# Conditional probability and probability rules

02/03/2021

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## Outline

• The following topics will be covered in this lecture:

• Axioms of probabilty
• Conditional probability
• Multiplication rule

## Axioms of Probability

• 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:
1. $$P(S) = 1$$ where $$S$$ is the sample space
2. $$0 ≤ P(E) ≤ 1$$ for any event $$E$$
3. 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 $P(S) + P(\emptyset) = P(S) \Leftrightarrow P(\emptyset) = 0.$

### Axioms of Probability – continued

• 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 $$P(E_1'\cap E_2) \geq 0$$.

## Unions of Events and Addition Rules

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 $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

## Probability of a union –example

• 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 $P(H \cap C)=\frac{112}{940}$

• We can use the addition rule to obtain \begin{align} P(H \cup C) & = P(H)+P(C)-P(H \cap C)\\ & = \frac{358}{940}+\frac{626}{940}-\frac{112}{940}= \frac{872}{940} \end{align}
• ## Two or more events

• 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…