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
    • Addition rule
    • 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

Venn diagram of events \( A \) and \( B \) with nontrivial intersection.

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.
Wafers in Semiconductor Manufacturing Classified by Contamination and Location

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…