Hypergeometric and Poisson Distributions | Malone Math

The Central Idea

The hypergeometric distribution counts successes in a fixed-size sample drawn without replacement from a finite population containing a known number of successes and failures.

The Poisson distribution counts the number of events in a fixed exposure interval when events occur according to a stable average rate.

Both are discrete distributions, but their modeling assumptions are very different.

Why the Hypergeometric Distribution Is Needed

The binomial distribution assumes independent trials with constant success probability.

Sampling without replacement violates those assumptions because every draw changes the composition of the remaining population.

The hypergeometric distribution handles this dependence exactly.

The Hypergeometric Setup

Suppose a finite population contains:

\(N\) total objects;
\(K\) objects classified as successes;
\(N-K\) objects classified as failures.

A sample of size \(n\) is drawn without replacement.

Let \(X\) be the number of successes in the sample.

Hypergeometric Notation

One common notation is

\[ \boxed{X\sim\operatorname{Hypergeometric}(N,K,n)}. \]

Here:

\(N\) is the population size;
\(K\) is the number of successes in the population;
\(n\) is the sample size.

Deriving the Hypergeometric Formula

To obtain exactly \(k\) successes in the sample:

choose \(k\) objects from the \(K\) successes;
choose \(n-k\) objects from the \(N-K\) failures.

The number of favorable samples is therefore

\[ \binom{K}{k}\binom{N-K}{n-k}. \]

The total number of samples of size \(n\) is

\[ \binom{N}{n}. \]

The Hypergeometric PMF

Therefore,

\[ \boxed{ P(X=k) = \frac{ \binom{K}{k}\binom{N-K}{n-k} }{ \binom{N}{n} } }. \]

This formula counts equally likely samples rather than multiplying independent trial probabilities.

Possible Hypergeometric Values

The number of successes \(k\) must satisfy all physical constraints of the population and sample.

In particular,

\[ \boxed{ \max(0,n-(N-K)) \le k\le \min(n,K) }. \]

These bounds ensure that the requested sample composition is possible.

Example: Sampling Without Replacement

A box contains \(10\) red balls and \(15\) blue balls. Five balls are drawn without replacement.

Let \(X\) be the number of red balls selected.

Then

\[ N=25,\qquad K=10,\qquad n=5. \]

Find

\[ P(X=2). \]

Complete the Hypergeometric Example

\[ P(X=2) = \frac{ \binom{10}{2}\binom{15}{3} }{ \binom{25}{5} }. \]

Since

\[ \binom{10}{2}=45,\qquad \binom{15}{3}=455,\qquad \binom{25}{5}=53130, \]

we obtain

\[ P(X=2) = \frac{20475}{53130} \approx \boxed{0.3854}. \]

Why the Draws Are Dependent

Suppose the first ball drawn is red.

Before the draw, the probability of red is

\[ \frac{10}{25}. \]

After drawing a red ball without replacement, the probability that the next ball is red becomes

\[ \frac{9}{24}. \]

The probability changes, so the draws are not independent.

Hypergeometric Mean

\[ X\sim\operatorname{Hypergeometric}(N,K,n), \]

then

\[ \boxed{ \mathbb{E}[X] = n\frac{K}{N} }. \]

The mean equals the sample size multiplied by the population proportion of successes.

Interpreting the Hypergeometric Mean

In the red-ball example,

\[ \mathbb{E}[X] = 5\left(\frac{10}{25}\right) = \boxed{2}. \]

Across many samples of five balls, the average number of red balls tends toward \(2\).

Hypergeometric Variance

The variance is

\[ \boxed{ \operatorname{Var}(X) = n\frac{K}{N} \left(1-\frac{K}{N}\right) \frac{N-n}{N-1} }. \]

The factor

\[ \boxed{\frac{N-n}{N-1}} \]

is called the finite population correction.

