---
title: "Poisson process; intersection accidents"
subject: "Probability and Statistics"
chapter: 2
key: "semester_5/probability_and_statistics:poisson-process-accidents-intersection"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/poisson-process-accidents-intersection
site: BCT Engineering
---

# Poisson process; intersection accidents

Probability and Statistics · Chapter 2: Discrete Probability Distribution

- Asked in: 2068 Jestha (old course), 5+6 marks
- Video: [An Introduction to the Poisson Distribution](https://www.youtube.com/watch?v=jmqZG6roVqU) — jbstatistics
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/poisson-process-accidents-intersection
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2

## Question

a) Explain the Poisson process. What are mean and standard deviation of Poisson distribution? Give one example of Poisson distribution.
b) Suppose that we are investigating the safety of a dangerous intersection. Past police records indicate a mean of 5 accidents per month at this intersection. Calculate the probability, in any month, of:
i) No accidents occurring
ii) Exactly one accident occurring
iii) At most one accident occurring

## Short answer

A Poisson process has independent events at constant average rate; count has Poisson distribution with mean lambda and standard deviation sqrt(lambda). For lambda=5 per month, probabilities are: no accidents 0.0067, exactly one 0.0337, at most one 0.0404.

## Answer

#### a) Poisson process, mean, SD and example
A **Poisson process** is a random process in which events occur one at a time, at a constant average rate, independently of each other, in a continuous interval of time, length or area. Its properties:
1. The number of events in non-overlapping intervals is independent.
2. The probability of exactly one event in a very short interval is proportional to the length of the interval.
3. The probability of two or more events in a very short interval is negligible.
4. The average rate is constant over the interval.

The number of events X in an interval with average λ follows the Poisson distribution P(X = x) = e⁻ˡ λˣ / x!, x = 0, 1, 2, …

- Mean = λ
- Variance = λ, so standard deviation = √λ

**Example:** Number of telephone calls arriving at an exchange per minute, or the number of printing errors per page.

#### b) Accidents at the intersection
λ = 5 per month, e⁻⁵ = 0.006738.

**i) No accident:** P(0) = e⁻⁵ = **0.0067**

**ii) Exactly one accident:** P(1) = e⁻⁵ × 5 = 0.033690 ≈ **0.0337**

**iii) At most one accident:** P(X ≤ 1) = 0.006738 + 0.033690 = 0.040428 ≈ **0.0404**

**Answer: (i) 0.0067, (ii) 0.0337, (iii) 0.0404**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A Poisson process has events that:
   A. Occur in groups at fixed times
   B. Depend on previous events
   C. Have a rate that rises with time
   D. Occur one at a time, independently, at a constant average rate
   Answer: D. Independence and constant rate define it. (Key point: Poisson process)

2. The standard deviation of a Poisson variable is:
   A. lambda
   B. sqrt(lambda)
   C. lambda^2
   D. 1/lambda
   Answer: B. Variance = lambda, so SD = sqrt(lambda). (Key point: SD)

3. With lambda = 5, P(0 accidents) is:
   A. e^-5 = 0.0067
   B. 5 e^-5
   C. 1 - e^-5
   D. e^5
   Answer: A. x = 0 gives e^-lambda. (Key point: No accidents)

4. P(X <= 1) is:
   A. P(1) = 0.0337
   B. 1 - P(0)
   C. P(0) + P(1) = 0.0404
   D. P(0) = 0.0067
   Answer: C. Add the probabilities for 0 and 1. (Key point: At most one)

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