Math Squad

Bayes' Theorem
Example 1: Quality Control
Example 2: False Positives
Example 3: Monty Hall Problem

Example 1: Quality Control

by Maliha Hossain, proud Member of the Math Squad.

keyword: probability, Bayes' Theorem, Bayes' Rule 

The following problem has been adapted from a few practice problems from chapter 2 of Probability, Statistics and Random Processes for Electrical Engineers by Alberto Leon-Garcia. The example illustrates how Bayes' Theorem plays a role in quality control.

A manufacturer produces a mix of "good" chips and "bad" chips. The proportion of good chips whose lifetime exceeds time $ t $ seconds decreases exponentially at the rate $ \alpha $. The proportion of bad chips whose lifetime exceeds t decreases much faster at a rate $ 1000\alpha $. Suppose that the fraction of bad chips is $ p $, and of good chips, $ 1 - p $.

Let $ C $ be the event that the chip is functioning after $ t $ seconds. Let $ G $ be the event that the chip is good. Let $ B $ be the event that the chip is bad.

Here's what we can infer from the problem statement thus far:

the probability that the lifetime of a good chip exceeds $ t $: $ P[C|G] = e^{-\alpha t} $

the probability that the lifetime of a bad chip exceeds $ t $: $ P[C|B] = e^{-1000\alpha t} $

So by the theorem of total probability, we have that

$ \begin{align} P[C] &= P[C|G]P[G] + P[C|B]P[B] \\ &= e^{-\alpha t}(1-p) + e^{-1000\alpha t}p \end{align} $

Now suppose that in order to weed out the bad chips, every chip is tested for t seconds prior to leaving the factory. the chips that fail are discarded and the remaining chips are sent out to customers. Can you find the value of $ t $ for which 99% of the chips sent out to customers are good?

The problem requires that we find the value of $ t $ such that

$ P[G|C] = .99 \ $

We find $ P[G|C] $ by applying Bayes' Theorem

$ \begin{align} P[G|C] &= \frac{P[C|G]P[G]}{P[C|G]P[G] + P[C|B]P[B]} \\ &= \frac{e^{-\alpha t}(1-p)}{e^{-\alpha t}(1-p) + e^{-1000\alpha t}} \\ &= \frac{1}{1 + \frac{pe^{-1000\alpha t}}{e^{-\alpha t}(1-p)}} = .99 \end{align} $

The above equation can be solved for $ t $

$ t = \frac{1}{999\alpha}ln(\frac{99p}{1-p}) $


  • Alberto Leon-Garcia, Probability, Statistics, and Random Processes for Electrical Engineering, Third Edition

Questions and comments

If you have any questions, comments, etc. please post them below:

  • Comment / question 1

Back to tutorial

Back to Math Squad page

The Spring 2013 Math Squad 2013 was supported by an anonymous gift to Project Rhea. If you enjoyed reading these tutorials, please help Rhea "help students learn" with a donation to this project. Your contribution is greatly appreciated.

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva