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LEARNING OBJECTIVES

• Understand and explain Bayes’ Theorem.

• Apply Bayes’ Theorem to calculate conditional probabilities.

• Solve engineering problems using Bayes’ Theorem in real-world contexts.

Bayes’ Theorem is a powerful statistical tool that allows us to update probabilities based on new information. Named after Reverend Thomas Bayes, this theorem plays a central role in areas such as reliability analysis, quality control, and risk assessment in engineering.

Bayes’ Theorem links conditional probabilities, providing a formal method for “reversing” known conditional probabilities.

BAYES’ THEOREM FORMULA

For two events, A and B, where P(B) > 0, Bayes’ Theorem is written as:

$$P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$$

P(A∣B)=P(B)P(B∣A)⋅P(A)​

Where:

• P(A|B) is the probability that event A occurs given that event B has occurred (posterior probability).

• P(B|A) is the probability that event B occurs given that event A has occurred (likelihood).

• P(A) is the probability that event A occurs (prior probability).

• P(B) is the total probability that event B occurs.

A DIAGRAM TO ILLUSTRATE BAYES’ THEOREM

A flow diagram illustrating Bayes’ Theorem:

• A rectangle represents the entire sample space.

• Circles within the rectangle represent events A and B.

• Overlapping regions represent scenarios where both A and B occur.

• Bayes’ Theorem helps us evaluate the likelihood of A given that B has occurred, using prior knowledge about A and B.

EXAMPLE: FAULT DETECTION IN A SENSOR SYSTEM

An engineer is testing a sensor system. The probability that the sensor is defective (D) is 0.05. If a sensor is defective, the probability that it will trigger a false alarm (F) is 0.98. If a sensor is not defective, the probability of a false alarm is 0.02.

What is the probability that a sensor is defective, given that a false alarm has occurred?

Solution

Let:

• D = defective sensor

• F = false alarm

We are asked to find P(D|F).

From the problem:

• P(D) = 0.05

• P(F|D) = 0.98

• P(F|Dᶜ) = 0.02 (where Dᶜ means “not defective”)

• P(Dᶜ) = 1 – P(D) = 0.95

Using the Law of Total Probability to find P(F):

$$P(F) = P(F|D) \cdot P(D) + P(F|Dᶜ) \cdot P(Dᶜ)$$

P(F)=P(F∣D)⋅P(D)+P(F∣Dc)⋅P(Dc)

$$P(F) = (0.98)(0.05) + (0.02)(0.95) = 0.049 + 0.019 = 0.068$$

P(F)=(0.98)(0.05)+(0.02)(0.95)=0.049+0.019=0.068

Now, applying Bayes’ Theorem:

$$P(D|F) = \frac{P(F|D) \cdot P(D)}{P(F)} = \frac{(0.98)(0.05)}{0.068} \approx 0.7206$$

P(D∣F)=P(F)P(F∣D)⋅P(D)​=0.068(0.98)(0.05)​≈0.7206

Answer: Approximately 72% chance that the sensor is defective given that a false alarm has occurred.

TRY IT

In a manufacturing plant, 2% of parts produced are defective. A quality control test detects defective parts 95% of the time. For non-defective parts, the test incorrectly flags them as defective 1% of the time.

If a part tests defective, what is the probability it is actually defective?

Problem Recap:

• 2% of parts are defective.

