Geometric Distribution

Core Concept

The Geometric distribution models the number of independent Bernoulli trials until the first success.

• A Bernoulli trial is an experiment with two outcomes: success (1) or failure (0).
• Trials are independent (e.g., coin flips, dice rolls).
• The distribution answers: “What is the probability that the first success occurs on the \(n\)-th trial?”

Definitions

Let:

• \(p\) = probability of success on a single trial
• \(q = 1 - p\) = probability of failure
• \(X\) = number of trials until the first success

Then:

\[ X \sim \text{Geometric}(p) \]

The Probability Mass Function (PMF) is:

\[ \begin{aligned} P(\text{first success on the } n\text{th trial}) &= P(F_1 \cap F_2 \cap \dots \cap F_{n-1} \cap S_n) \\ &= P(F)^{n - 1} \cdot P(S) \quad \\ &= (1 - p)^{n - 1} \cdot p \\ &= q^{n - 1} \cdot p \\ \text{where } q &= 1 - p, \\ \text{and each } F_i &\sim \text{Bernoulli(0), and } S_n \sim \text{Bernoulli(1)}) \end{aligned} \]

Intuition

To succeed for the first time on the \(n\)-th trial:

• First \(n - 1\) trials must be failures (\(q^{n - 1}\))
• The \(n\)-th trial must be a success (\(p\))

So:

\[ P(X = n) = q^{n - 1} \cdot p = (1 - p)^{n - 1} \cdot p \]

Type	Formula
Support	\(x \in \{1, 2, 3, \dots\}\)
Mean	\(\mathbb{E}[X] = \frac{1}{p}\)
Variance	\(\text{Var}(X) = \frac{1 - p}{p^2}\)
Standard Deviation	\(\sigma = \frac{\sqrt{1 - p}}{p}\)
PMF (Probability Mass Function)	\(\mathbb{P}(X = k) = (1 - p)^{k - 1} p\)
CDF (Cumulative Distribution Function)	\(\mathbb{P}(X \leq k) = 1 - (1 - p)^k\)
Memoryless Property	\(\mathbb{P}(X > m + n \mid X > m) = \mathbb{P}(X > n)\)

Example

Question: What is the probability that it takes at least 10 rolls of a fair die to roll a 5?

• Success = rolling a 5
• \(p = \frac{1}{6}, q = \frac{5}{6}\)
• \(X \sim \text{Geometric}\left(\frac{1}{6}\right)\)

We want: \(P(X \geq 10)\)

Method 1: Summing Infinite Tail

\[ P(X \geq 10) = \sum_{k=10}^{\infty} P(X = k) = \sum_{k=10}^{\infty} q^{k - 1} \cdot p \]

Factor out \(q^9 \cdot p\):

\[ P(X \geq 10) = q^9 \cdot p \cdot \left(1 + q + q^2 + \dots \right) \]

This is a geometric series, sum of an infinite geometric series where:

• The first term is 1,
• The common ratio is \(q\),
• And \(∣q∣<1\) (so the series converges)

\[ p\sum_{k=0}^{\infty} q^k = \frac{a}{1 - r} = \frac{1}{1 - q} = \frac{1}{p} \]

So:

\[ P(X \geq 10) = q^9 \cdot p \cdot \frac{1}{p} = q^9 \]

Final answer:

\[ P(X \geq 10) = \left( \frac{5}{6} \right)^9 \]

Method 2: No Success in First 9 Trials

\[ P(X \geq 10) = P(\text{no success in first 9 trials}) = q^9 = \left( \frac{5}{6} \right)^9 \]

Same result, simpler reasoning.

When to Use the Poisson Distribution

• Number of independent trials until the first success.
• Each trial has the same success probability \(p\).
• Trials are independent Bernoulli (success/failure) events.