---
title: "Poisson: accidents among 1000 taxi drivers"
subject: "Probability and Statistics"
chapter: 2
key: "semester_5/probability_and_statistics:taxi-accidents-poisson-1000"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/taxi-accidents-poisson-1000
site: BCT Engineering
---

# Poisson: accidents among 1000 taxi drivers

Probability and Statistics · Chapter 2: Discrete Probability Distribution

- Asked in: 2070 Chaitra, 5 marks
- Video: [Poisson Distribution | Accidents to Taxi Drivers | VTU 1BMATCS301, 1BMATEE301, 1BMATCV301](https://www.youtube.com/watch?v=slERrYlCaIc) — Mathematics Tutor
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/taxi-accidents-poisson-1000
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2

## Question

The number of accident in a year attributes to taxi drivers in a city follows Poisson distribution with mean 3. Out of 1000 taxi driver, find the approximately the number of driver with:
i) No accidents in a year
ii) More than 3 accident in a year

## Short answer

For Poisson with mean 3, use P(0)=e^-3 and complement. About 1000*0.0498 gives approximately 50 drivers with no accidents; about 1000*0.3528 gives approximately 353 drivers with more than 3 accidents.

## Answer

The number of accidents per driver per year X ~ Poisson with λ = 3: P(X = x) = e⁻³ 3ˣ / x!, with e⁻³ = 0.049787.

```text
P(0) = 0.049787
P(1) = 0.149361
P(2) = 0.224042
P(3) = 0.224042
```

#### i) No accidents
P(X = 0) = 0.0498, so the expected number of drivers = 1000 × 0.0498 ≈ **50 drivers**

#### ii) More than 3 accidents
P(X > 3) = 1 − [P(0) + P(1) + P(2) + P(3)] = 1 − (0.049787 + 0.149361 + 0.224042 + 0.224042) = 1 − 0.647232 = 0.3528

Expected number of drivers = 1000 × 0.3528 ≈ **353 drivers**

**Answer: (i) about 50 drivers, (ii) about 353 drivers**

## Check yourself

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

1. Accidents per driver per year follow:
   A. Poisson with lambda = 3
   B. Binomial with n = 1000
   C. Poisson with lambda = 1000
   D. Normal with mean 3
   Answer: A. Given as Poisson with mean 3. (Key point: Model)

2. P(X = 0) is:
   A. 3 e^-3
   B. 1 - e^-3
   C. e^-3 = 0.0498
   D. e^3
   Answer: C. With x = 0, the formula gives e^-lambda. (Key point: No accidents)

3. P(X > 3) is found as:
   A. P(0)+P(1)+P(2)+P(3) = 0.6472
   B. 1 - P(3)
   C. 1 - [P(0)+P(1)+P(2)+P(3)] = 0.3528
   D. P(4) only
   Answer: C. Use the complement of X <= 3. (Key point: More than 3)

4. Number of drivers expected is:
   A. 1000 x probability
   B. Probability / 1000
   C. Probability x 3
   D. 1000 / probability
   Answer: A. Expected count = N x P; for no accidents about 50, for more than 3 about 353. (Key point: Converting to drivers)

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