Why the Finite Population Correction Appears

In a binomial model with success probability \(K/N\), the variance would be

\[ n\frac{K}{N}\left(1-\frac{K}{N}\right). \]

Sampling without replacement introduces dependence that reduces variability.

The factor

\[ \frac{N-n}{N-1} \]

adjusts the binomial-like variance downward.

Hypergeometric Versus Binomial

Hypergeometric	Binomial
Finite population	Repeated Bernoulli trials
Sampling without replacement	Independent trials or replacement-like sampling
Success probability changes	Success probability stays constant
Dependent draws	Independent trials
Combinatorial sampling formula	\(\binom nkp^k(1-p)^{n-k}\)

When the Binomial Can Approximate the Hypergeometric

If the sample is small compared with the population, removing one object changes the remaining proportions only slightly.

In such cases, the hypergeometric distribution may be approximated by a binomial distribution with

\[ p=\frac{K}{N}. \]

This is an approximation, not an exact equality.

A Common Sampling Guideline

A frequently used practical guideline is that the sample size should be no more than a small fraction of the population before treating draws as approximately independent.

The exact adequacy of the approximation depends on the required precision and the application.

Hypergeometric Cumulative Probabilities

To find

\[ P(X\le k), \]

add the relevant hypergeometric probabilities:

\[ \boxed{ P(X\le k) = \sum_j \frac{ \binom{K}{j}\binom{N-K}{n-j} }{ \binom{N}{n} } }. \]

The sum includes only feasible values \(j\le k\).

Hypergeometric Complement Problems

The complement rule is often efficient.

For example,

\[ P(X\ge1) = 1-P(X=0). \]

This avoids summing many possible positive values.

The Poisson Distribution

The Poisson distribution models the number of events occurring in a fixed interval of time, distance, area, volume, or another exposure measure.

The model is characterized by an average event count

\[ \lambda>0. \]

Examples of Poisson-Type Counts

Depending on whether the assumptions are reasonable, Poisson models may be used for counts such as:

calls arriving at a service center during one minute;
defects appearing along a length of material;
accidents occurring at an intersection during a month;
particles detected during a fixed observation period;
customers arriving during a short time interval;
rare events occurring across a large number of opportunities.

Poisson Notation

If \(X\) counts events with average count \(\lambda\), we write

\[ \boxed{ X\sim\operatorname{Poisson}(\lambda) }. \]

The possible values are

\[ \boxed{ X\in\{0,1,2,3,\ldots\} }. \]

The Poisson PMF

The probability of exactly \(k\) events is

\[ \boxed{ P(X=k) = e^{-\lambda}\frac{\lambda^k}{k!} } \]

for

\[ k=0,1,2,\ldots. \]

Interpreting the Poisson Parameter

The parameter \(\lambda\) represents the expected number of events in the specified interval.

For example, if a service desk receives an average of \(6\) requests per hour, then a one-hour Poisson model would use

\[ \lambda=6. \]

Example: Exactly Four Events

Suppose

\[ X\sim\operatorname{Poisson}(3). \]

Find

\[ P(X=4). \]

Apply the PMF:

\[ P(X=4) = e^{-3}\frac{3^4}{4!}. \]

Complete the Poisson Example

Since

\[ 3^4=81 \]

and

\[ 4!=24, \]

we have

\[ P(X=4) = e^{-3}\frac{81}{24}. \]

Therefore,

\[ \boxed{P(X=4)\approx0.1680}. \]

Poisson Mean

\[ X\sim\operatorname{Poisson}(\lambda), \]

then

\[ \boxed{\mathbb{E}[X]=\lambda}. \]

The parameter itself is the expected count.

Poisson Variance

A distinctive property of the Poisson distribution is

\[ \boxed{\operatorname{Var}(X)=\lambda}. \]

Therefore,

\[ \boxed{\sigma_X=\sqrt{\lambda}}. \]

Mean and variance are equal in the ideal Poisson model.

