Geometric and Poisson Distributions

The binomial distribution counts successes in a fixed number of trials. But what if you want to know how many trials until the first success? Or how many events occur in a fixed time period? These questions require two other discrete distributions: the geometric distribution and the Poisson distribution. Together with the binomial, they form the essential toolkit for modeling count data in statistics.

The Geometric Distribution

The geometric distribution models the number of trials until the first success. Instead of fixing the number of trials in advance (as binomial does), you keep going until success occurs and ask: how long did it take?

Conditions

The geometric distribution applies when:

• Each trial has exactly two outcomes (success or failure)
• The probability of success $p$ is the same on every trial
• Trials are independent
• You count the trial number of the first success (there is no fixed $n$)

Formula

The probability that the first success occurs on trial $k$:

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

The logic: you need $k - 1$ failures in a row (each with probability $1 - p$), followed by one success (probability $p$).

Mean (expected number of trials until first success):

$\mu = \frac{1}{p}$

Standard deviation:

$\sigma = \frac{\sqrt{1-p}}{p}$

Example 1: Free Throw Shooter

A basketball player makes free throws with probability $p = 0.70$. What is the probability that the first successful free throw comes on the 3rd attempt?

This means the player misses the first two shots, then makes the third:

$P(X = 3) = (1 - 0.70)^{3-1} \times 0.70 = (0.30)^2 \times 0.70 = 0.09 \times 0.70 = 0.063$

Answer: There is a 6.3% probability that the first make comes on the 3rd attempt. This is relatively unlikely because a 70% shooter usually makes it on the first or second try.

Example 2: Expected Number of Trials

For the same 70% free throw shooter, how many attempts do we expect before the first make?

$\mu = \frac{1}{p} = \frac{1}{0.70} \approx 1.43$

Answer: On average, the first successful free throw comes in about 1.4 attempts. This makes intuitive sense — a 70% shooter usually makes the first one.

Notice how the geometric distribution is always right-skewed. The most likely outcome is success on the very first trial, and the probabilities decrease by a factor of $(1 - p)$ with each additional trial.

Example 3: Quality Control — Defect Detection

A quality inspector examines items on an assembly line. The defect rate is $p = 0.02$ (2% of items are defective). What is the probability that the first defective item is found within the first 10 inspections?

Rather than calculating $P(X = 1) + P(X = 2) + \cdots + P(X = 10)$, use the complement:

$P(X \leq 10) = 1 - P(X > 10) = 1 - (1 - p)^{10} = 1 - (0.98)^{10}$

Computing $(0.98)^{10}$: first find $(0.98)^5 = 0.9039$, then square it: $(0.9039)^2 = 0.8170$.

$P(X \leq 10) = 1 - 0.8170 = 0.1830$

Answer: There is about an 18.3% chance of finding at least one defective item in the first 10 inspections. Even though each item has a 2% defect rate, looking at 10 items gives a nearly one-in-five chance of catching a defect.

Cumulative Geometric Probability

The general formula for the cumulative probability (first success by trial $k$):

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

This shortcut avoids summing individual probabilities. It works because $P(X > k) = (1-p)^k$ — the probability of $k$ consecutive failures.

The Poisson Distribution

The Poisson distribution models the count of events in a fixed interval of time or space, when events occur randomly and independently at a known average rate. It is named after French mathematician Simeon Denis Poisson.

Conditions

The Poisson distribution applies when:

• Events occur independently (one event does not affect the probability of the next)
• The average rate $\lambda$ (lambda) is constant across the interval
• Two events cannot occur at the exact same instant
• You are counting discrete events in a continuous interval

Formula

The probability of observing exactly $k$ events:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

Where $\lambda$ is the average number of events per interval, $e \approx 2.71828$, and $k! = k \times (k-1) \times \cdots \times 1$ is the factorial.

Mean: $\mu = \lambda$

Standard deviation: $\sigma = \sqrt{\lambda}$

Notice that the Poisson distribution has a single parameter $\lambda$ that determines both the center and the spread.

Example 4: Hospital Emergency Room

An emergency room receives an average of $\lambda = 4$ patients per hour. What is the probability of receiving exactly 6 patients in a given hour?

$P(X = 6) = \frac{e^{-4} \times 4^6}{6!}$

Computing each piece:

• $e^{-4} = 0.01832$
• $4^6 = 4{,}096$
• $6! = 720$

$P(X = 6) = \frac{0.01832 \times 4096}{720} = \frac{75.04}{720} \approx 0.1042$

Answer: There is about a 10.4% probability of exactly 6 patients arriving in one hour when the average rate is 4 per hour.

