---
title: "Bayes' theorem; DVD players warranty"
subject: "Probability and Statistics"
chapter: 1
key: "semester_5/probability_and_statistics:bayes-dvd-warranty"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bayes-dvd-warranty
site: BCT Engineering
---

# Bayes' theorem; DVD players warranty

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

- Asked in: 2073 Chaitra, 2+4 marks
- Video: [Probability example three machines produces and what is the probability that it is defective](https://www.youtube.com/watch?v=l8MO5t6OAd0) — EASY MATHS EASY TRICKS
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/bayes-dvd-warranty
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1

## Question

Explain Baye's theorem. A chain of video stores sells three different brands of DVD players; Of its DVD players sales, 50% are brand 1 (the least expensive), 30% are brand 2, and 20% are brand 3. Each manufacturer offers a 1-year warranty on parts and labor. It is known that 25% of brand 1's DVD players require warranty repair work, where the corresponding percentages for brands 2 and 3 are 20% and 10%, respectively.
a) What is the probability that a randomly selected purchaser has bought a brand 1 DVD players that will need repair while under warranty?
b) What is the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty?

## Short answer

P(brand 1 and repair) = 0.50 × 0.25 = 0.125. Total probability of repair = 0.50×0.25 + 0.30×0.20 + 0.20×0.10 = 0.205.

## Answer

#### Bayes' theorem
Let E₁, E₂, …, Eₙ be mutually exclusive and exhaustive events with P(Eᵢ) > 0, and let A be any event that can occur only with one of them. Then

```text
P(Ei | A) = P(Ei) P(A | Ei) / Σ P(Ej) P(A | Ej)
```

It revises a prior probability P(Eᵢ) into a posterior probability after the event A is known to have occurred. The denominator is the total probability of A.

#### Problem
Let B₁, B₂, B₃ be the brands and R the event that the player needs warranty repair.

```text
P(B1) = 0.50   P(R|B1) = 0.25
P(B2) = 0.30   P(R|B2) = 0.20
P(B3) = 0.20   P(R|B3) = 0.10
```

**a) Brand 1 and needs repair**

```text
P(B1 ∩ R) = P(B1) P(R|B1) = 0.50 × 0.25 = 0.125
```

**b) Needs repair (total probability)**

```text
P(R) = (0.50)(0.25) + (0.30)(0.20) + (0.20)(0.10)
     = 0.125 + 0.060 + 0.020
     = 0.205
```

**Answer:** (a) 0.125, (b) 0.205.

## Check yourself

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

1. What does Bayes' theorem do?
   A. Adds probabilities of exclusive events
   B. Finds independence
   C. Revises prior P(Ei) into posterior P(Ei | A)
   D. Computes mean
   Answer: C. Updates after A occurs. (Key point: Bayes statement)

2. What is P(brand 1 and repair)?
   A. 0.25
   B. 0.50
   C. 0.75
   D. 0.50 x 0.25 = 0.125
   Answer: D. Multiplication rule. (Key point: Joint probability)

3. What is P(repair)?
   A. 0.125 + 0.060 + 0.020 = 0.205
   B. 0.25 + 0.20 + 0.10
   C. 0.125
   D. 0.50
   Answer: A. Sum over all brands. (Key point: Total probability)

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