---
title: "Event types; airport weapon detection Bayes"
subject: "Probability and Statistics"
chapter: 1
key: "semester_5/probability_and_statistics:airport-weapon-bayes"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/airport-weapon-bayes
site: BCT Engineering
---

# Event types; airport weapon detection Bayes

Probability and Statistics · Chapter 1: Descriptive Statistics and Basic Probability

- Asked in: 2075 Chaitra, 5 marks
- Video: [Bayes' Theorem EXPLAINED with Examples](https://www.youtube.com/watch?v=cqTwHnNbc8g) — Ace Tutors
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/airport-weapon-bayes
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1

## Question

What do you mean by mutually exclusive, exhaustive and complementary events? Explain with examples. In a particular city, airport A handles 50% of all airlines traffic, airport B handles 30% and airport C handles 20%. The detection rates for weapons at the three airports are 0.9, 0.5 and 0.4 respectively. A passenger is randomly selected at one of the airports. Then (i) what is the probability that he/she carrying a weapon? (ii) If he/she is found to be carrying a weapon, what is the probability that airport A is being used?

## Short answer

Mutually exclusive events cannot occur together (e.g., even and odd in a die toss). Exhaustive events cover the whole sample space (e.g., head or tail in a coin toss). Complementary events are both mutually exclusive and exhaustive (e.g., bulb working or not). (i) P(weapon) = 0.5×0.9+0.3×0.5+0.2×0.4 = 0.68. (ii) P(A|weapon) = 0.5×0.9/0.68 ≈ 0.662.

## Answer

#### Types of events
- **Mutually exclusive events:** cannot occur together, so P(A ∩ B) = 0. Example: getting an even number and an odd number in one throw of a die.
- **Exhaustive events:** together they cover the whole sample space, so at least one must occur. Example: getting a head or a tail in one toss. Events "even" and "odd" are exhaustive for a die.
- **Complementary events:** A and its complement A′ are both mutually exclusive and exhaustive, with P(A) + P(A′) = 1. Example: a bulb is working (A) or not working (A′).

#### Airport problem
The detection rates are read as the probability that a weapon is detected (found) at each airport.

Let A, B, C be the events that the airport used is A, B, C, and W be the event that a weapon is found.

```text
P(A) = 0.5   P(W|A) = 0.9
P(B) = 0.3   P(W|B) = 0.5
P(C) = 0.2   P(W|C) = 0.4
```

**(i) Probability of a weapon being found (total probability)**

```text
P(W) = (0.5)(0.9) + (0.3)(0.5) + (0.2)(0.4)
     = 0.45 + 0.15 + 0.08
     = 0.68
```

**(ii) Airport A, given a weapon is found (Bayes' theorem)**

```text
P(A|W) = P(A) P(W|A) / P(W) = 0.45 / 0.68 = 0.662
```

**Answer:** (i) 0.68, (ii) 0.662.

## Check yourself

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

1. What makes A and its complement A' complementary?
   A. Only mutually exclusive
   B. Only exhaustive
   C. Mutually exclusive and exhaustive, so P(A) + P(A') = 1
   D. Independent of each other
   Answer: C. Complementary events are both exclusive and exhaustive. (Key point: Complementary events)

2. What is P(weapon found)?
   A. 0.9 + 0.5 + 0.4 = 1.8
   B. 0.9 x 0.5 x 0.4 = 0.18
   C. 0.45
   D. 0.5x0.9 + 0.3x0.5 + 0.2x0.4 = 0.68
   Answer: D. Weight each detection rate by the airport's traffic share and add. (Key point: Total probability)

3. What is P(airport A | weapon found)?
   A. 0.45/0.68 = 0.662
   B. 0.45
   C. 0.5
   D. 0.68/0.45
   Answer: A. Bayes divides the A contribution by the total probability. (Key point: Bayes)

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