# The Geometric Distribution

> Mathematics · CIE A-Level 9709
> Source: https://www.owlsprep.com/study/cie-9709-u4-the-geometric-distribution/

This guide covers all CIE 9709 S1 requirements for the geometric distribution: model conditions, probability formulae, tail results, mode, and expectation, with no out-of-scope content.

**Prerequisites:** [Discrete random variables basics](https://www.owlsprep.com/study/cie-9709-u4-discrete-random-variables/); [Binomial distribution fundamentals](https://www.owlsprep.com/study/cie-9709-u4-the-binomial-distribution/)

## Learning objectives

- Identify practical scenarios where the geometric distribution is a valid model and use the X ~ Geo(p) notation
- Calculate individual geometric probabilities using the MF19 formula
- Evaluate cumulative tail probabilities P(X > r) and P(X ≤ r) for geometric distributions
- State the mode and compute the expected value of any geometric distribution

## Conditions for a Valid Geometric Model

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.

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

*Notation:* $X \sim \text{Geo}(p)$

> **warning**
>
> CIE explicitly defines X as the trial number of the first success, not the number of failures before the first success. Using the alternate definition will give incorrect results for all standard formulae.

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

## Core Individual Probability Formula

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 $X \sim \text{Geo}(0.25)$, 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

## Tail Probabilities and Mode

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:** 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. $$P(X > r) = q \times q \times ... \times q \quad (r \text{ times})$$

*Conclusion:* $P(X > r) = q^r = (1-p)^r$

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

> **info**
>
> Every geometric distribution has a mode of 1, because the probability values $P(X=r)$ decrease strictly as r increases. No calculation is required to state the mode.

**Worked example:** For $X \sim \text{Geo}(0.1)$, 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. $$P(X \leq 3) = 1 - (1-0.1)^3 = 1 - 0.729 = 0.271$$

## Expectation of the Geometric Distribution

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.

**Check your understanding**

Test your understanding of core geometric rules:

1. What is the mode of X ~ Geo(0.7)?

   - 0.7
   - 1
   - 1.43
   - 7

   *Why:* All geometric distributions have a mode of 1, no calculation needed.

2. For X ~ Geo(0.2), what is E(X)?

   - 0.2
   - 2
   - 5
   - 0.8

   *Why:* E(X) = 1/p = 1/0.2 = 5.

## Common pitfalls

- **Wrong:** Defining X as the number of failures before the first success
  - Why it fails: 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: Always confirm X counts trials up to and including the first success before applying any geometric formula
- **Wrong:** Attempting to calculate the variance of a geometric distribution
  - Why it fails: Geometric variance is explicitly out of S1 syllabus scope, and irrelevant working will lose you answer marks
  - Correct: Only ever calculate the expectation E(X) = 1/p for geometric distribution questions
- **Wrong:** Using P(X = r) = p(1-p)^r instead of p(1-p)^{r-1}
  - Why it fails: The exponent counts failures before the first success, which is r-1, not r
  - Correct: Double check the exponent is one less than the trial number r when substituting values
- **Wrong:** Applying geometric formulae to scenarios where trials are not independent
  - Why it fails: All geometric results rely on the assumption that trial outcomes do not affect each other
  - Correct: Verify all four geometric model conditions before using any probability or expectation formula
- **Wrong:** Stating the mode of a geometric distribution is not 1
  - Why it fails: All geometric distributions have strictly decreasing probability values, so the highest probability always occurs at r=1
  - Correct: Memorise that the mode of any Geo(p) is 1, no calculation is required

## 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 |

## 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](https://www.owlsprep.com/study/cie-9709-u4-the-binomial-distribution/)

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