Difference between revisions of "Bayes theorem eg1 S13" - Rhea

Latest revision as of 12:09, 25 November 2013

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}) $

References

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

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