What is geometric distribution used for?

It models the number of trials until the first success in a sequence of independent experiments.

What’s the difference between geometric and hyper geometric distribution?

Geometric involves independent trials with replacement; hyper geometric involves dependent trials without replacement.

Can the geometric distribution start at 0?

Yes, in some definitions, it starts at 0 (when counting the number of failures before success), but commonly it starts at 1.

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Geometric Distribution

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Geometric Distribution

The Geometric Distribution is a discrete probability distribution that models the number of Bernoulli trials needed for a success to occur. It's widely used in statistics, especially in situations involving repeated independent trials like flipping coins or rolling dice until a desired outcome is achieved. Unlike the Hyper Geometric Distribution, which deals with selection without replacement, the geometric distribution focuses on independent trials with constant success probability.

1.0What is Geometric Distribution?

The Geometric distribution gives the probability that the first success will occur on the k-th trial in a sequence of independent Bernoulli trials, each with the same probability p of success.

Formula for Geometric Distribution

The Probability Mass Function (PMF) of the geometric distribution is:

$P (X = k) = (1 - p)^{k - 1} \cdot p, k = 1, 2, 3, \dots$

Where:

p is the probability of success,
(1 - p) is the probability of failure,
k is the trial number on which the first success occurs.

Example of Geometric Distribution

Example 1: Suppose the probability of getting a heads when flipping a fair coin is p = 0.5. What is the probability that the first heads appear on the 3rd toss?

Solution:

$P (X = 3) = (1 - 0.5)^{2} \cdot 0.5 \Rightarrow P (X = 3) = 0.25 \cdot 0.5 \Rightarrow P (X = 3) = 0.125$ So, the probability is 0.125.

2.0Mean and Variance of Geometric Distribution

$Mean μ = \frac{1}{p} Variance σ^{2} = \frac{1 - p}{p ^{2}}$

Geometric distribution mean tells us the expected number of trials until the first success.

3.0Geometric Distribution CDF (Cumulative Distribution Function)

The Geometric distribution CDF gives the probability that the first success occurs on or before the k-th trial:

$P (X \leq k) = 1 - (1 - p)^{k}$

4.0Geometric Distribution PDF

In discrete distributions, we often use PMF (Probability Mass Function) instead of PDF (Probability Density Function). So:

Geometric distribution pdf typically refers to the PMF:

$P (X = k) = (1 - p)^{k - 1} \cdot p$

Graph:

Here's the graph of the Cumulative Distribution Function (CDF) for a Geometric Distribution with p = 0.3. It shows the probability that the first success occurs on or before trial number kk.

5.0PMF of Geometric Distribution

The Probability Mass Function (PMF) of the Geometric Distribution gives the probability that the first success occurs on the k-th trial.

PMF Formula:

$P (X = k) = (1 - p)^{k - 1} \cdot p for k = 1, 2, 3, \dots$

Where:

X = number of trials until the first success
p = probability of success on each trial
(1 - p) = probability of failure on each trial
k = trial number of first success

Graph:

Here’s the graph of the Probability Mass Function (PMF) for a Geometric Distribution with p=0.3p = 0.3. The x-axis represents the trial number kk, and the y-axis gives the probability that the first success occurs on the trial.

Here’s the graph of the Probability Mass Function (PMF) for a Geometric Distribution with p=0.3p=0.3. The x-axis represents the trial number k, and the y-axis gives the probability that the first success occurs on the $k^{th}$ trial.

Example:

If a coin has a probability of heads p = 0.5, then the probability that the first head appears on the 3rd toss is:

$P (X = 3) = (1 - 0.5)^{2} \cdot 0.5 = 0.25 \cdot 0.5 = 0.125$

6.0Probability Generating Function of Geometric Distribution

The probability generating function (PGF) of a geometric distribution is:

$G_{X} (t) = \frac{p t}{1 - ( 1 - p ) t}, ∣ t ∣ < \frac{1}{1 - p}$

This function helps in finding moments like mean and variance.

7.0Geometric Distribution vs Hyper Geometric Distribution

Feature	Geometric Distribution	Hyper Geometric Distribution
Replacement	With replacement	Without replacement
Independence	Trials are independent	Trials are dependent
Use case	Wait for the first success	Probability of k successes in n draws

8.0Solved Example on Geometric Distribution

Example 1: A fair die is rolled repeatedly until a “4” appears. What is the probability that the number “4” appears for the first time on the 3rd roll?

Solution:

$This follows a geometric distribution with p = \frac{1}{6} (probability of getting a 4) We need: P (X = 3) = (1 - p)^{3 - 1} \cdot p \Rightarrow P (X = 3) = (\frac{5}{6})^{2} \cdot \frac{1}{6} = \frac{25}{216} Answer: \frac{25}{216}$

Example 2: The probability of success in a single trial is p=0.2p = 0.2. Find the expected number of trials until the first success.

