Poisson Probability Distribution: Formulas & Solved Examples

5 fully solved examples and 3 practice problems on the Poisson distribution

What is the Poisson Distribution?

The Poisson probability distribution models the number of times a discrete event occurs randomly and independently within a fixed interval of time or space. Because it counts occurrences, the random variable \( X \) can only take non-negative integer values (\( 0, 1, 2, \dots \)).

Conditions for a Poisson Process

  1. Independence — events are random and independent.
  2. Constant rate — the average rate \( \lambda \) does not change over the interval.
  3. Proportionality — the probability of an event in a short interval is proportional to the interval length.
  4. No simultaneous events — two events cannot occur at the exact same instant.

Real-World Examples

Poisson Probability Formula

If an event occurs at an average rate \( \lambda \) per interval, the probability it occurs exactly \( x \) times in the same interval is:

\[ P(X = x) = \frac{e^{-\lambda}\, \lambda^x}{x!} \]
Symbol Meaning
\( \lambda \)Average rate of occurrence (the mean) in the interval
\( e \)Euler's number, \( \approx 2.71828 \)
\( x \)Number of actual occurrences (\( 0, 1, 2, \dots \))
\( x! \)Factorial of \( x \) (e.g., \( 3! = 6 \))

Mean and variance. For a Poisson distribution: \( \mu = \lambda \) and \( \sigma^2 = \lambda \). The standard deviation is \( \sqrt{\lambda} \).

Several Poisson probability mass functions plotted for different λ values, showing how larger λ shifts the peak right and broadens the distribution
Figure 1 — Poisson PMFs for different \( \lambda \) values.

Poisson vs. Binomial Distribution

Both are discrete distributions, but they model very different scenarios.

Distribution Parameters Support
Binomial Fixed number of trials \( n \), fixed per-trial success probability \( p \) Finite: \( x = 0, 1, \dots, n \)
Poisson Average rate \( \lambda \) over a continuous interval Infinite: \( x = 0, 1, 2, \dots \) (in practice bounded by the interval)

Connection. The Poisson is the limit of the binomial as \( n \to \infty \) and \( p \to 0 \) with \( np = \lambda \) fixed. So when \( n \) is large and \( p \) is small (typically \( p \le 0.05 \)), the binomial is well approximated by a Poisson with \( \lambda = np \).

Poisson Distribution Examples with Step-by-Step Solutions

Example 1 — Computer crashes (λ = 1)

A computer crashes on average once every 4 months. Find the probability that over a 4-month span it will:

  1. not crash at all;
  2. crash exactly once;
  3. crash exactly twice;
  4. crash exactly three times.
View solution

Setup. The interval is 4 months, and the average rate for that interval is \( \lambda = 1 \).

  1. \( P(X = 0) = \dfrac{e^{-1} \cdot 1^0}{0!} = e^{-1} \approx 0.36788 \)
  2. \( P(X = 1) = \dfrac{e^{-1} \cdot 1^1}{1!} = e^{-1} \approx 0.36788 \)
  3. \( P(X = 2) = \dfrac{e^{-1} \cdot 1^2}{2!} = \dfrac{e^{-1}}{2} \approx 0.18394 \)
  4. \( P(X = 3) = \dfrac{e^{-1} \cdot 1^3}{3!} = \dfrac{e^{-1}}{6} \approx 0.06131 \)

Example 2 — Call center (λ = 3.5)

A customer help center receives an average of 3.5 calls per hour. Find the probability that in a randomly selected 1-hour period it will:

  1. receive at most 4 calls;
  2. receive at least 5 calls.
View solution

Setup. \( \lambda = 3.5 \) for a 1-hour interval.

(a) At most 4 calls.

xP(X = x)
00.03020
10.10569
20.18496
30.21579
40.18881
\[ P(X \le 4) \approx 0.72545 \]

(b) At least 5 calls. Use the complement:

\[ P(X \ge 5) = 1 - P(X \le 4) \approx 1 - 0.72545 = 0.27455 \]

Example 3 — Emails over 2 hours (λ = 6)

A person receives an average of 3 emails per hour. Find the probability that over a continuous 2-hour period they will:

  1. receive exactly 5 emails;
  2. receive more than 2 emails.
View solution

Setup. Rescale to the 2-hour window: \( \lambda = 3 \times 2 = 6 \).

