---
title: "Poisson arrival probabilities in 15 minutes"
subject: "Transportation Engineering II"
chapter: 1
key: "semester_7/transportation_engineering_ii:poisson-arrivals-probability-15-minute"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/poisson-arrivals-probability-15-minute
site: BCT Engineering
---

# Poisson arrival probabilities in 15 minutes

Transportation Engineering II · Chapter 1: Traffic Engineering

- Asked in: 2076 Chaitra, 4 marks
- Video: [Poisson Probability Distribution: Number of cars passing a point in a five-minute period](https://www.youtube.com/watch?v=Pl-F0QNmH_U) — MATHalino
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/poisson-arrivals-probability-15-minute
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1

## Question

The vehicle arrivals at the section of road are assumed to be Poisson distributed with an average arrival rate of 1 vehicle every 5 minutes. What is the probability of
(i) Exactly 3 vehicles arrive in a 15 minute interval
(ii) less than 3 vehicles arrive in a 15 minute interval?
(iii) More than 3 vehicles arrive in a 15 minute intervals?

## Short answer

With mean m = 3 vehicles in 15 minutes, exactly 3 has probability 0.2240, less than 3 has probability 0.4232, and more than 3 has probability 0.3528.

## Answer

For Poisson arrivals, the probability of $n$ arrivals in an interval is $P(n) = \dfrac{m^n e^{-m}}{n!}$ where $m = \lambda t$ is the mean number of arrivals in the interval.

Rate: 1 vehicle per 5 min, so in 15 min $m = 15/5 = 3$ vehicles.

#### (i) Exactly 3 vehicles
$$
P(3) = \frac{3^3 e^{-3}}{3!} = \frac{27(0.04979)}{6} = 0.2240
$$

#### (ii) Less than 3 vehicles
$$
\begin{aligned} P(n<3) &= P(0)+P(1)+P(2)\\ &= 0.0498 + 0.1494 + 0.2240 = 0.4232 \end{aligned}
$$

#### (iii) More than 3 vehicles
$$
P(n>3) = 1 - [P(0)+P(1)+P(2)+P(3)] = 1 - (0.4232 + 0.2240) = 0.3528
$$

**Answer: (i) 0.2240 (22.4 %), (ii) 0.4232 (42.3 %), (iii) 0.3528 (35.3 %).**

## Check yourself

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

1. Arrivals average 1 vehicle per 5 min. What is the mean m for a 15 minute interval?
   A. 0.2
   B. 1
   C. 3
   D. 5
   Answer: C. m = lambda t = 15/5 = 3 vehicles. (Key point: Mean in interval)

2. What is P(3) for m = 3?
   A. 0.0498
   B. 0.2240
   C. 0.4232
   D. 0.3528
   Answer: B. P(3) = 3^3 e^-3 / 3! = 27(0.04979)/6 = 0.2240. (Key point: Exactly 3 vehicles)

3. How is P(n < 3) calculated?
   A. P(3) only
   B. P(0) + P(1) + P(2) = 0.4232
   C. 1 - P(3)
   D. P(1) + P(2) + P(3)
   Answer: B. Less than 3 means 0, 1 or 2 arrivals: 0.0498 + 0.1494 + 0.2240 = 0.4232. (Key point: Less than 3)

4. What is P(n > 3)?
   A. 0.4232
   B. 0.2240
   C. 0.6472
   D. 0.3528
   Answer: D. P(n>3) = 1 - [P(0..2) + P(3)] = 1 - (0.4232 + 0.2240) = 0.3528. (Key point: More than 3)

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