---
title: "Mutually exclusive events; insurance accident Bayes"
subject: "Probability and Statistics"
chapter: 1
key: "semester_5/probability_and_statistics:bayes-insurance-engineers"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bayes-insurance-engineers
site: BCT Engineering
---

# Mutually exclusive events; insurance accident Bayes

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

- Asked in: 2076 Chaitra, 6 marks
- Video: [Probability of a disease given a postiive test: Bayes Theorem ex2](https://www.youtube.com/watch?v=_c3xZvHto3k) — OCLPhase2
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bayes-insurance-engineers
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1

## Question

Define mutually exclusive and independent events. An insurance company insured 2000 Civil engineers, 4000 Electrical engineers and 6000 Mechanical engineers. The probability of an accident involving Civil engineer, Electrical engineer and Mechanical engineer during their jobs is 0.01, 0.03 and 0.15 respectively. One of the insured engineers meets with an accident. What is the probability that he is civil Engineer?

## Short answer

Mutually exclusive events cannot happen together (P(A∩B)=0). Independent events have P(A∩B)=P(A)P(B). Using Bayes: P(Civil|accident) = (1/6 × 0.01) / [(1/6×0.01)+(1/3×0.03)+(1/2×0.15)] = 0.001667/0.086667 ≈ 0.0192 (about 1.9%).

## Answer

#### Mutually exclusive events
Two events that cannot occur together are mutually exclusive: A ∩ B = ∅, so P(A ∩ B) = 0 and P(A ∪ B) = P(A) + P(B). Example: getting a head and getting a tail in one toss.

#### Independent events
Events are independent if P(A ∩ B) = P(A) P(B), that is, one event does not affect the probability of the other. Example: a head on a coin and an even number on a die.

#### Insurance problem (Bayes' theorem)
Let C, E, M be the events that the engineer is Civil, Electrical, Mechanical, and let X be the event of an accident.

Total insured = 2000 + 4000 + 6000 = 12000.

```text
P(C) = 2000/12000 = 1/6    P(X|C) = 0.01
P(E) = 4000/12000 = 1/3    P(X|E) = 0.03
P(M) = 6000/12000 = 1/2    P(X|M) = 0.15
```

Total probability of an accident:

```text
P(X) = (1/6)(0.01) + (1/3)(0.03) + (1/2)(0.15)
     = 0.001667 + 0.01 + 0.075
     = 0.086667
```

Bayes' theorem:

```text
P(C|X) = P(C) P(X|C) / P(X)
       = 0.001667 / 0.086667
       = 0.0192
```

The same result follows from expected counts: accidents are 20 (civil), 120 (electrical), 900 (mechanical), total 1040, and 20/1040 = 0.0192.

**Answer:** P(Civil | accident) ≈ 0.0192 (about 1.9 %).

## Check yourself

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

1. Which describes mutually exclusive events?
   A. P(A ∩ B) = P(A)P(B)
   B. They always occur together
   C. One changes the other's probability
   D. They cannot occur together: P(A ∩ B) = 0
   Answer: D. Mutually exclusive means A ∩ B is empty. (Key point: Mutually exclusive)

2. What is P(Civil) among the insured?
   A. 2000/12000 = 1/6
   B. 1/3
   C. 1/2
   D. 0.01
   Answer: A. Civil engineers are 2000 of 12000 insured. (Key point: Prior probabilities)

3. What is P(X), the probability of an accident?
   A. 0.01 + 0.03 + 0.15 = 0.19
   B. (1/6)(0.01) + (1/3)(0.03) + (1/2)(0.15) = 0.0867
   C. (0.01)(0.03)(0.15)
   D. 0.15
   Answer: B. Weight each accident rate by its group's share and add. (Key point: Total probability of accident)

4. What is P(Civil | accident)?
   A. About 0.01
   B. About 0.167
   C. P(C)P(X|C)/P(X) = about 0.0192
   D. About 0.865
   Answer: C. 0.001667/0.086667 = 0.0192, or 20/1040 from counts. (Key point: Bayes result)

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