Geometric Distribution

Calculate probabilities related to the number of trials until first success or the number of failures before the first success

Geometric Distribution Calculator

Distribution Parameters

Probability of success on a single trial (0 < p ≤ 1)

Calculation Type

Calculate P(X = k), the probability of getting the first success on the kth trial

Display Options

Geometric Distribution Visualization

p = 0.3, Trials until first success

Probability Table

k P(X = k) P(X ≤ k)

Geometric Distribution Overview

The geometric distribution models the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials, each with the same probability of success.

Probability Mass Function (PMF)

Trials until first success (X ≥ 1):

\[ P(X = k) = p(1-p)^{k-1} \]

Failures before first success (X ≥ 0):

\[ P(X = k) = p(1-p)^k \]

Cumulative Distribution Function (CDF)

Trials until first success:

\[ P(X \leq k) = 1 - (1-p)^k \]

Failures before first success:

\[ P(X \leq k) = 1 - (1-p)^{k+1} \]

Expected Value

Trials until first success:

\[ E[X] = \frac{1}{p} \]

Failures before first success:

\[ E[X] = \frac{1-p}{p} \]

Variance

For both versions:

\[ Var(X) = \frac{1-p}{p^2} \]

Key Properties

  • The geometric distribution is memoryless: P(X > m+n | X > m) = P(X > n)
  • It is the only discrete distribution with this property
  • The distribution is always positively skewed (right-tailed)
  • As p approaches 0, both the mean and variance increase rapidly
  • The mode is always at k = 1 (for trials version) or k = 0 (for failures version)

Understanding the Geometric Distribution

Definition & Characteristics

The geometric distribution models the number of trials required to achieve the first success in a sequence of independent Bernoulli trials, where each trial has the same probability of success p.

There are two common formulations of the geometric distribution:

  1. Trials Until First Success (X ≥ 1): The random variable X represents the number of trials needed to get the first success.
  2. Failures Before First Success (X ≥ 0): The random variable X represents the number of failures before the first success is observed.

The difference between these formulations is simply a shift of 1 in the random variable.

Requirements for a Geometric Distribution

  1. Binary outcomes: Each trial has exactly two possible outcomes: "success" or "failure".
  2. Independence: The outcome of each trial is independent of all other trials.
  3. Constant probability: The probability of success p is the same for each trial.
  4. Counting until first success: We're interested in the trial number when the first success occurs.

Mathematical Formulation

A random variable X follows a geometric distribution with parameter p, written as X ~ Geo(p), if:

  • X represents the number of trials until the first success (or failures before first success, depending on the version)
  • The probability of success in any single trial is p
  • All trials are independent of each other

Properties & Applications

Memoryless Property

The geometric distribution is memoryless, which means that the probability of success in future trials is not affected by what has happened in past trials. Mathematically:

\[ P(X > m+n | X > m) = P(X > n) \]

This property is unique among discrete distributions and has important applications in reliability theory and stochastic processes.

Key Properties

Expected Value (Mean)
E[X] = 1/p or (1-p)/p
The average number of trials needed to get the first success
Variance
Var(X) = (1-p)/p²
Measures the spread or dispersion of the distribution
Standard Deviation
σ = √((1-p)/p²)
Typical deviation from the mean
Skewness
(2-p)/√(1-p)
Always positive, indicating right-skewed distribution
Mode
1 (trials version) or 0 (failures version)
The most likely outcome

Real-World Applications

  • Quality Control: Modeling the number of items inspected until finding the first defective item
  • Reliability Theory: Time until failure for components with constant failure rates
  • Gambling: Number of bets until the first win
  • Medicine: Number of patients screened until finding one with a rare disease
  • Marketing: Number of sales calls until the first successful sale
  • Epidemiology: Number of individuals tested until finding the first infected person

Example Problems

Example 1: Quality Control

A manufacturing process produces items with a 5% defect rate. Quality control inspectors examine items one at a time until they find a defective item. What is the probability that they need to examine exactly 10 items to find the first defective item?

This is a geometric distribution where:

  • p = 0.05 (probability of finding a defective item)
  • We want P(X = 10), the probability of finding the first defective item on the 10th inspection
\begin{align} P(X = 10) &= p(1-p)^{10-1} \\ &= 0.05 \times (1-0.05)^9 \\ &= 0.05 \times 0.95^9 \\ &= 0.05 \times 0.6302 \\ &= 0.0315 \end{align}

The probability of finding the first defective item on exactly the 10th inspection is 0.0315 or about 3.15%.

Example 2: Sales Calls

A salesperson has a 20% success rate on sales calls. What is the probability that they will need to make at most 3 calls to make their first sale?

This is a geometric distribution where:

  • p = 0.2 (probability of success on a single call)
  • We want P(X ≤ 3), the probability of making the first sale within the first 3 calls
\begin{align} P(X \leq 3) &= 1 - (1-p)^3 \\ &= 1 - (1-0.2)^3 \\ &= 1 - 0.8^3 \\ &= 1 - 0.512 \\ &= 0.488 \end{align}

The probability that the salesperson will make their first sale within 3 calls is 0.488 or about 48.8%.

Example 3: Expected Number of Attempts

A basketball player makes free throws with 75% accuracy. On average, how many free throws must the player attempt before making one?

This is a geometric distribution where:

  • p = 0.75 (probability of making a free throw)
  • We want E[X], the expected number of attempts until the first success
\begin{align} E[X] &= \frac{1}{p} \\ &= \frac{1}{0.75} \\ &= 1.333... \end{align}

On average, the player will need to attempt about 1.33 free throws before making one. In other words, they will usually make a free throw on the first or second attempt.

Example 4: Memoryless Property

A person rolls a fair six-sided die repeatedly until they roll a 6. If they've already rolled 10 times without getting a 6, what is the probability that they will need at least 5 more rolls to get a 6?

This problem demonstrates the memoryless property of the geometric distribution.

  • p = 1/6 (probability of rolling a 6)
  • We want P(X > 10+4 | X > 10) = P(X > 4), the probability of needing at least 5 more rolls
\begin{align} P(X > 4) &= (1-p)^4 \\ &= (1-1/6)^4 \\ &= (5/6)^4 \\ &= 0.482 \end{align}

Due to the memoryless property, the probability is the same as if they were just starting to roll. The probability that they will need at least 5 more rolls is about 0.482 or 48.2%.

Test Your Knowledge

Quick Quiz: Geometric Distribution

1. A fair coin is flipped until a head appears. What is the probability that exactly 4 flips will be needed?

A. 1/16
B. 1/8
C. 3/16
D. 1/4

2. If the probability of success on a single trial is 0.4, what is the expected number of trials until the first success?

A. 2.5
B. 0.4
C. 1.5
D. 4

3. Which of the following statements about the geometric distribution is FALSE?

A. It models the number of trials until the first success
B. It has the memoryless property
C. The variance is always less than the mean
D. The distribution is always right-skewed