Lecture 05. Introduction to Probability Web

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    Statistics

    1

    ST 361: Introduction to Statistics

    Introduction to Probability

    Kimberly Weems

    [email protected]

    5260 SAS Hall

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    Statistics

    Outline

    Probability Trees Probability Models

    Sample Spaces, Events, Venn Diagrams

    Axioms of Probability Probability Rules (Laws)

    Addition Rule

    Multiplication Rule

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    Statistics

    Example: Southwest Energy

    A Southwest Energy Company pipeline has 3 safety shutoffvalves in case the line starts to leak.

    The valves are designed to operate independently of one

    another:

    7% chance that valve 1 will fail 10% chance that valve 2 will fail

    5% chance that valve 3 will fail

    If there is a leak in the line, find the following probabilities:

    a. That all three valves operate correctly

    b. That all three valves fail

    c. That only one valve operates correctly

    d. That at least one valve operates correctly

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    Statistics

    Probability Tree Approach

    A probability tree is a useful way to visualize

    this problem and to find the desiredprobability.

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    A: P(all three valves operate correctly)

    P(all three valves work)

    = .93*.90*.95

    = .79515

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    Statistics

    B: P(all three valves fail)

    P(all three valves fail)

    = .07*.10*.05

    = .00035

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    D: P(at least one valve operates correctly)

    P(at least one valve operatescorrectly

    = 1P(no valves operate correctly)

    = 1 - .00035 = .99965

    7 paths

    1 path

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    Statistics

    Example: AIDS Testing

    V={person has HIV}; CDC: P(V)=.006

    +: test outcome is positive (test indicates

    HIV present)

    -: test outcome is negative

    clinical reliabilities for a new HIV test:

    1. If a person has the virus, the test result will be

    positive with probability .999

    2. If a person does not have the virus, the test result

    will be negative with probability .990

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    Statistics

    Question 1

    What is the probability that a randomlyselected person will test positive?

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    Probability Tree

    clinical

    reliability

    clinical

    reliability

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    Probability Tree

    Multiply

    branch probsclinical

    reliability

    clinical

    reliability

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    Statistics

    Question 1 Answer

    What is the probability that a randomlyselected person will test positive?

    P(+) = .00599 + .00994 = .01593

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    Statistics

    Question 2

    If your test comes back positive, what is theprobability that you have HIV?

    (Remember: we know that if a person has thevirus, the test result will be positive with

    probability .999; if a person does not have thevirus, the test result will be negative withprobability .990).

    Looks very reliable

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    Statistics

    Question 2 Answer

    Answertwo sequences of branches lead to positive test;

    only 1 sequence represented people who have

    HIV.P(person has HIV given that test is positive)

    =.00599/(.00599+.00994) = .376

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    Statistics

    Summary

    Question 1:

    P(+) = .00599 + .00994 = .01593

    Question 2: two sequences of branches lead to

    positive test; only 1 sequence representedpeople who have HIV.

    P(person has HIV given that test is positive)

    =.00599/(.00599+.00994) = .376

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    Recap

    We have a test with very high clinical reliabilities:1. If a person has the virus, the test result will be positivewith probability .999

    2. If a person does not have the virus, the test result will benegative with probability .990

    But we have extremely poor performance when thetest is positive:

    P(person has HIV given that test is positive) =.376

    In other words, 62.4% of the positives are falsepositives! Why?

    When the characteristic the test is looking for israre, most positives will be false.

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    Statistics

    Probability models

    Aprobability modelis a mathematical representationof a random phenomenon. It is defined by its

    sample space,

    events within the sample space, and

    probabilities associated with each event.

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    Sample Space

    Random experiments have unique outcomes.

    The set of all possible outcome of a random experiment is

    called the sample space, S.

    Sis discrete if it consists of a finite or countable infinite set ofoutcomes.

    Sis continuous if it contains an interval (either a finite or

    infinite width) of real numbers.

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    Example: Sample Spaces

    [Sis continuous] Randomly select and measure the thickness

    of a part.

    S=R+ = {x|x > 0}, the positive real line. Negative or zero

    thickness is not possible.

    [Sis continuous, finite width] It is known that the thickness is

    between 10 and 11 mm. We have S= {x|10

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    Example: Sample Spaces (contd)

    [Sis continuous] Two parts are randomly selected & measured.

    S=R+ xR+, Sis continuous.

    [Sis discrete] Do the 2 parts conform to specifications?

    S= {yy, yn, ny, nn}, Sis discrete.

    [Sis discrete] Number of conforming parts?

    S= {0, 1, 2}, Sis discrete.

    [Sis discrete, countable infinite ] Parts are randomly selected

    until a non-conforming part is found.

    S= {n, yn, yyn, yyyn, },

    Sis countably infinite.

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    Statistics

    Events Are Sets of Outcomes

    An event (E) is a subset of the sample space of a random

    experiment, i.e., one or more outcomes of the sample space.

    Event combinations are:

    Union of 2 events = the event consisting of all outcomes

    that are contained in either of two events, E1

    U E2. Called

    E1 or E2.

    Intersection of 2 events = the event consisting of all

    outcomes that contained in both of two events, E1 E2.

    Called E1 and E2.

    Complement of an event = the set of outcomes that are not

    contained in the event, E ornot E, or Ec .

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    Example: Discrete Event Algebra

    Recall the sample space from Example 2, S= {yy, yn, ny, nn}

    concerning conformance to specifications.

    Let E1 denote the event that at least one part does conform

    to specifications, E1 = {yy, yn, ny}

    Let E2 denote the event that no part conforms to

    specifications, E2 = {nn}

    Let E3 = , the null or empty set.

