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