What is wrong exponent in the geometric formula?

It is (1 - p)^x-1p: x - 1 failures then a success, not (1-p)^x p.

What is wrong combination in the negative binomial?

Use y - 1r - 1, since the rth success is fixed on the last trial and only the first y - 1 trials are arranged.

What is mean of geometric as p?

The mean is 1p, not p; a small success probability means many trials on average.

Write the geometric probability P(X = x) for success probability p. [1 mark]

SingaporeFurther MathsSyllabus dot point

What are the geometric and negative binomial distributions, and when does each model a counting situation?

Recognise and apply the geometric and negative binomial distributions, including their probabilities, expectations and variances

A focused answer to the H2 Further Mathematics outcome on the geometric and negative binomial distributions. Their probability formulae, when each applies, the expectation and variance of each, and the link to the binomial distribution.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to recognise when a counting situation follows the geometric distribution (number of trials to the first success) or the negative binomial distribution (number of trials to the $r$ th success), to write down and use their probability formulae, and to state and apply their expectations and variances. Both arise from independent trials with a constant success probability.

The answer

The setting: independent trials

Both distributions assume a sequence of independent trials, each a success (probability $p$ ) or failure (probability $1 - p$ ), with $p$ constant. The difference is what is counted.

The geometric distribution

If $X$ is the number of trials up to and including the first success, then $X \sim \text{Geometric}(p)$ with

\mathrm{P}(X = x) = (1 - p)^{x - 1}p, \qquad x = 1, 2, 3, \dots

There must be $x - 1$ failures followed by one success. Its mean and variance are

\mathrm{E}(X) = \frac{1}{p}, \qquad \operatorname{Var}(X) = \frac{1 - p}{p^2}.

The mean $\tfrac{1}{p}$ matches intuition: a success with probability $\tfrac{1}{6}$ takes on average $6$ trials.

The negative binomial distribution

If $Y$ is the number of trials up to and including the $r$ th success, then $Y$ is negative binomial with parameters $r$ and $p$ :

\mathrm{P}(Y = y) = \binom{y - 1}{r - 1}p^{r}(1 - p)^{y - r}, \qquad y = r, r + 1, \dots

The combination counts the ways to place $r - 1$ successes among the first $y - 1$ trials, with the $r$ th success on trial $y$ . Its mean and variance are

\mathrm{E}(Y) = \frac{r}{p}, \qquad \operatorname{Var}(Y) = \frac{r(1 - p)}{p^2}.

The relationships between the distributions

The geometric distribution is the negative binomial with $r = 1$ . And the negative binomial counting trials to the $r$ th success is the sum of $r$ independent geometric variables, which is why its mean $\tfrac{r}{p}$ and variance $\tfrac{r(1-p)}{p^2}$ are $r$ times the geometric values.

Worked example

A quality inspector tests items one at a time; each is defective independently with probability $0.2$ . Find the probability that the first defective is the third item tested, and the expected number tested to find the first defective.

Step 1: Identify the distribution

"Number of trials to the first success (defective)" with $p = 0.2$ is geometric: $X \sim \text{Geometric}(0.2)$ .

Step 2: Apply the geometric probability

\mathrm{P}(X = 3) = (1 - 0.2)^{3 - 1}(0.2) = (0.8)^2(0.2) = 0.64\times 0.2 = 0.128.

Step 3: Interpret the result

There is a $12.8\%$ chance the first defective is exactly the third item: two good items then a defective.

Step 4: Find the expected number tested

\mathrm{E}(X) = \frac{1}{p} = \frac{1}{0.2} = 5.

Step 5: State the answer

The probability is $0.128$ , and on average $5$ items are tested to find the first defective.

Common traps

Confusing with the binomial: The binomial fixes the number of trials and counts successes; the geometric and negative binomial fix the number of successes and count trials. Read which quantity is fixed.
Wrong exponent in the geometric formula: It is $(1 - p)^{x-1}p$ : $x - 1$ failures then a success, not $(1-p)^x p$ .
Wrong combination in the negative binomial: Use $\binom{y - 1}{r - 1}$ , since the $r$ th success is fixed on the last trial and only the first $y - 1$ trials are arranged.
Mean of geometric as $p$: The mean is $\tfrac{1}{p}$ , not $p$ ; a small success probability means many trials on average.
Forgetting the support: Geometric $X \geq 1$ and negative binomial $Y \geq r$ ; the count cannot be smaller than the number of successes sought.

Examples in context

Example 1. Reliability testing. The number of components tested before the first failure follows a geometric distribution; its mean $\tfrac{1}{p}$ estimates how many units a test run will consume, central to quality-control planning.

Example 2. Sales conversions. If each sales call succeeds with a fixed probability, the number of calls needed to close $r$ deals is negative binomial; its mean $\tfrac{r}{p}$ forecasts the effort required to hit a target, a standard model in operations.

Try this

Q1. Write the geometric probability $\mathrm{P}(X = x)$ for success probability $p$ . [1 mark]

Cue. $(1 - p)^{x-1}p$ for $x = 1, 2, 3, \dots$ .

Q2. State the mean of a geometric distribution with parameter $p$ . [1 mark]

Cue. $\dfrac{1}{p}$ .

Q3. What distribution counts the number of trials to the $r$ th success? [1 mark]

Cue. The negative binomial distribution with parameters $r$ and $p$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original6 marksA fair die is rolled repeatedly until a six appears. Let

X

be the number of rolls needed. State the distribution of

X

, and find

\mathrm{P}(X = 4)

and

\mathrm{E}(X)

Show worked answer →

Each roll is an independent trial with $\mathrm{P}(\text{six}) = p = \tfrac{1}{6}$ , and $X$ counts the trial on which the first success occurs, so $X \sim \text{Geometric}(p)$ with $p = \tfrac{1}{6}$ .

The geometric probability is $\mathrm{P}(X = x) = (1 - p)^{x-1}p$ . For $x = 4$ :

\mathrm{P}(X = 4) = \left(\frac{5}{6}\right)^{3}\cdot\frac{1}{6} = \frac{125}{216}\cdot\frac{1}{6} = \frac{125}{1296} \approx 0.0965.

The mean is

\mathrm{E}(X) = \dfrac{1}{p} = \dfrac{1}{1/6} = 6

Markers reward identifying the geometric distribution with $p = \tfrac{1}{6}$ , the probability formula giving $\tfrac{125}{1296}$ , and the mean $\tfrac{1}{p} = 6$ .

Original7 marksA basketball player has probability

0.4

of scoring each free throw, independently. Let

Y

be the number of throws needed to score the third basket. State the distribution of

Y

and find

\mathrm{P}(Y = 5)