Solution:

In geometric distribution:

$Mean (Expected value) = 1 p = 10.2 = 5 Mean (Expected value) = \frac{1}{p} = \frac{1}{0.2} = 5$

Example 3:

Solution:

$Using the memoryless property: P (X > 4 ∣ X > 2) = P (X > 2) = (1 - p)^{2} = (0.6)^{2} = 0.36 Answer: 0.36$

Example 4: A light bulb has a 10% chance of failing on any given day. What is the probability that it survives at least 5 days?

Solution:

$Let X be the day the first failure occurs. We want P (X > 5) = (1 - p)^{5} = (0.9)^{5} = 0.59049 Answer: 0.59049$

Example 5: A man tries to start his bike each morning. The probability of success on each try is 0.3. He pays ₹1 for each attempt. What is the expected cost to start the bike successfully?

Solution:

$L e tX \sim Geometric (0.3) . Expected number of attempts = \frac{1}{p} = \frac{1}{0.3} = \frac{10}{3} \approx 3.33 Expected cost = \frac{10}{3} Answer: \frac{10}{3}$

Example 6: A biased coin has a probability of heads p = 0.3. What is the probability that the first head appears on the 4th toss?

Solution:

$P (X = 4) = (1 - 0.3)^{3} \cdot 0.3 \Rightarrow P (X = 4) = (0.7)^{3} \cdot 0.3 \Rightarrow P (X = 4) = 0.343 \cdot 0.3 \Rightarrow P (X = 4) = 0.1029 Answer: 0.1029$

9.0Practice Questions on Geometric Distribution

If the probability of a machine failing is 0.1, what is the probability it fails for the first time on the 5th day?
A basketball player has a 0.7 chance of making a free throw. What is the expected number of attempts needed for the first success?
What is the variance of a geometric distribution where p = 0.3?
$Compute P (X \leq 4) when p = 0.25.$
Find the PGF of a geometric distribution with p = 0.4.

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Geometric Distribution

The Geometric Distribution is a discrete probability distribution that models the number of Bernoulli trials needed for a success to occur. It's widely used in statistics, especially in situations involving repeated independent trials like flipping coins or rolling dice until a desired outcome is achieved. Unlike the Hyper Geometric Distribution, which deals with selection without replacement, the geometric distribution focuses on independent trials with constant success probability.

1.0What is Geometric Distribution?

The Geometric distribution gives the probability that the first success will occur on the k-th trial in a sequence of independent Bernoulli trials, each with the same probability p of success.

Formula for Geometric Distribution

The Probability Mass Function (PMF) of the geometric distribution is:

$P (X = k) = (1 - p)^{k - 1} \cdot p, k = 1, 2, 3, \dots$

Where:

p is the probability of success,
(1 - p) is the probability of failure,
k is the trial number on which the first success occurs.

Example of Geometric Distribution

Example 1: Suppose the probability of getting a heads when flipping a fair coin is p = 0.5. What is the probability that the first heads appear on the 3rd toss?

Solution:

$P (X = 3) = (1 - 0.5)^{2} \cdot 0.5 \Rightarrow P (X = 3) = 0.25 \cdot 0.5 \Rightarrow P (X = 3) = 0.125$ So, the probability is 0.125.

2.0Mean and Variance of Geometric Distribution

$Mean μ = \frac{1}{p} Variance σ^{2} = \frac{1 - p}{p ^{2}}$

Geometric distribution mean tells us the expected number of trials until the first success.

3.0Geometric Distribution CDF (Cumulative Distribution Function)

The Geometric distribution CDF gives the probability that the first success occurs on or before the k-th trial:

$P (X \leq k) = 1 - (1 - p)^{k}$

4.0Geometric Distribution PDF

In discrete distributions, we often use PMF (Probability Mass Function) instead of PDF (Probability Density Function). So:

Geometric distribution pdf typically refers to the PMF:

$P (X = k) = (1 - p)^{k - 1} \cdot p$

Graph:

Here's the graph of the Cumulative Distribution Function (CDF) for a Geometric Distribution with p = 0.3. It shows the probability that the first success occurs on or before trial number kk.

5.0PMF of Geometric Distribution

The Probability Mass Function (PMF) of the Geometric Distribution gives the probability that the first success occurs on the k-th trial.

PMF Formula:

$P (X = k) = (1 - p)^{k - 1} \cdot p for k = 1, 2, 3, \dots$

Where:

X = number of trials until the first success
p = probability of success on each trial
(1 - p) = probability of failure on each trial
k = trial number of first success

Graph:

Here’s the graph of the Probability Mass Function (PMF) for a Geometric Distribution with p=0.3p = 0.3. The x-axis represents the trial number kk, and the y-axis gives the probability that the first success occurs on the trial.

Here’s the graph of the Probability Mass Function (PMF) for a Geometric Distribution with p=0.3p=0.3. The x-axis represents the trial number k, and the y-axis gives the probability that the first success occurs on the $k^{th}$ trial.

