The Geometric Distribution

Mathematics· 9709 S1 Section 5.4, Coursebook Ch 7.2· 12 min read

1. Conditions for a Valid Geometric Model★★☆☆☆⏱ 3 min

A geometric distribution is only a suitable model if all four of these rules hold: every trial is independent of all other trials, each trial has exactly two mutually exclusive outcomes (success or failure), the probability of success $p$ is constant across all trials, and you are counting the number of trials up to and including the first success.

📘 Definition

Geometric Distribution

$X \sim Geo (p)$

Discrete distribution for the trial number on which the first success occurs, where $r = 1, 2, 3, ...$

📐 Worked Example

State whether a geometric distribution is a suitable model for the scenario: 'A fair 6-sided die is rolled repeatedly until a 6 is rolled, X is the number of total rolls.'

1
Check the four conditions one by one:
2
1. Each die roll is independent of all previous rolls: satisfied.
3
1. Each trial has two outcomes: roll a 6 (success) or not roll a 6 (failure): satisfied.
4
1. Probability of success $p = 1/6$ is constant: satisfied.
5
1. X counts total rolls up to and including the first 6: satisfied.
6
Conclusion: X ~ Geo(1/6) is a valid model.

2. Core Individual Probability Formula★★☆☆☆⏱ 3 min

The probability that the first success occurs exactly on the $r$ -th trial is given by the standard MF19 formula. This comes from having $r - 1$ consecutive failures before the single success on the $r$ -th trial.

P(X = r) = p(1-p)^{r-1} \quad \text{for } r = 1, 2, 3, ...

📐 Worked Example

For $X \sim Geo (0.25)$ , calculate $P (X = 4)$ , giving your answer to 3 significant figures.

1
Identify values: $p = 0.25$ , $r = 4$
2
Substitute into the formula:
3
$P(X=4) = 0.25 \times (1-0.25)^{4-1} = 0.25 \times 0.75^3$
4
$0.25 \times 0.421875 = 0.10546875$
5
Round to 3 significant figures: 0.105

3. Tail Probabilities and Mode★★★☆☆⏱ 3 min

The most commonly tested non-individual probability result for geometric distributions is the tail probability $P (X > r)$ . This describes the event that the first $r$ trials are all failures, with no success in the first $r$ attempts.

🔬 Derivation

Goal:

Prove $P (X > r) = q^{r}$

Starting from:

X > r means the first success does not occur in trials 1 to r

1
For the first success to be after trial r, every one of the first r trials must be a failure
2
Trials are independent, so multiply the probability of failure for each trial
3
$P(X > r) = q \times q \times ... \times q \quad (r \text{ times})$

Result:

$P (X > r) = q^{r} = (1 - p)^{r}$

P(X \leq r) = 1 - P(X > r) = 1 - (1-p)^r

📐 Worked Example

For $X \sim Geo (0.1)$ , calculate $P (X > 5)$ and $P (X \leq 3)$ to 3 s.f.

1
First calculate $P (X > 5)$ :
2
$P(X > 5) = (1-0.1)^5 = 0.9^5 = 0.59049 \approx 0.590$
3
Next calculate $P (X \leq 3)$ :
4
$P(X \leq 3) = 1 - (1-0.1)^3 = 1 - 0.729 = 0.271$

4. Expectation of the Geometric Distribution★★☆☆☆⏱ 3 min

The expected (mean) number of trials required to get the first success for a geometric distribution is a simple result given directly on the MF19 formula sheet. Note that variance of the geometric distribution is explicitly out of the CIE 9709 S1 syllabus, and you will never be asked to calculate it.

E(X) = \frac{1}{p}

📐 Worked Example

A fair coin is flipped repeatedly until heads appears. Find the expected number of total flips required.

1
Identify the success probability: $p = 0.5$ for heads on a fair coin
2
$E(X) = \frac{1}{p} = \frac{1}{0.5} = 2$
3
The expected number of flips to get the first head is 2.

✓ Quick check

Test your understanding of core geometric rules:

What is the mode of X ~ Geo(0.7)?
- 0.7
- 1
- 1.43
- 7
Reveal answer
1 —
All geometric distributions have a mode of 1, no calculation needed.
For X ~ Geo(0.2), what is E(X)?
- 0.2
- 2
- 5
- 0.8
Reveal answer
5 —
E(X) = 1/p = 1/0.2 = 5.

5. Common Pitfalls

Wrong move:

Defining X as the number of failures before the first success

Why:

CIE's standard notation defines X as the trial number of the first success, so all formulae will return incorrect values for this alternate definition

Correct move:

Always confirm X counts trials up to and including the first success before applying any geometric formula

Wrong move:

Attempting to calculate the variance of a geometric distribution

Why:

Geometric variance is explicitly out of S1 syllabus scope, and irrelevant working will lose you answer marks

Correct move:

Only ever calculate the expectation E(X) = 1/p for geometric distribution questions

Wrong move:

Using P(X = r) = p(1-p)^r instead of p(1-p)^{r-1}

Why:

The exponent counts failures before the first success, which is r-1, not r

Correct move:

Double check the exponent is one less than the trial number r when substituting values

Wrong move:

Applying geometric formulae to scenarios where trials are not independent

Why:

All geometric results rely on the assumption that trial outcomes do not affect each other

Correct move:

Verify all four geometric model conditions before using any probability or expectation formula

Wrong move:

Stating the mode of a geometric distribution is not 1

Why:

All geometric distributions have strictly decreasing probability values, so the highest probability always occurs at r=1

Correct move:

Memorise that the mode of any Geo(p) is 1, no calculation is required

6. Quick Reference Cheatsheet

Quantity	Formula (Geo(p))	Notes
P(X = r)	p(1-p)^{r-1}	r ≥ 1, first success on trial r
P(X > r)	(1-p)^r	First r trials are all failures
P(X ≤ r)	1 - (1-p)^r	Complement of all first r trials failing
Mode	1	True for all valid values of p
E(X)	1/p	Given on the MF19 formula sheet

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 52
Geometric probability and expectation
2024 · 51
Identify geometric model context
2023 · 53
Tail probability calculation

What's Next

Now that you have mastered the geometric distribution, you can move on to compare it against the binomial distribution, the other core discrete named model in CIE 9709 S1. Many exam questions ask you to select the correct model from a context, so practicing distinguishing between binomial (fixed number of trials, count total successes) and geometric (stop at first success) scenarios will immediately boost your Paper 5 score. The geometric distribution is a low-effort high-mark topic that appears in almost every other S1 paper, so revisiting this guide before your exam will help you secure full marks on any related question.

→The Binomial Distribution