Poisson Rate and Interval Length

Suppose events occur at average rate

\[ r \]

per unit of exposure.

Over an interval of length \(t\), the expected count is

\[ \boxed{\lambda=rt}. \]

The Poisson model for that interval therefore uses

\[ X\sim\operatorname{Poisson}(rt). \]

Example: Changing the Time Interval

Suppose calls arrive at an average rate of \(4\) per hour.

Over \(30\) minutes,

\[ t=\frac12\text{ hour}. \]

Therefore,

\[ \lambda = 4\left(\frac12\right) = \boxed{2}. \]

The half-hour count can be modeled as

\[ X\sim\operatorname{Poisson}(2). \]

Poisson Cumulative Probabilities

To find the probability of at most \(k\) events,

\[ P(X\le k), \]

sum the PMF values:

\[ \boxed{ P(X\le k) = \sum_{j=0}^{k} e^{-\lambda}\frac{\lambda^j}{j!} }. \]

At Least One Poisson Event

The complement rule gives a particularly simple formula:

\[ P(X\ge1) = 1-P(X=0). \]

Since

\[ P(X=0)=e^{-\lambda}, \]

we obtain

\[ \boxed{ P(X\ge1)=1-e^{-\lambda} }. \]

Example: At Least One Event

Suppose

\[ X\sim\operatorname{Poisson}(0.5). \]

Then

\[ P(X\ge1) = 1-e^{-0.5}. \]

Therefore,

\[ \boxed{P(X\ge1)\approx0.3935}. \]

Poisson Process Assumptions

A Poisson count model is most natural when:

events occur one at a time;
the average rate is stable across the interval being modeled;
counts in disjoint intervals behave independently;
the probability of multiple events in a sufficiently tiny interval is negligible relative to one event;
the event count is the quantity of interest.

Rate Stability Matters

A constant-rate Poisson model may be inappropriate when the event rate changes systematically over the interval.

For example, customer arrivals may vary strongly by hour of day.

In that case, one constant \(\lambda\) across the entire day may oversimplify the process.

Independence Across Disjoint Intervals

In an ideal Poisson process, the number of events occurring in one interval does not affect the number occurring in a disjoint interval.

This independent-increments property is a central structural assumption of the classical Poisson process.

Adding Independent Poisson Variables

\[ X\sim\operatorname{Poisson}(\lambda_1) \]

and

\[ Y\sim\operatorname{Poisson}(\lambda_2) \]

are independent, then

\[ \boxed{ X+Y \sim \operatorname{Poisson}(\lambda_1+\lambda_2) }. \]

Independent Poisson counts combine by adding their rates.

Example: Combining Two Independent Sources

Suppose one source generates events according to

\[ X\sim\operatorname{Poisson}(2) \]

and another independent source generates

\[ Y\sim\operatorname{Poisson}(5). \]

Then the total count is

\[ \boxed{ X+Y\sim\operatorname{Poisson}(7) }. \]

The Poisson Approximation to the Binomial

When

\[ n \]

is large and

\[ p \]

is small, a binomial distribution may sometimes be approximated by a Poisson distribution with

\[ \boxed{\lambda=np}. \]

Why the Approximation Makes Sense

A large number of independent opportunities with a small event probability creates a rare-event counting process.

The expected number of events is

\[ np. \]

The Poisson approximation preserves that mean through

\[ \lambda=np. \]

Example: Rare Binomial Events

Suppose

\[ X\sim\operatorname{Binomial}(1000,0.002). \]

Then

\[ np=2. \]

A Poisson approximation would use

\[ \boxed{ Y\sim\operatorname{Poisson}(2) }. \]

Hypergeometric Versus Poisson

Hypergeometric	Poisson
Finite population sampling	Event counts over exposure
Sampling without replacement	Rate-based event process
Parameters \(N,K,n\)	Parameter \(\lambda\)
Dependent draws	Independent counts across disjoint intervals in the ideal process
Finite support	Support \(0,1,2,\ldots\)

Recognizing the Correct Distribution

Ask what mechanism is generating the count.