Example:

If a coin has a probability of heads p = 0.5, then the probability that the first head appears on the 3rd toss is:

$P (X = 3) = (1 - 0.5)^{2} \cdot 0.5 = 0.25 \cdot 0.5 = 0.125$

6.0Probability Generating Function of Geometric Distribution

The probability generating function (PGF) of a geometric distribution is:

$G_{X} (t) = \frac{p t}{1 - ( 1 - p ) t}, ∣ t ∣ < \frac{1}{1 - p}$

This function helps in finding moments like mean and variance.

7.0Geometric Distribution vs Hyper Geometric Distribution

Feature	Geometric Distribution	Hyper Geometric Distribution
Replacement	With replacement	Without replacement
Independence	Trials are independent	Trials are dependent
Use case	Wait for the first success	Probability of k successes in n draws

8.0Solved Example on Geometric Distribution

Example 1: A fair die is rolled repeatedly until a “4” appears. What is the probability that the number “4” appears for the first time on the 3rd roll?

Solution:

$This follows a geometric distribution with p = \frac{1}{6} (probability of getting a 4) We need: P (X = 3) = (1 - p)^{3 - 1} \cdot p \Rightarrow P (X = 3) = (\frac{5}{6})^{2} \cdot \frac{1}{6} = \frac{25}{216} Answer: \frac{25}{216}$

Example 2: The probability of success in a single trial is p=0.2p = 0.2. Find the expected number of trials until the first success.

Solution:

In geometric distribution:

$Mean (Expected value) = 1 p = 10.2 = 5 Mean (Expected value) = \frac{1}{p} = \frac{1}{0.2} = 5$

Example 3:

Solution:

$Using the memoryless property: P (X > 4 ∣ X > 2) = P (X > 2) = (1 - p)^{2} = (0.6)^{2} = 0.36 Answer: 0.36$

Example 4: A light bulb has a 10% chance of failing on any given day. What is the probability that it survives at least 5 days?

Solution:

$Let X be the day the first failure occurs. We want P (X > 5) = (1 - p)^{5} = (0.9)^{5} = 0.59049 Answer: 0.59049$

Example 5: A man tries to start his bike each morning. The probability of success on each try is 0.3. He pays ₹1 for each attempt. What is the expected cost to start the bike successfully?

Solution:

$L e tX \sim Geometric (0.3) . Expected number of attempts = \frac{1}{p} = \frac{1}{0.3} = \frac{10}{3} \approx 3.33 Expected cost = \frac{10}{3} Answer: \frac{10}{3}$

Example 6: A biased coin has a probability of heads p = 0.3. What is the probability that the first head appears on the 4th toss?

Solution:

$P (X = 4) = (1 - 0.3)^{3} \cdot 0.3 \Rightarrow P (X = 4) = (0.7)^{3} \cdot 0.3 \Rightarrow P (X = 4) = 0.343 \cdot 0.3 \Rightarrow P (X = 4) = 0.1029 Answer: 0.1029$

9.0Practice Questions on Geometric Distribution

If the probability of a machine failing is 0.1, what is the probability it fails for the first time on the 5th day?
A basketball player has a 0.7 chance of making a free throw. What is the expected number of attempts needed for the first success?
What is the variance of a geometric distribution where p = 0.3?
$Compute P (X \leq 4) when p = 0.25.$
Find the PGF of a geometric distribution with p = 0.4.

Frequently Asked Questions

What is geometric distribution used for?

Is geometric distribution discrete or continuous?

What’s the difference between geometric and hyper geometric distribution?

Can the geometric distribution start at 0?

Frequently Asked Questions

What is geometric distribution used for?

Is geometric distribution discrete or continuous?

What’s the difference between geometric and hyper geometric distribution?

Can the geometric distribution start at 0?

Join ALLEN!

Geometric Distribution

1.0What is Geometric Distribution?

Formula for Geometric Distribution

Example of Geometric Distribution

2.0Mean and Variance of Geometric Distribution

3.0Geometric Distribution CDF (Cumulative Distribution Function)

4.0Geometric Distribution PDF

5.0PMF of Geometric Distribution

6.0Probability Generating Function of Geometric Distribution

7.0Geometric Distribution vs Hyper Geometric Distribution

8.0Solved Example on Geometric Distribution

9.0Practice Questions on Geometric Distribution

Table of Contents

Join ALLEN!

Geometric Distribution

1.0What is Geometric Distribution?

Formula for Geometric Distribution

Example of Geometric Distribution

2.0Mean and Variance of Geometric Distribution

3.0Geometric Distribution CDF (Cumulative Distribution Function)

4.0Geometric Distribution PDF

5.0PMF of Geometric Distribution

6.0Probability Generating Function of Geometric Distribution

7.0Geometric Distribution vs Hyper Geometric Distribution

8.0Solved Example on Geometric Distribution

9.0Practice Questions on Geometric Distribution

Table of Contents