02/08/2021

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The following topics will be covered in this lecture:

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

- 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:**\[ P(A \cap B) = P(B\vert A) P(A) = P(A\vert B) P(B) \]

**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.

- 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 \( B \)- \[ \begin{align} P(B)=P(B\cap A)+P(B\cap A')=P(B|A)P(A)+P(B|A')P(A') \end{align} \]

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 \[ P(F)=0.10(0.20)+0.005(0.80)=0.024 \]