    Let E4 = S, the universal set.

    Let E5

    = {yn, ny, nn}, at least one part does not conform.

    Then E1 U E5 = S

    Then E1 E5 = {yn, ny}

    Then E1= {nn}

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    Example: Continuous Event Algebra

    Measurements of the thickness of a part are modeled with the

    sample space: S= R+.

    Let E1 = {x|10 x < 12}, show on the real line below.

    Let E2 = {x|11

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    Venn Diagrams Show Event Relationships

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    Events A & B contain their respective outcomes. The shaded

    regions indicate the event relation of each diagram.

    Venn diagrams

    i f ll l i

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    Statistics

    Venn Diagram of Mutually Exclusive

    Events

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    Events A & B are mutually exclusive because they share no

    common outcomes.

    The occurrence of one event precludes the occurrence of the

    other.

    Symbolically, A B =

    Mutually exclusive events

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    Statistics

    What is Probability?

    Probability is the likelihood or chance that a particular

    outcome or event from a random experiment will occur.

    Here, only finite sample spaces ideas apply.

    Probability is a number in the [0,1] interval.

    May be expressed as a:

    proportion (0.15)

    percent (15%)

    fraction (3/20)

    Generally speaking, a probability of: 1 indicates highly likely

    0 indicates highly unlikely

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    P b bili B d E ll Lik l

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    Probability Based on Equally-Likely

    Outcomes Whenever a sample space consists ofNpossible outcomes that

    are equally likely, the probability of each outcome is 1/N.

    Example: Consider an unbiased die with 10 faces labeled

    1,2,10. Unbiased means that no face is favored, when

    throwing the die; thus each face has an equal chance of being

    shown. We throw the die; the probability that the die shows

    face 10 is 1/10 or 0.1, because each outcome in the sample

    space is equally likely.

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    Probability of an Event

    For a discrete sample space, theprobability of an event E,

    denoted byP(E), equals the sum of the probabilities of the

    outcomes inE.

    The discrete sample space may be:

    A finite set of outcomes

    A countably infinite set of outcomes.

    Further explanation is necessary to describe probability with

    respect to continuous sample spaces.

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    Equally likely outcomes (uniform model)

    Equally likely outcomes (uniform model):

    If a random phenomenon had kpossible outcomes, all equally

    likely , then each individual outcome has probability 1/k. The

    probability of any event A is:

    P(A) = {count of outcomes in A} / {count of outcomes in S}

    = |A| / k.

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    Statistics

    Example: Probabilities of Events

    A random experiment has a sample space {w,x,y,z}. These

    outcomes are not equally-likely; their probabilities are: 0.1,

    0.3, 0.5, 0.1.

    Event A ={w,x}, event B = {x,y,z}, event C = {z}

    P(A) = _______

    P(B) = _______

    P(C) = _______

    P(A) = _____ and P(B) = ____ and P(C) = ______

    Since event AB = {x}, then P(AB) = _____ Since event AUB = {w,x,y,z}, then P(AUB) = ______

    Since event AC = {null}, then P(AC ) = ________

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    Axioms 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:

    1. P(S) = 1

    2. 0 P(E) 1, for any event E

    3. For each two eventsE1andE2 withE1E2 = ,

    P(E1UE2) =P(E1) +P(E2)addition rule

    These imply that: P() =0 ;P(E) = 1P(E)complement rule

    IfE1is contained inE2, thenP(E1)P(E2).

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    Probability Rules

    Joint events are generated by applying basic set operations to

    individual events, specifically:

    Unions of events,A UB

    Intersections of events,AB

    Complements of events,A

    Probabilities of joint events can often be determined from the

    probabilities of the individual events that comprise it. And

    conversely.

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    Statistics

    Birthday Problem

    What is the smallest number of people youneed in a group so that the probability of2 or

    morepeople having the same birthday is

    greater than 1/2?

    Answer: 23

    No. of people 23 30 40 60

    Probability .507 .706 .891 .994

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    Birthday Problem

    A={at least 2 people in the group have acommon birthday}

    A = {no one has common birthday}

    502.498.1)'(1)(

    498.365

    343

    365

    363

    365

    364)'(

    :23

    365

    363

    365

    364)'(:3

    APAPso

    AP

    people

    APpeople

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    Statistics

    Probability rules

    Addition Rule for Mutually Exclusive Events: Recall that

    two events A and B are mutually exclusive (or disjoint) events

    if they have no outcomes from S in common

    If A and B are mutually exclusive events, then

    P(A or B) = P(A) + P(B)

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    Probability rules

    General Addition Rule: For any two events A and B

    P(A or B) = P(A) + P(B)P(A and B)

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    Probability rules

    Multiplication Rule for independent events. Two events, A

    and B are independent if the occurrence of one does not affectthe probability that the other one will occur.

    If A and B are independent then

    P(A andB) = P(A) P(B).

    Remark on independence: The fact that a coin tossed with my

    left hand comes up T rather than H, does not influence the

    outcome of a coin tossed with my right hand.

    The probability of falling on the street is NOT independent of

    whether it has snowed. These events are dependent.

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    St ti ti

    Some examples

    Example1: We roll a 6-sided die.

    The sample space (set of all possible outcomes) S = ____

    The simple events are:___________.

    The event A that the outcome is {3} is _______.

    Example 2. We roll a die, and the event of interest,E, is

    obtaining an odd number. That is . What is the

    probability of this event ?

    Let . What is the probability ofF?

    1,3,5E

    2,4,6F