# Total Probability, Independence and Bayes' theorem

02/08/2021

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## Outline

• The following topics will be covered in this lecture:

• Multiplication Rule
• Total Probability Rule
• Independence
• Bayes' theorem

## Intersections of Events and Multiplication Rule

• Recall that last time we discussed the probability of the intersection of two events.
• Let us suppose that $$A$$ and $$B$$ are events for which $$P(A)\neq 0$$ and $$P(B)\neq 0$$.
• Using the definition of conditional probability $P(B|A)=\frac{P(A\cap B)}{P(A)}$
• We can solve for the intersection of events $P(A \cap B) = P(B\vert A) P(A)$
• Similarly, from the same definition of conditional probability we have $P(A|B)=\frac{P(A\cap B)}{P(B)}$
• which implies that $P(A \cap B) = P(A\vert B) P(B)$
• We can provide a formula known as the multiplication rule
Probability of an Intersection: $P(A \cap B) = P(B\vert A) P(A) = P(A\vert B) P(B)$

## Intersections of Events – example

• EXAMPLE: The probability that the first stage of a numerically controlled machining operation for high-rpm pistons meets specifications is 0.90.

• Failures are due to metal variations, fixture alignment, cutting blade condition, vibration, and ambient environmental conditions.

• Given that the first stage meets specifications, the probability that a second stage of machining meets specifications is 0.95.

• Question: Using the multiplication rule,

$P(A \cap B) = P(B\vert A) P(A) = P(A\vert B) P(B)$ what is the probability that both stages meet specifications?

• Let the events be $$A=$$"first stage meets specifications" and $$B=$$"second stage meets specifications".
• The probability requested is $$P(A \text{ and }B)$$
• where $$P(A)=0.90$$
• and $$P(B|A)=0.95$$
• Using the multiplication rule, we get \begin{align} P(A \cap B) &= P(B\vert A) P(A) \\ &= 0.95(0.90) = 0.855 \end{align}
• Note: although it is also true that $$P(A \cap B) = P(A | B)P(B)$$, the information provided in the problem does not match this second formulation.

## Total Probability Rule

• So far we have described probabilities of events in terms of the probabilities of union or intersections of events, i.e.:
• Addition rule: $$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$;
• Conditional probability: $$P(A\vert B) = \frac{P(A\cap B)}{P(B)}$$; and the
• Multiplication rule: $$P(A\cap B) = P(A\vert B)P(B)$$.
• What if we want to recover the probability of single event $$P(B)$$ given several conditions? Courtesy of Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th edition

• For any event $$B$$, we can write $$B$$ as the union of the part of $$B$$ in $$A$$ and the part of $$B$$ in $$A′$$. That is $B=(A\cap B)\cup(A'\cap B)$
• Because $$A$$ and $$A′$$ are mutually exclusive
• $$A \cap B$$ and $$A' \cap B$$ are mutually exclusive
• So we can use the addition rule for mutually exclusive events as \begin{align} P(B)&=P((A\cap B)\cup(A'\cap B))\\ &=P(A\cap B)+P(A'\cap B)\end{align}
• Using the multiplication rule on each term of $$P(B)$$ we get $P(B)=P(B|A)P(A)+P(B|A')P(A')$
• Total Probability Rule (Two Events) For any two events $$A$$ and $$B$$
• \begin{align} P(B)=P(B\cap A)+P(B\cap A')=P(B|A)P(A)+P(B|A')P(A') \end{align}

## Total Probability Rule – example Courtesy of Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th edition

• Suppose that in semiconductor manufacturing, the probability is 0.10 that a chip subjected to high levels of contamination during manufacturing has a product failure.
• The probability is 0.005 that a chip not subjected to high contamination levels during manufacturing has a product failure.
• In a particular production run, 20% of the chips are subject to high levels of contamination.
• Question: using the total probability rule \begin{align} P(B)=P(B\cap A)+P(B\cap A')=P(B|A)P(A)+P(B|A')P(A'), \end{align}
• what is the total probability of a chip failure?
• Lets denote the events be $$F=$$"product fails" and $$H=$$"chip is exposed to high levels of contamination".
• From the table we can extract some pieces of information
• $$P(H)=0.2$$ and $$P(H')=0.8$$
• $$P(F|H)=0.10$$ and $$P(F|H')=0.005$$
• We can use the total probability rule on $$P(F)$$ in terms of conditional probabilities $P(F)=P(F|H)P(H)+P(F|H')P(H')$
• Then the probability that a product fails is $P(F)=0.10(0.20)+0.005(0.80)=0.024$