• Detection test:

  • True Positive: 95% (P(Positive|Defective) = 0.95)

  • False Positive: 1% (P(Positive|Not Defective) = 0.01)

• Find: P(Defective|Positive Test)

Solution:

Let:

• D = defective

• T+ = test positive

Given:

• P(D) = 0.02

• P(T+|D) = 0.95

• P(T+|Dᶜ) = 0.01

• P(Dᶜ) = 0.98

Using Law of Total Probability:

$$P(T+) = (0.95)(0.02) + (0.01)(0.98) = 0.019 + 0.0098 = 0.0288$$

P(T+)=(0.95)(0.02)+(0.01)(0.98)=0.019+0.0098=0.0288

Applying Bayes’ Theorem:

$$P(D|T+) = \frac{(0.95)(0.02)}{0.0288} = \frac{0.019}{0.0288} \approx 0.6597$$

P(D∣T+)=0.0288(0.95)(0.02)​=0.02880.019​≈0.6597

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

USING BAYES’ THEOREM IN ENGINEERING

In engineering contexts, Bayes’ Theorem is commonly used for:

• System reliability assessments.

• Failure prediction and risk analysis.

• Updating failure probabilities as test results are gathered.

In real-time monitoring systems, engineers can use Bayes’ Theorem to improve diagnostic accuracy as new evidence (sensor readings, test outcomes) becomes available.

KEY TAKEAWAYS

• Bayes’ Theorem updates existing probabilities based on new evidence.

• It is particularly useful when direct measurement of a conditional probability is difficult.

• In engineering, Bayes’ Theorem assists in decision-making under uncertainty.

EXERCISES

1. A machine produces 5% defective bolts. A sensor detects defective bolts correctly 90% of the time but falsely labels good bolts as defective 3% of the time. If a bolt is flagged defective, what is the probability it is actually defective?

Solution:

• P(D) = 0.05

• P(T+|D) = 0.90

• P(T+|Dᶜ) = 0.03

• P(Dᶜ) = 0.95

$$P(T+) = (0.90)(0.05) + (0.03)(0.95) = 0.045 + 0.0285 = 0.0735$$

P(T+)=(0.90)(0.05)+(0.03)(0.95)=0.045+0.0285=0.0735

$$P(D|T+) = \frac{(0.90)(0.05)}{0.0735} = \frac{0.045}{0.0735} \approx 0.6122$$

P(D∣T+)=0.0735(0.90)(0.05)​=0.07350.045​≈0.6122

Answer: ~61% chance the bolt is actually defective if flagged.

2. A diagnostic tool tests for overheating in circuits. Suppose 1% of circuits overheat. If overheating is present, the test detects it 99% of the time. If no overheating is present, it gives a false positive 4% of the time. If a circuit is flagged as overheating, what is the probability it actually is overheating?

Solution:

• P(O) = 0.01

• P(T+|O) = 0.99

• P(T+|Oᶜ) = 0.04

• P(Oᶜ) = 0.99

$$P(T+) = (0.99)(0.01) + (0.04)(0.99) = 0.0099 + 0.0396 = 0.0495$$

P(T+)=(0.99)(0.01)+(0.04)(0.99)=0.0099+0.0396=0.0495

$$P(O|T+) = \frac{(0.99)(0.01)}{0.0495} = \frac{0.0099}{0.0495} = 0.2$$

P(O∣T+)=0.0495(0.99)(0.01)​=0.04950.0099​=0.2

Answer: Only 20% chance the circuit is actually overheating if flagged.

3. In a project, 20% of the sensors are known to be faulty. Faulty sensors report incorrect measurements 95% of the time, while good sensors report incorrect measurements 5% of the time. If a sensor reports incorrect data, what is the probability it is faulty?

Solution:

• P(F) = 0.20

• P(I|F) = 0.95

• P(I|Fᶜ) = 0.05

• P(Fᶜ) = 0.80

$$P(I) = (0.95)(0.20) + (0.05)(0.80) = 0.19 + 0.04 = 0.23$$

P(I)=(0.95)(0.20)+(0.05)(0.80)=0.19+0.04=0.23

$$P(F|I) = \frac{(0.95)(0.20)}{0.23} = \frac{0.19}{0.23} \approx 0.8261$$

P(F∣I)=0.23(0.95)(0.20)​=0.230.19​≈0.8261

Answer: Approximately 83% chance the sensor is faulty if incorrect data is reported.