---
title: "Poisson: injection reaction with distribution"
subject: "Probability and Statistics"
chapter: 2
key: "semester_5/probability_and_statistics:poisson-injection-with-pdf"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/poisson-injection-with-pdf
site: BCT Engineering
---

# Poisson: injection reaction with distribution

Probability and Statistics · Chapter 2: Discrete Probability Distribution

- Asked in: 2075 Asoj, 6 marks
- Video: [Poisson Distribution Example | Injection Reaction Out of 2000 People](https://www.youtube.com/watch?v=MZA0Str3wIE) — Mathematics Tutor
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/poisson-injection-with-pdf
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2

## Question

If the probability that an individual suffers a bad reaction from a certain injection is 0.001, among 2000 individual
a) obtain probability distribution function for suffering bad reaction
b) determine the probability that
(i) exactly 3 individuals will suffer bad reaction
(ii) more than 2, individuals will suffer bad reaction

## Short answer

The number of reactions is approximately Poisson with λ = 2000*0.001 = 2, so pmf is e^{-2}2^x / x!. Probability of exactly 3 is 0.1804, and more than 2 is 0.3233.

## Answer

n = 2000 and p = 0.001, so the number of bad reactions X is binomial. Since n is large and p is small, it is approximated by a **Poisson** distribution with

λ = np = 2000 × 0.001 = 2.

#### a) Probability distribution function
P(X = x) = e^(−λ) λˣ / x! = **e⁻² 2ˣ / x!**, x = 0, 1, 2, …

Here mean = variance = λ = 2 and e⁻² = 0.135335.

#### b) Probabilities

```text
P(0) = e⁻²            = 0.1353
P(1) = e⁻² · 2        = 0.2707
P(2) = e⁻² · 2²/2!    = 0.2707
P(3) = e⁻² · 2³/3!    = 0.1804
```

**(i) Exactly 3:** P(X = 3) = 0.135335 × 8/6 = **0.1804**

**(ii) More than 2:**
P(X > 2) = 1 − [P(0) + P(1) + P(2)] = 1 − (0.1353 + 0.2707 + 0.2707) = 1 − 0.6767 = **0.3233**

**Answer: (i) 0.1804, (ii) 0.3233**

## Check yourself

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

1. Why can the binomial with n = 2000, p = 0.001 be approximated by a Poisson?
   A. n is small and p is large
   B. n is large and p is small
   C. p is exactly 0.5
   D. The trials are dependent
   Answer: B. Large n and small p with moderate np allows the Poisson approximation. (Key point: Why Poisson)

2. What is the Poisson parameter here?
   A. lambda = np = 2
   B. lambda = p = 0.001
   C. lambda = n = 2000
   D. lambda = npq = 1.998 used with mean 2000
   Answer: A. lambda = 2000 x 0.001 = 2, which equals both mean and variance. (Key point: Parameter)

3. P(X = 3) is found as:
   A. e^-2 x 2^3 = 1.083
   B. e^-2 x 3^2 / 2! = 0.609
   C. e^-2 x 2^3 / 3! = 0.1804
   D. 1 - e^-2 = 0.8647
   Answer: C. Substitute x = 3 and lambda = 2 into e^-lambda lambda^x / x!. (Key point: P(X=3))

4. How is P(X > 2) obtained?
   A. P(0) + P(1) + P(2) = 0.6767
   B. P(3) only = 0.1804
   C. 1 - P(2) = 0.7293
   D. 1 - [P(0) + P(1) + P(2)] = 0.3233
   Answer: D. The complement of X <= 2 is X > 2. (Key point: P(X>2))

## Close variants set in other exams

- [Poisson: bad reaction among 2000 (a, b)](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apoisson-injection-reaction-2000) — Probability and Statistics, chapter 2, asked 2×