02/03/2021

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The following topics will be covered in this lecture:

- Axioms of probabilty
- Addition rule
- Conditional probability
- Multiplication rule

- Now that we are more familiar with the notions of probability,
we can collect the assumptions into a set of axioms of probability that must be satisfied in
*any random experiment*. Probability is a number that is assigned to each member of a collection of events from a random experiment that satisfies the following properties:

- \( P(S) = 1 \) where \( S \) is the sample space
- \( 0 ≤ P(E) ≤ 1 \) for any event \( E \)
- For two events \( E_1 \) and \( E_2 \) with \( E_1 \cap E_2=\emptyset \) \[ P(E_1\cup E_2)=P(E_1)+P(E_2) \]

- These axioms
*do not determine probabilities*; the probabilities are assigned based on our knowledge of the system under study. - These axioms imply some important results
- Probability of the empty set is zero \[ P(\emptyset)=0 \]
- We can recognize this as \[ P(S \cup \emptyset ) = P(S) \] and \[ P(S \cup \emptyset ) = P(S) + P(\emptyset) \]
- Putting the above together, we have \[ P(S) + P(\emptyset) = P(S) \Leftrightarrow P(\emptyset) = 0. \]

- Probability that event \( E \) does not occur \[ P(E')=1-P(E) \]
- Notice that \[ P(E' \cup E) = P(S) = 1 \] and \[ P(E' \cup E) = P(E') + P(E) \]
- Therefore, we have \[ P(E') + P(E) = 1 \Leftrightarrow P(E') = 1 - P(E). \]
- If the event \( E_1 \) is contained in the event \( E_2 \), \[ P(E_1)\leq P(E_2) \]
- Notice that \( E_1 \cup E_2 = E_2 \) and \( E_1 \cap E_2 = E_1 \) because of the set containment.
- Therefore, \[ P(E_2) = P(E_1 \cup E_2 ) = P\left((E_1\cap E_2) \cup (E_1'\cap E_2)\right) = P(E_1)+ P(E_1'\cap E_2) \]
- Considering the above, we have \[ P(E_1) = P(E_2)- P(E_1'\cap E_2) \] where \( P(E_1'\cap E_2) \geq 0 \).

Courtesy of Bin im Garten CC via Wikimedia Commons

- More generally, suppose we want to
**compute the probability of two events \( A \) and \( B \) joined**by the compound operation**“or”**that**are not disjoint**. - We read the statement, \[ P(A \text{ or } B) \] as the probability of event:
- \( A \) occuring,
- event \( B \) occuring, or
- both \( A \) and \( B \) ocurring.
- Intuitively, we can express the probability in terms of all the ways \( A \) can occur and all the ways \( B \) can occur,
*if we don’t double count.* - Consider, if there is an
**overlap where both \( A \) and \( B \) occur simultaneously**, \[ P(A \cap B)\neq \emptyset \] then summing the total of all ways \( A \) occurs and the total of all ways \( B \) occurs**double counts the the cases where both \( A \) and \( B \) occur**.

- Therefore, the
**addition rule**for compound events is given as **Probability of a union**\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

**EXAMPLE:**the table below lists the history of 940 wafers in a semiconductor manufacturing process.

Courtesy of Montgomery & Runger, *Applied Statistics and Probability for Engineers*, 7th edition

**Question:**whats is the probability that a wafer is from the center of the sputtering tool or contains high levels of contamination ?- Suppose that 1 wafer is selected at random.
- Total number of outcomes is \( 626+314=940 \).
- Let \( H \) denote the event that the wafer contains
**high**levels of contamination. Then, \[ P(H) = \frac{358}{940} \] - Let \( C \) denote the event that the wafer is in the
**center**of a sputtering tool. Then \[ P(C) = \frac{626}{940} \] - The event \( H \cap C \) is the event that the wafer is from the center of the sputtering tool
**and**contains high levels of contamination.Then \[ P(H \cap C)=\frac{112}{940} \]

- More complicated probabilities, such as \( P(A \cup B \cup C) \), can be determined by repeated use of the
**addition rule**: \[ P(A \cup B \cup C) = P[(A \cup B) \cup C] = P(A \cup B) + P(C) − P[(A \cup B) \cap C]. \] - We can apply the
**addition rule**again on \( P(A \cup B)= P(A)+P(B)-P(A \cap B) \) - and we can use the distributed rule for set operations on \[ P[(A \cup B) \cap C]=P[(A\cap C)\cup (B\cap C)] \]
- We apply the
**addition rule**on the right-hand side of the expression above \[ P[(A\cap C)\cup (B\cap C)] = P(A\cap C)+P(B\cap C )-P(A\cap B \cap C) \] - Finally we put everything together \[ \begin{align} P(A \cup B \cup C) & = P(A \cup B) + P(C) − P[(A \cup B) \cap C]\\ & = P(A)+P(B)-P(A \cap B) + P(C) - P(A\cap C) -P(B\cap C )+P(A\cap B \cap C) \end{align} \]
- If the events are
**mutually exclusive**, the results simplify considerably…