Math Core

If you flip a coin and it lands heads, does that change what a number cube will do next? Of course not. The two results have nothing to do with each other, and that makes the probability of getting both easy to find: just multiply. This lesson makes "nothing to do with each other" precise and shows you how to test for it.

What independence means

Definition

Independent events

Two events $A$ and $B$ are independent if knowing that one of them happened does not change the probability of the other. Otherwise they are dependent.

Some examples:

Flipping a coin and rolling a cube: independent. The coin has no effect on the cube.
Drawing a marble, putting it back, and drawing again: independent. The bag is the same for both draws.
Drawing a marble, keeping it, and drawing again: dependent. The first draw changes what's left in the bag.

The multiplication rule for independent events

Think about flipping a coin and rolling a cube. The sample space has $2 \times 6 = 12$ equally likely outcomes, and exactly one of them is (heads, $5$ ). So $P(\text{heads and } 5) = \dfrac{1}{12}$ . Notice that

\frac{1}{12} = \frac{1}{2} \cdot \frac{1}{6} = P(\text{heads}) \cdot P(5).

That's no accident. Heads happens in half the outcomes, and among those outcomes a $5$ happens one-sixth of the time. Half of one-sixth is one-twelfth.

Multiplication rule (independent events)

If $A$ and $B$ are independent, then

P(A \cap B) = P(A) \cdot P(B).

This extends to more events: if $A$ , $B$ and $C$ are independent, $P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)$ .

Worked example: Drawing with replacement

A bag holds $4$ green marbles and $6$ yellow marbles. You draw one, note its color, put it back, and draw again. What is the probability that both marbles are green?

Because the first marble is replaced, the bag is the same for the second draw, so the draws are independent.

P(\text{green, then green}) = \frac{4}{10} \cdot \frac{4}{10} = \frac{16}{100} = \frac{4}{25}.

Testing for independence

Sometimes it isn't obvious whether two events are related. The multiplication rule works in reverse as a test.

Independence test

Events $A$ and $B$ are independent exactly when $P(A \cap B) = P(A) \cdot P(B)$ . Compute both sides. If they're equal, the events are independent; if not, they're dependent.

Worked example: Two tests with one number cube

Roll one number cube. Let $A$ = "even," $B$ = "at most $4$ ," and $C$ = "at most $3$ ." Is $A$ independent of $B$ ? Of $C$ ?

$A$ and $B$ . $P(A) = \dfrac{3}{6} = \dfrac{1}{2}$ and $P(B) = \dfrac{4}{6} = \dfrac{2}{3}$ . The outcomes in both are $\{2, 4\}$ , so $P(A \cap B) = \dfrac{2}{6} = \dfrac{1}{3}$ .

P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} = P(A \cap B).

The two sides match, so $A$ and $B$ are independent. Knowing the roll is at most $4$ leaves the chance of even at exactly one-half ( $2$ of the $4$ outcomes).

$A$ and $C$ . $P(C) = \dfrac{3}{6} = \dfrac{1}{2}$ . The only outcome in both is $\{2\}$ , so $P(A \cap C) = \dfrac{1}{6}$ . But

P(A) \cdot P(C) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \ne \frac{1}{6}.

So $A$ and $C$ are dependent. Knowing the roll is at most $3$ drops the chance of even to $\dfrac{1}{3}$ .

Common mistake

Independent does not mean mutually exclusive. In fact, mutually exclusive events (with nonzero probabilities) are always dependent: if $A$ happens, then $B$ definitely didn't, so knowing about $A$ changes the probability of $B$ all the way to $0$ . For independent events, use $P(A \cap B) = P(A) \cdot P(B)$ . For mutually exclusive events, $P(A \cap B) = 0$ .

"At least one" problems

Questions that ask for "at least one" success over several independent trials have many cases: exactly one, exactly two, and so on. The complement has only one case: no successes at all.

Worked example: At least one six

You roll a number cube $4$ times. What is the probability of rolling at least one $6$ ?

The complement of "at least one $6$ " is "no $6$ on any roll." Each roll misses $6$ with probability $\dfrac{5}{6}$ , and the rolls are independent, so

P(\text{no } 6) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296}.

P(\text{at least one } 6) = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518.

It's slightly better than a coin flip.

Tip

When you see "at least one," reach for the complement: $P(\text{at least one}) = 1 - P(\text{none})$ .

Practice

Practice 1

You flip a coin and roll a number cube. What is the probability of getting tails and a number greater than $4$ ?