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MANE 3332.05

Lecture 3

Agenda

  • Continue Chapter 2 Lecture
  • Search function in course website
  • Single Event Practice Problems

Handouts

Examples of Random Experiments, Sample Space, Events

  • Consider the bead bowl

  • Consider the Texas Lottery's Pick Three game (I am not encouraging gambling)

Pick 3 Front Page

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Pick 3 Back Page

Tree Diagrams

  • Tree diagrams are a useful tool for understanding sample spaces and events. Apply to Pick Three game.

Probability

  • The probability of an event is the likelihood that it occurs

  • Probability is expressed as a number between 0 and 1

  • Probability of an event can be found by dividing the number of outcomes of the desired events divided by the total number of outcomes in the sample space (if all events are equally likely)

Counting Techniques

  • Consider ordered versus unordered subsets

  • Ordered subsets (Permutations)

    \[ P^n_r=\frac{n!}{\left(n-r\right)!} \]

  • Unordered subsets (Combinations)

    \[ C^n_r=\frac{n!}{r!\left(n-r\right)!} \]

  • Good idea to do a calculator check

Axioms (Rules) of Probability

Probability is a number that is assigned to each member of a collection of events from a random experiment that satisfies the following properties:

If \(S\) is the sample space and \(E\) is any event in a random experiment,

  1. \(P(S)=1\)

  2. \(0\leq P(E)\leq 1\)

  3. For two events \(E_1\) and \(E_2\) with \(E_1\cap E_2=\emptyset\)

\[ P\left(E_1\cup E_2\right)=P\left(E_1\right)+P\left(E_2\right) \]
  • Consider problem 2-70

Practice Problems - Single Event

A Word of Warning

  • It usually looks very easy when I work a problem

  • I have been using statistics for almost 40 years

  • This is something you MUST practice

  • Rework class room examples and textbook examples

Probability of Multiple Events

Intersection:

\(P\left(A\cap B\right)\) is "the probability of \(A\) and \(B\) occurring

Union:

\(P\left(A\cup B\right)\) is "the probability of \(A\) or \(B\) (or both)"

Complement:

\(P\left(A^{\prime}\right)\) is "the probability of not \(A\)"

  • Venn diagrams are a very useful tool for understanding multiple events and calculating probabilities

Addition Rules

  • Used to calculate the union of two events

    \[ P(A\cup B)=P(A)+P(B)-P(A\cap B) \]

  • If two events are mutually exclusive (\(A\cap B=\emptyset\))

\[ P(A\cup B)=P(A)+P(B) \]
  • Consider problems 2-82 and 2-85

Addition Rule for 3 or More Events

  • For three events
\[ \begin{aligned} P(A\cup B\cup C)&=&P(A)+P(B)+P(C)\\ &&-P(A\cap B)-P(A\cap C)-P(B\cap C)\\ && + P(A\cap B\cap C)\\ \end{aligned} \]

  • For a set of events to mutually exclusive all pairs of variables must satisfy \(E_1\cap E_2=\emptyset\)

  • For a collection of mutually exclusive events,

\[ P(E_1\cup E_2\cup \ldots\cup E_k)=P(E_1)+P(E_2)+\cdots+P(E_k) \]

Conditional Probability

  • Hayter (2002) states that "For experiments with two or more events of interest, attention is often directed not only at the probabilities of individual events but also at the probability of an event occurring conditional on the knowledge that another event has occurred."

  • The conditional probability of an event \(B\) given an event \(A\), denoted \(P(B|A)\) is

\[ P(B|A)=\frac{P(A\cap B)}{P(A)} \]

for \(P(A)>0\)

  • Consider problems 2-99

Multiplication Rules

  • This rule provides another method for calculating \(P(A\cap B)\) $$ \begin{aligned} P(A\cap B)&=&P(A|B)P(B)=P(B|A)P(A) \end{aligned} $$
  • This leads to the total probability rule $$ \begin{aligned} P(B)&=&P(B\cap A)+P(B\cap A^{\prime})\ &=&P(B|A)P(A)+P(B|A^{\prime})P(A^{\prime})\ \end{aligned} $$
  • Consider problems from 3rd edition (next slide) and 2-129

Example Problem 2-76

problem 2-76

Independent Events

  • Two events are independent if any one of the following is true:

    1. \(P(A|B)=P(A)\)

    2. \(P(B|A)=P(B)\)

    3. \(P(A\cap B)=P(A)P(B)\)

  • Consider problem 2-146

Reliability Analysis

  • Reliability is the application of statistics and probability to determine the probability that a system will be working properly

  • Components can be arranged in series. All components must work for the system to work.

\[ P(\mbox{system works})=P(A\mbox{ works})P(B\mbox{ works}) \]
  • Components can be arranged in parallel. As long as one component works, the system works.
\[ P(\mbox{system works})=1-(1-P(A\mbox{ works}))\times 1-P(B\mbox{ works})) \]
  • Consider problem 2-157