Fixed number of independent success-failure trials: binomial.
Sampling without replacement from a finite population: hypergeometric.
Waiting for the first success: geometric.
Waiting for the \(r\)-th success: negative binomial.
Counting events over a fixed exposure interval at a stable average rate: Poisson.

Common Error: Using the Binomial Formula for Sampling Without Replacement

Without replacement, the success probability changes after each draw.

Therefore a standard binomial model is generally not exact.

Use the hypergeometric distribution when the finite population composition is known.

Common Error: Forgetting the Failure Combination

In the hypergeometric formula,

\[ \binom{K}{k} \]

chooses the successes, but the failures must also be selected:

\[ \binom{N-K}{n-k}. \]

Both factors are required.

Common Error: Using the Wrong Denominator

The denominator of the hypergeometric PMF is

\[ \boxed{\binom{N}{n}}, \]

the number of possible samples of size \(n\) from the full population.

Common Error: Treating \(\lambda\) as a Probability

The Poisson parameter \(\lambda\) is an expected count, not a probability.

It can be larger than \(1\).

For example,

\[ \lambda=12 \]

simply means the expected count is \(12\) events per modeled interval.

Common Error: Forgetting to Rescale the Rate

If a rate is given per hour but the question asks about \(15\) minutes, the Poisson parameter must be adjusted to the shorter interval.

Always use

\[ \lambda=(\text{rate})(\text{exposure length}). \]

Common Error: Assuming Every Count Is Poisson

The fact that a random variable counts events does not automatically make it Poisson.

The underlying rate stability and independence assumptions must be reasonable.

Common Error: Confusing Poisson Mean and Standard Deviation

For a Poisson random variable,

\[ \mathbb{E}[X]=\lambda \]

and

\[ \operatorname{Var}(X)=\lambda. \]

But the standard deviation is

\[ \boxed{\sigma_X=\sqrt{\lambda}}, \]

not \(\lambda\).

A Reliable Procedure for Hypergeometric Problems

Identify the population size \(N\).
Identify the number of population successes \(K\).
Identify the sample size \(n\).
Confirm that sampling is without replacement.
Define \(X\) as the number of successes in the sample.
Check that the requested value \(k\) is feasible.
Use \(\binom{K}{k}\binom{N-K}{n-k}/\binom{N}{n}\).
Use complements or cumulative sums when appropriate.

A Reliable Procedure for Poisson Problems

Identify the event being counted.
Identify the exposure interval.
Determine the average rate.
Convert the rate to the requested interval to obtain \(\lambda\).
Check whether the Poisson assumptions are reasonable.
Use \(e^{-\lambda}\lambda^k/k!\) for an exact count.
Use sums or complements for cumulative events.
Interpret the answer in context.

Practice 1: Hypergeometric Exact Probability

A population contains \(8\) successes and \(12\) failures. A sample of \(5\) objects is drawn without replacement.

Find the probability of exactly \(2\) successes.

Practice 2: Hypergeometric Mean

Using Practice 1, find \(\mathbb{E}[X]\).

Practice 3: Hypergeometric Variance

Using Practice 1, find \(\operatorname{Var}(X)\).

Practice 4: At Least One Success

A box contains \(4\) defective and \(16\) nondefective parts. Three parts are sampled without replacement.

Find the probability of selecting at least one defective part.

Practice 5: Poisson Exact Probability

Let

\[ X\sim\operatorname{Poisson}(4). \]

Find

\[ P(X=2). \]

Practice 6: Poisson Mean and Variance

Let

\[ X\sim\operatorname{Poisson}(7). \]

Find the mean, variance, and standard deviation.

Practice 7: Rescale the Rate

Events occur at an average rate of \(12\) per hour.

Find the Poisson parameter for a \(10\)-minute interval.

