Geometric Distribution

AS · S1 · 3 min

The geometric distribution counts how many trials are needed to get the first success, when each trial succeeds independently with probability $p$ . Unlike the binomial, the number of trials is not fixed, and the variable has no upper limit.

The model

Key result

$X \sim \operatorname{Geo}(p)$ when independent trials each succeed with probability $p$ and $X$ is the number of the trial on which the first success occurs.

P(X = r) = (1 - p)^{r - 1}p, \qquad r = 1, 2, 3, \ldots

P(X > r) = (1 - p)^r, \qquad P(X \leq r) = 1 - (1 - p)^r, \qquad E(X) = \frac{1}{p}

The variance is not in the syllabus.

The formula reads " $r - 1$ failures, then a success". $P(X > r)$ is just "the first $r$ trials all fail", which makes cumulative probabilities a one-line calculation.

Basic probabilities

A fair die is thrown until a six appears. Find the probability that the first six is (a) on the 4th throw; (b) after the 3rd throw; (c) within the first 5 throws.

Solution

$X \sim \operatorname{Geo}\left(\tfrac{1}{6}\right)$ .

(a) $\left(\tfrac{5}{6}\right)^3 \cdot \tfrac{1}{6} = 0.0965$ . (b) $\left(\tfrac{5}{6}\right)^3 = 0.579$ . (c) $1 - \left(\tfrac{5}{6}\right)^5 = 0.598$ .

Expected number of trials

A darts player hits the bullseye with probability $0.15$ on each throw. How many throws are expected to be needed to hit it, and what is the probability it takes more than the expected number?

Solution

$E(X) = \dfrac{1}{0.15} = 6.67$ throws. $P(X > 6) = 0.85^6 = 0.377$ (using the whole number $6$ ; $P(X > 7) = 0.321$ ).

Between two values

$X \sim \operatorname{Geo}(0.3)$ . Find $P(3 \leq X \leq 6)$ and $P(X = 4 \mid X > 2)$ .

Solution

$P(3 \leq X \leq 6) = P(X > 2) - P(X > 6) = 0.7^2 - 0.7^6 = 0.49 - 0.1176 = 0.372$ .

$P(X = 4 \mid X > 2) = \dfrac{P(X = 4)}{P(X > 2)} = \dfrac{0.7^3 \times 0.3}{0.7^2} = 0.21$ . (This is $P(X = 2)$ : the process has no memory.)

Finding p

For $X \sim \operatorname{Geo}(p)$ , $P(X = 2) = 0.21$ . Find the possible values of $p$ , and the value of $E(X)$ in each case.

Solution

$(1 - p)p = 0.21 \Rightarrow p^2 - p + 0.21 = 0 \Rightarrow (p - 0.3)(p - 0.7) = 0$ .

$p = 0.3$ : $E(X) = 3.33$ . $p = 0.7$ : $E(X) = 1.43$ .

Binomial or geometric?

Question asks	Model
how many successes in $n$ trials	$B(n, p)$
on which trial the first success occurs	$\operatorname{Geo}(p)$
whether the first success is within $r$ trials	$\operatorname{Geo}(p)$ , $P(X \leq r)$

Choosing the model

Calls to a helpline are answered within 10 seconds with probability $0.4$ , independently. (a) Find the probability that of 8 calls exactly 3 are answered in time. (b) Find the probability that the first call answered in time is the 3rd call.

Solution

(a) $B(8, 0.4)$ : $\binom{8}{3}(0.4)^3(0.6)^5 = 56 \times 0.064 \times 0.07776 = 0.279$ .

(b) $\operatorname{Geo}(0.4)$ : $0.6^2 \times 0.4 = 0.144$ .

Watch out

$X$ starts at $1$ , not $0$ : the first success cannot happen on the "zeroth" trial. $P(X \leq r) = 1 - (1 - p)^r$ , with the power $r$ , not $r - 1$ .

Exam tip

State the model and parameter, $X \sim \operatorname{Geo}(0.15)$ , then use $P(X > r) = (1 - p)^r$ for cumulative questions rather than adding terms. Examiners accept either, but the single-power method avoids arithmetic slips.

Practice

Question

$X \sim \operatorname{Geo}(0.25)$ . Find $P(X = 3)$ , $P(X \leq 3)$ and $P(X > 5)$ .
A coin is biased so that $P(\text{head}) = 0.6$ . Find the expected number of tosses to get the first head, and the probability that the first head is on an even-numbered toss.
$P(X > 2) = 0.64$ for a geometric variable. Find $p$ and $P(X = 1)$ .
Seeds germinate independently with probability $0.8$ . Find the probability that the first seed to fail is the 5th seed planted.

Answers

$0.75^2 \times 0.25 = 0.141$ ; $1 - 0.75^3 = 0.578$ ; $0.75^5 = 0.237$ .
$\tfrac{1}{0.6} = 1.67$ ; $P(\text{even}) = 0.4(0.6) + 0.4^3(0.6) + \cdots = \dfrac{0.24}{1 - 0.16} = \tfrac{2}{7}$ .
$(1 - p)^2 = 0.64 \Rightarrow p = 0.2$ ; $P(X = 1) = 0.2$ .
"Success" is failing to germinate, $p = 0.2$ : $0.8^4 \times 0.2 = 0.0819$ .

The model

Binomial or geometric?

Practice

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