---
title: "Binomial: green traffic light mornings"
subject: "Probability and Statistics"
chapter: 2
key: "semester_5/probability_and_statistics:traffic-light-green-binomial"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/traffic-light-green-binomial
site: BCT Engineering
---

# Binomial: green traffic light mornings

Probability and Statistics · Chapter 2: Discrete Probability Distribution

- Asked in: 2072 Kartik, 5 marks
- Video: [Binomial Probability: Traffic Lights](https://www.youtube.com/watch?v=UKoAGbNAyDU) — ASA Amity Maths
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/traffic-light-green-binomial
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2

## Question

A particularly long traffic light on your morning commute is green 20% of the time that you approach it. Assume that each morning represents as independent trial.
i) Over five mornings, what is the probability that the light is green on exactly one day?
ii) Over 20 mornings, what is the probability that the light is green on exactly four days?

## Short answer

Using binomial p=0.2: probability exactly one green in 5 mornings=0.4096; exactly four in 20 mornings=0.2182.

## Answer

Each morning the light is green with probability p = 0.2 (q = 0.8), independently, so the number of green days X is binomial.

#### i) n = 5, exactly one green day
P(X = 1) = C(5, 1)(0.2)¹(0.8)⁴ = 5 × 0.2 × 0.4096 = **0.4096**

#### ii) n = 20, exactly four green days
P(X = 4) = C(20, 4)(0.2)⁴(0.8)¹⁶
= 4845 × 0.0016 × 0.028147
= **0.2182**

**Answer: (i) 0.4096, (ii) 0.2182**

## Check yourself

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

1. The number of green days follows:
   A. Binomial with p = 0.8
   B. Binomial with p = 0.2
   C. Poisson with lambda 0.2
   D. Hypergeometric
   Answer: B. Independent mornings, fixed n, constant p = 0.2 of green. (Key point: Model)

2. P(exactly 1 green in 5) is:
   A. C(5,1)(0.2)(0.8)^4 = 0.4096
   B. (0.2)(0.8)^4 = 0.0819
   C. C(5,1)(0.8)(0.2)^4
   D. 1 - 0.8^5
   Answer: A. The factor C(5,1) = 5 counts which day is green. (Key point: Part i)

3. P(exactly 4 green in 20) uses:
   A. C(20,4)(0.2)^16(0.8)^4
   B. C(4,20)(0.2)^4(0.8)^16
   C. C(20,4)(0.2)^4(0.8)^16
   D. (0.2)^4(0.8)^16
   Answer: C. x = 4 successes, n - x = 16 failures. (Key point: Part ii)

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