Example 5: Typos in a Document

A document averages $\lambda = 2$ typos per page. What is the probability of a page having zero typos?

$P(X = 0) = \frac{e^{-2} \times 2^0}{0!} = \frac{0.1353 \times 1}{1} = 0.1353$

Answer: About 13.5% of pages will have no typos at all. So roughly 1 in 7 or 8 pages will be error-free.

The Poisson distribution is slightly right-skewed for small $\lambda$ and becomes more symmetric as $\lambda$ increases. When $\lambda$ is an integer, the mode occurs at both $\lambda$ and $\lambda - 1$ (as shown above with $\lambda = 4$).

Comparing the Three Discrete Distributions

Understanding when to use each distribution is one of the most important skills in applied statistics. Here is a side-by-side comparison:

When to Use Each Distribution

Choosing the right distribution comes down to identifying what you are counting and what is fixed:

Use the binomial distribution when:

• You have a fixed number of trials $n$
• Each trial has two outcomes (success/failure)
• You are counting the total number of successes

Use the geometric distribution when:

• Trials are independent with constant probability $p$
• There is no fixed number of trials
• You are counting the number of trials until the first success

Use the Poisson distribution when:

• Events occur randomly in a continuous interval (time, area, volume)
• The average rate $\lambda$ is known and constant
• You are counting the number of events in a fixed interval

A helpful test: ask yourself, “Is there a fixed number of trials?” If yes, use binomial. If no, ask “Am I waiting for one event, or counting events in an interval?” Waiting for one event points to geometric; counting events in an interval points to Poisson.

Real-World Application: Nursing — Patient Call Rates

A nursing station receives an average of $\lambda = 3$ call-light requests per 30-minute period during the evening shift. The charge nurse needs to plan staffing. What is the probability of receiving 5 or more calls in a 30-minute window?

This is a Poisson problem: we are counting events (call-light requests) in a fixed time interval (30 minutes), with a known average rate.

Step 1: Find $P(X \geq 5) = 1 - P(X \leq 4)$.

Step 2: Calculate $P(X \leq 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$.

Using $\lambda = 3$ and $e^{-3} = 0.0498$:

$P(X=0) = \frac{e^{-3} \times 3^0}{0!} = 0.0498$

$P(X=1) = \frac{e^{-3} \times 3^1}{1!} = 0.1494$

$P(X=2) = \frac{e^{-3} \times 3^2}{2!} = \frac{0.0498 \times 9}{2} = 0.2241$

$P(X=3) = \frac{e^{-3} \times 3^3}{3!} = \frac{0.0498 \times 27}{6} = 0.2241$

$P(X=4) = \frac{e^{-3} \times 3^4}{4!} = \frac{0.0498 \times 81}{24} = 0.1681$

Step 3: Sum and subtract.

$P(X \leq 4) = 0.0498 + 0.1494 + 0.2241 + 0.2241 + 0.1681 = 0.8155$

$P(X \geq 5) = 1 - 0.8155 = 0.1845$

Answer: There is about an 18.5% probability of receiving 5 or more call-light requests in any given 30-minute period. This means that roughly one out of every five or six half-hour windows will have a higher-than-expected volume.

Staffing implication: If the nursing station can comfortably handle up to 4 calls per 30 minutes with current staffing, then about 18.5% of the time the staff will be stretched beyond capacity. The charge nurse might consider adding a float nurse during peak evening hours when the consequences of delayed response (patient falls, missed medications) are most serious.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The geometric distribution models the number of trials until the first success: $P(X = k) = (1-p)^{k-1} \cdot p$, with mean $\mu = 1/p$
• The Poisson distribution models the count of events in a fixed interval: $P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}$, with mean $\mu = \lambda$ and standard deviation $\sigma = \sqrt{\lambda}$
• Geometric answers “How many trials until the first success?” while Poisson answers “How many events in this time period?”
• The binomial requires a fixed number of trials; the geometric does not fix $n$; the Poisson does not involve trials at all
• For cumulative geometric probabilities, use the shortcut $P(X \leq k) = 1 - (1-p)^k$
• For cumulative Poisson probabilities, sum individual terms: $P(X \leq k) = \sum_{i=0}^{k} \frac{e^{-\lambda}\lambda^i}{i!}$
• In healthcare and staffing, the Poisson distribution helps predict demand (patient arrivals, call volumes) for resource planning

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