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So by the theorem of total probability, we have that | So by the theorem of total probability, we have that | ||
| − | <math>P[C] = P[C|G]P[G] + P[C|B]P[B]</math> | + | <math>P[C] = P[C|G]P[G] + P[C|B]P[B] </math> |
| − | <math> = e^{-\alpha t}(1-p) + e^{-1000\alpha t}p</math> | + | <math>= e^{-\alpha t}(1-p) + e^{-1000\alpha t}p </math> |
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 <math>t</math> for which 99% of the chips sent out to customers are good? | 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 <math>t</math> for which 99% of the chips sent out to customers are good? | ||
Revision as of 15:57, 16 March 2013
Example 1: Quality Control
by Maliha Hossain
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
$ P[C] = P[C|G]P[G] + P[C|B]P[B] $
$ = e^{-\alpha t}(1-p) + e^{-1000\alpha t}p $
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
$ 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 $
The above equation can be solved for $ t $
$ t = \frac{1}{999\alpha}ln(\frac{99p}{1-p}) $
Questions and comments
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