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3.8 Solving Bayes’ Theorem Problems Using a Grid System

LEARNING OBJECTIVES

  • Use a structured grid (table) method to solve conditional probability problems.

  • Visually organize outcomes and probabilities in Bayes’ Theorem scenarios.

  • Apply the grid method to real-world engineering problems.


Bayes’ Theorem can be solved algebraically, but many engineers find it helpful to organize probabilities in a grid (table). This grid system breaks down possible outcomes systematically, making it easier to track conditional and total probabilities.

The grid divides possible outcomes (such as defective/non-defective, test positive/test negative) into categories. Each cell shows a probability or weighted probability, helping visualize the entire scenario.


GRID SYSTEM STRUCTURE

In typical problems:

  • One dimension represents the true state (e.g., Defective vs. Not Defective).

  • The other dimension represents the observed result (e.g., Test Positive vs. Test Negative).

Each cell shows the joint probability: the probability that both the true state and the observed result occur together.

The total probability of the observed result (e.g., test positive) is found by adding the relevant joint probabilities.


EXAMPLE: SENSOR FAULT DETECTION

An engineer knows:

  • 5% of sensors are defective.

  • If defective, the false alarm probability is 98%.

  • If not defective, false alarms occur 2% of the time.

What is P(Defective | False Alarm)?

STEP 1: BUILD THE GRID

True State False Alarm (F) No False Alarm (Fᶜ) Row Total
Defective (D) (0.05)(0.98) = 0.0490 (0.05)(0.02) = 0.0010 0.05
Not Defective (Dᶜ) (0.95)(0.02) = 0.0190 (0.95)(0.98) = 0.9310 0.95
Column Total 0.0680 1.00

STEP 2: SOLVE

  • P(D and F) = 0.0490

  • P(Dᶜ and F) = 0.0190

  • Total P(F) = 0.0680

 

P(DF)=P(DandF)P(F)=0.04900.06800.7206P(D|F) = \frac{P(D \text{ and } F)}{P(F)} = \frac{0.0490}{0.0680} \approx 0.7206

Answer: Approximately 72% chance the sensor is defective if a false alarm occurs.


TRY IT: SOLVE WITH A GRID

In a factory:

  • 2% of parts are defective.

  • Detection test:

    • Detects defective parts: 95%.

    • False alarms on good parts: 1%.

If a part tests positive, what is the probability it’s defective?

Solution Grid:

True State Test Positive (T⁺) Test Negative (T⁻) Row Total
Defective (D) (0.02)(0.95) = 0.0190 (0.02)(0.05) = 0.0010 0.0190
Not Defective (Dᶜ) (0.98)(0.01) = 0.0098 (0.98)(0.99) = 0.9702 0.98
Column Total 0.0288 1.00

 

P(DT+)=0.01900.02880.6597P(D|T⁺) = \frac{0.0190}{0.0288} \approx 0.6597

Answer: Approximately 66% chance the part is defective if it tests positive.

WHY THE GRID METHOD HELPS ENGINEERS

  • Encourages clear, structured thinking.

  • Reduces mistakes in complex scenarios.

  • Makes visual sense of conditional relationships.

 

EXERCISES: USE THE GRID METHOD

Try completing the same problems from Section 3.7, just using the grid method!

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Introduction to Statistics - Second Edition Copyright © 2025 by Aimee Birdsall and Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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