The geometric distribution

By the end of this lesson

By the end you can recognise a geometric situation, write it as $X \sim Geo(p)$ , use $P(X = r) = p(1-p)^{r-1}$ and $E(X) = \dfrac{1}{p}$ , and handle questions about "more than $r$ trials".

Waiting for the first success

The geometric distribution answers a different question from the binomial one.

As in the binomial lesson, a success is the outcome you are counting or waiting for. The binomial counts successes in a fixed number of trials. The geometric distribution has no fixed number: you keep going until the first success, then stop, and $X$ records how many trials that took.

Typical situations: rolling a die until the first six appears, testing components until the first defective one, or phoning numbers until someone answers.

The number of trials is not known in advance, so $X$ has no upper limit. Its possible values are $1, 2, 3, \dots$ and they continue indefinitely.

Definition

The geometric distribution

Suppose independent trials each succeed with constant probability $p$ . Let $X$ be the number of trials up to and including the first success. Then $X$ has the geometric distribution, written $X \sim Geo(p)$ , with

$P(X = r) = p(1-p)^{r-1} \qquad \text{for } r = 1, 2, 3, \dots$

and expectation

$E(X) = \dfrac{1}{p}$

Reading the formula as a sentence

The formula describes exactly one sequence of events.

For the first success to happen on trial $r$ , the first $r - 1$ trials must all fail and trial $r$ must succeed. Independence lets us multiply:

$\underbrace{(1-p) \times (1-p) \times \cdots \times (1-p)}_{r-1 \text{ failures}} \times\ p = p(1-p)^{r-1}$

There is no $^{n}C_{r}$ factor, and that is the key contrast with the binomial distribution: the successes and failures are not free to be in any order. The failures must all come first. There is exactly one arrangement.

The expectation $1/p$ also passes a sanity check. If a trial succeeds one time in five, you would expect to wait about $5$ trials, and indeed $1/0.2 = 5$ .

Worked example: Example 1: the first success on a particular trial

A biased die shows a six with probability $0.2$ on each roll, independently. The die is rolled until a six appears. Find the probability that the first six appears on the fourth roll.

Have a go first?

Check 1 — using the formula

CalculateQuestion 1 · Apply

Multiple choiceQuestion 2 · Recall

Answer the 2 questions above to continue.