(a) Exactly 5 emails.

\[ P(X = 5) = \frac{e^{-6} \cdot 6^5}{5!} \approx 0.16062 \]

(b) More than 2 emails. Use the complement:

\[ P(X \le 2) = P(0) + P(1) + P(2) \approx 0.00248 + 0.01487 + 0.04462 = 0.06197 \] \[ P(X > 2) = 1 - 0.06197 = 0.93803 \]

Example 4 — Football goals (λ estimated from data)

Goals scored by a football player across 35 matches:

Goals \( x \) 0 1 2 3 > 3
Frequency \( f \) 12 15 6 2 0

Assuming a Poisson model, find the probability that in the next match the player:

  1. scores exactly 1 goal;
  2. scores at least 1 goal.
View solution

Setup. Estimate \( \lambda \) from the data (all 35 matches):

\[ \lambda = \frac{(12 \times 0) + (15 \times 1) + (6 \times 2) + (2 \times 3)}{35} = \frac{33}{35} \approx 0.94286 \]

(a) Exactly 1 goal.

\[ P(X = 1) = e^{-33/35} \cdot \frac{33}{35} \approx 0.36726 \]

(b) At least 1 goal. Use the complement:

\[ P(X = 0) = e^{-33/35} \approx 0.38951 \] \[ P(X \ge 1) = 1 - 0.38951 = 0.61049 \]

Example 5 — Quality control (λ = 1)

Defective items returned daily over 100 days:

Defective items \( x \) 0 1 2 3 4 > 4
Frequency \( f \) 50 20 15 10 5 0

Assuming a Poisson distribution, find the probability that on any future day:

  1. no defective item is returned;
  2. three or more defective items are returned.
View solution

Setup. Estimate \( \lambda \) from the data:

\[ \lambda = \frac{(50 \times 0) + (20 \times 1) + (15 \times 2) + (10 \times 3) + (5 \times 4)}{100} = \frac{100}{100} = 1 \]

(a) No defective items.

\[ P(X = 0) = e^{-1} \approx 0.36788 \]

(b) Three or more defective items.

\[ P(X \le 2) = e^{-1} + e^{-1} + \frac{e^{-1}}{2} = 2.5\,e^{-1} \approx 0.91970 \] \[ P(X \ge 3) = 1 - 0.91970 = 0.08030 \]

Practice Problems

Practice 1 — Exact count

A radioactive sample emits an average of 2 particles per second. What is the probability of observing exactly 4 particles in a given second?

Answer
\[ P(X = 4) = \frac{e^{-2} \cdot 2^4}{4!} = \frac{0.135335 \times 16}{24} \approx 0.09022 \]

Practice 2 — Interval rescaling

A website gets 5 new signups per hour on average. What is the probability of getting 12 signups over a 2-hour period?

Answer

Rescale: \( \lambda = 5 \times 2 = 10 \).

\[ P(X = 12) = \frac{e^{-10} \cdot 10^{12}}{12!} \approx 0.09478 \]

Practice 3 — Mean, variance, and tail

A call center receives an average of 6 calls per minute.

  1. What are the mean and variance of the per-minute call count?
  2. What is the probability of receiving more than 8 calls in a given minute?
Answer

(a) Mean = \( \lambda = 6 \), Variance = \( \lambda = 6 \).

(b) \( P(X > 8) = 1 - P(X \le 8) \):

\[ P(X \le 8) \approx 0.84724 \] \[ P(X > 8) \approx 0.15276 \]

Frequently Asked Questions

What is the Poisson distribution formula?

\(P(X = x) = \dfrac{e^{-\lambda} \lambda^x}{x!}\), where \(\lambda\) is the average rate per interval and \(x\) is the number of occurrences.

What are the conditions for a Poisson process?

Four conditions: (1) independent events; (2) constant rate λ over the interval; (3) probability of an event in a short interval is proportional to interval length; (4) no two events occur simultaneously.

What are the mean and variance of a Poisson distribution?

Both equal \(\lambda\): \(\mu = \lambda\), \(\sigma^2 = \lambda\). Standard deviation: \(\sqrt{\lambda}\).

How is the Poisson distribution related to the binomial distribution?

The Poisson is the limit of the binomial as \(n \to \infty\) and \(p \to 0\) with \(np = \lambda\) fixed. So when \(n\) is large and \(p\) is small (typically \(p \le 0.05\)), the binomial is well approximated by a Poisson with \(\lambda = np\).

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Last updated: October 20, 2026