Practice 8: At Least One Poisson Event

Let

\[ X\sim\operatorname{Poisson}(1.5). \]

Find

\[ P(X\ge1). \]

Practice 9: Choose the Model

Identify the appropriate model.

Number of red cards in five cards drawn without replacement from a standard deck.
Number of calls received during the next minute at a stable average rate.
Number of successes in \(20\) independent trials with constant success probability.

Practice 10: Identify the Error

A student says that a Poisson parameter must satisfy

\[ 0\le\lambda\le1 \]

because probabilities cannot exceed \(1\).

Explain the error.

Practice Answers

Practice 1

Here

\[ N=20,\qquad K=8,\qquad n=5,\qquad k=2. \]

Therefore,

\[ P(X=2) = \frac{ \binom82\binom{12}{3} }{ \binom{20}{5} }. \] \[ = \frac{28\cdot220}{15504} = \boxed{\frac{6160}{15504}\approx0.3973}. \]

Practice 2

\[ \mathbb{E}[X] = n\frac{K}{N} = 5\left(\frac8{20}\right) = \boxed{2}. \]

Practice 3

\[ \operatorname{Var}(X) = 5\left(\frac8{20}\right) \left(\frac{12}{20}\right) \left(\frac{15}{19}\right). \] \[ = \boxed{\frac{18}{19}\approx0.9474}. \]

Practice 4

Use the complement:

\[ P(X\ge1) = 1-P(X=0). \] \[ P(X=0) = \frac{\binom40\binom{16}{3}}{\binom{20}{3}} = \frac{560}{1140}. \]

Therefore,

\[ \boxed{ P(X\ge1) = 1-\frac{560}{1140} = \frac{580}{1140} \approx0.5088 }. \]

Practice 5

\[ P(X=2) = e^{-4}\frac{4^2}{2!}. \] \[ = 8e^{-4} \approx \boxed{0.1465}. \]

Practice 6

\[ \boxed{\mathbb{E}[X]=7}, \] \[ \boxed{\operatorname{Var}(X)=7}, \]

and

\[ \boxed{\sigma_X=\sqrt7\approx2.646}. \]

Practice 7

Ten minutes is

\[ \frac16 \]

of an hour. Therefore,

\[ \lambda = 12\left(\frac16\right) = \boxed{2}. \]

Practice 8

\[ P(X\ge1) = 1-e^{-1.5}. \]

Therefore,

\[ \boxed{P(X\ge1)\approx0.7769}. \]

Practice 9

Hypergeometric.
Poisson.
Binomial.

Practice 10

The parameter \(\lambda\) is an expected count, not a probability. It may be any positive real number. Individual Poisson probabilities are still between \(0\) and \(1\).

What to Remember

For sampling without replacement from a finite population,

\[ \boxed{ X\sim\operatorname{Hypergeometric}(N,K,n) }. \]

Its probability mass function is

\[ \boxed{ P(X=k) = \frac{ \binom{K}{k}\binom{N-K}{n-k} }{ \binom{N}{n} } }. \]

Its mean is

\[ \boxed{ \mathbb{E}[X]=n\frac{K}{N} }. \]

For event counts with average count \(\lambda\),

\[ \boxed{ X\sim\operatorname{Poisson}(\lambda) }. \]

Its PMF is

\[ \boxed{ P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!} }. \]

Its mean and variance are

\[ \boxed{\mathbb{E}[X]=\lambda} \]

and

\[ \boxed{\operatorname{Var}(X)=\lambda}. \]

The key modeling distinction is: use the hypergeometric distribution for finite-population sampling without replacement, and use the Poisson distribution for rate-based event counts over a fixed exposure interval.

Where We Go Next

Hypergeometric and Poisson distributions are important discrete models. The next group of distributions describes continuous quantities such as bounded measurements, bell-shaped variation, and waiting times.

The next lesson, Uniform, Normal, and Exponential Distributions, introduces three foundational continuous probability families.