence with phrases like "put it back," "replace it," or by using objects that cannot affect each other.

The multiplication rule

Once the events are independent, the work is three short steps: find the probability of the first event, find the probability of the second event, and multiply. For a die and a coin, $P(6) = \dfrac{1}{6}$ and $P(\text{heads}) = \dfrac{1}{2}$ , so $P(6 \text{ and heads}) = \dfrac{1}{6} \cdot \dfrac{1}{2} = \dfrac{1}{12}$ .

Notice the answer is smaller than either single probability. That is a built-in sanity check: requiring both things to happen is harder than requiring one, so the combined probability must shrink. If your "and" answer comes out bigger than one of the pieces, you added when you should have multiplied.

The rule extends to any number of independent events — three events means multiplying three probabilities.

Checking with a sample space

The multiplication rule is not magic; it is the counting principle applied to probability. Flip a coin twice: the sample space is $HH$ , $HT$ , $TH$ , $TT$ , and exactly one of those $4$ equally likely outcomes is heads-then-heads, so $P(\text{heads, then heads}) = \dfrac{1}{4}$ . The rule gives the same thing directly: $\dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{4}$ . The tree diagram lays out all four paths.

When a problem feels shaky, build the small sample space and count — it will match the multiplication every time the events are independent.

Worked examples

Example 1: a die and a coin

You roll a standard number cube and flip a coin. Find the probability of rolling a $6$ and flipping heads.

The roll does not affect the flip — independent events

P(A \text{ and } B) = P(A) \cdot P(B)

Find each probability

P(6) = \dfrac{1}{6}, \quad P(\text{heads}) = \dfrac{1}{2}

Multiply

\dfrac{1}{6} \cdot \dfrac{1}{2} = \dfrac{1}{12}

Answer: $P(6 \text{ and heads}) = \dfrac{1}{12}$

Example 2: two spins

A spinner has $4$ equal sections, and one section is blue. You spin twice. Find the probability of landing on blue both times.

Probability of blue on one spin

P(\text{blue}) = \dfrac{1}{4}

The spins are independent, so multiply

\dfrac{1}{4} \cdot \dfrac{1}{4} = \dfrac{1}{16}

Answer: $P(\text{blue, then blue}) = \dfrac{1}{16}$

Example 3: drawing with replacement

A bag holds $5$ marbles, and $3$ of them are red. You draw one marble, put it back, and draw again. Find the probability that both draws are red.

Probability of red on one draw

P(\text{red}) = \dfrac{3}{5}

Replacing the marble keeps the second draw identical to the first

P(\text{red, then red}) = \dfrac{3}{5} \cdot \dfrac{3}{5}

Multiply

\dfrac{3}{5} \cdot \dfrac{3}{5} = \dfrac{9}{25}

Answer: $P(\text{red, then red}) = \dfrac{9}{25}$

Try one yourself

Common questions

Why do I multiply instead of add?

"And" means both events must happen, which is a stricter requirement than either alone — so the probability must get smaller, and multiplying fractions between $0$ and $1$ makes them smaller. Adding would say the combined event is more likely than each piece, which cannot be right for "and."

How can I tell if two events are independent?

Ask whether the first result changes the setup for the second. Separate objects (a die and a coin), repeated actions that reset (two spins, two flips), and draws with replacement are independent. Drawing without putting the object back is the classic dependent case, because the pool of outcomes changes.

What does "with replacement" actually change?

Putting the object back restores the bag to its original state, so the second draw has exactly the same probabilities as the first and the multiplication rule applies directly. Without replacement, the totals shrink after the first draw, so the second probability has to be recomputed from the smaller bag.

Does the rule work for more than two events?

Yes. As long as every event is independent of the others, multiply all the probabilities. Flipping three heads in a row is $\dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{8}$ .

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