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  • 7/24/2019 kondicionalna vjerovatnoa

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    Conditional probability

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 1 / 6

    http://find/
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    Conditional probability

    Multiplication rule for all A, B: P(A and B) = P(A) P(B|A)

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 1 / 6

    http://find/
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    Conditional probability

    Multiplication rule for all A, B: P(A and B) = P(A) P(B|A)Rewrite:

    Definition of conditional probability

    P(B|A) =

    P(Aand B)

    P(A)

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 1 / 6

    http://find/
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    Conditional probability

    Multiplication rule for all A, B: P(A and B) = P(A) P(B|A)Rewrite:

    Definition of conditional probability

    P(B|A) =

    P(Aand B)

    P(A)

    denominator: restricts the set of outcomes to what is givennumerator: the event of interest, only on the restricted set that is given

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 1 / 6

    http://find/http://goback/
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    Conditional probability

    Multiplication rule for all A, B: P(A and B) = P(A) P(B|A)Rewrite:

    Definition of conditional probability

    P(B|A) =

    P(Aand B)

    P(A)

    denominator: restricts the set of outcomes to what is givennumerator: the event of interest, only on the restricted set that is given

    Exampleof use: Bayes Rule

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 1 / 6

    http://find/
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    Conditional probability

    Multiplication rule for all A, B: P(A and B) = P(A) P(B|A)Rewrite:

    Definition of conditional probability

    P(B|A) =

    P(Aand B)

    P(A)

    denominator: restricts the set of outcomes to what is givennumerator: the event of interest, only on the restricted set that is given

    Exampleof use: Bayes RuleBut many conditional probabilities can be found without this.

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 1 / 6

    http://find/
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    Conditional probability

    Multiplication rule for all A, B: P(A and B) = P(A) P(B|A)Rewrite:

    Definition of conditional probability

    P(B|A) =

    P(Aand B)

    P(A)

    denominator: restricts the set of outcomes to what is givennumerator: the event of interest, only on the restricted set that is given

    Exampleof use: Bayes RuleBut many conditional probabilities can be found without this.Example: Conditional probabilities for the second stage, given the result ofthe first stage

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 1 / 6

    http://find/
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    Relation between draws

    Two draws at random from R R R G G P(second draw is R )

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 2 / 6

    http://find/
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    Relation between draws

    Two draws at random from R R R G G P(second draw is R ) = 3/5

    regardless of whether the draws are with replacement or without

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 2 / 6

    http://find/http://goback/
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    Relation between draws

    Two draws at random from R R R G G P(second draw is R ) = 3/5

    regardless of whether the draws are with replacement or without

    P(second draw is R | first draw is R )

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 2 / 6

    http://find/
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    Relation between draws

    Two draws at random from R R R G G P(second draw is R ) = 3/5

    regardless of whether the draws are with replacement or without

    P(second draw is R | first draw is R ) withreplacement withoutreplacement

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 2 / 6

    http://find/
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    Relation between draws

    Two draws at random from R R R G G P(second draw is R ) = 3/5

    regardless of whether the draws are with replacement or without

    P(second draw is R | first draw is R ) withreplacement withoutreplacement

    = 3/5 = 2/4

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 2 / 6

    http://find/
  • 7/24/2019 kondicionalna vjerovatnoa

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    Relation between draws

    Two draws at random from R R R G G P(second draw is R ) = 3/5

    regardless of whether the draws are with replacement or without

    P(second draw is R | first draw is R ) withreplacement withoutreplacement

    = 3/5 = 2/4

    P(second draw is R | first draw is G ) withreplacement withoutreplacement= 3/5 = 3/4

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 2 / 6

    http://find/
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    Relation between draws

    Two draws at random from R R R G G P(second draw is R ) = 3/5

    regardless of whether the draws are with replacement or without

    P(second draw is R | first draw is R ) withreplacement withoutreplacement

    = 3/5 = 2/4

    P(second draw is R | first draw is G ) withreplacement withoutreplacement= 3/5 = 3/4

    independent

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 2 / 6

    http://find/
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    Relation between draws

    Two draws at random from R R R G G P(second draw is R ) = 3/5

    regardless of whether the draws are with replacement or without

    P(second draw is R | first draw is R ) withreplacement withoutreplacement

    = 3/5 = 2/4

    P(second draw is R | first draw is G ) withreplacement withoutreplacement= 3/5 = 3/4

    independent dependent

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 2 / 6

    http://find/
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    In practice

    independent trials

    tosses of a coin rolls of a die

    drawswith replacement

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 3 / 6

    http://find/
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    In practice

    independent trials

    tosses of a coin rolls of a die

    drawswith replacement

    trials that arenot independent (that is, dependent trials)

    cards dealt from a deck

    drawswithout replacement

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 3 / 6

    http://find/
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    Independence

    Rough definition

    Two random quantities areindependentif knowing how one of themturned out does not change chances for the other.

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 4 / 6

    http://find/
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    Independence

    Rough definition

    Two random quantities areindependentif knowing how one of themturned out does not change chances for the other.

    Independent events

    Two events Aand Bare independent ifP(B|A) = P(B|not A) = P(B)

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 4 / 6

    http://find/
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    Independence

    Rough definition

    Two random quantities areindependentif knowing how one of themturned out does not change chances for the other.

    Independent events

    Two events Aand Bare independent ifP(B|A) = P(B|not A) = P(B)

    Recall multiplication rule for all A, B: P(A and B) = P(A) P(B|A)

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 4 / 6

    http://find/
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    Independence

    Rough definition

    Two random quantities areindependentif knowing how one of themturned out does not change chances for the other.

    Independent events

    Two events Aand Bare independent ifP(B|A) = P(B|not A) = P(B)

    Recall multiplication rule for all A, B: P(A and B) = P(A) P(B|A)Special case:

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 4 / 6

    http://find/
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    Independence

    Rough definition

    Two random quantities areindependentif knowing how one of themturned out does not change chances for the other.

    Independent events

    Two events Aand Bare independent ifP(B|A) = P(B|not A) = P(B)

    Recall multiplication rule for all A, B: P(A and B) = P(A) P(B|A)Special case:ifA and B are independent, P(Aand B) = P(A) P(B)

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 4 / 6

    http://find/
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    Independence and multiplication

    Two draws at random from R R R G G

    P(both draws are R )

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 5 / 6

    http://find/
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    Independence and multiplication

    Two draws at random from R R R G G

    P(both draws are R ) withreplacement withoutreplacement

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 5 / 6

    http://find/
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    Independence and multiplication

    Two draws at random from R R R G G

    P(both draws are R ) withreplacement withoutreplacement= (3/5)(3/5) = (3/5)(2/4)

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 5 / 6

    http://find/
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    Independence and multiplication

    Two draws at random from R R R G G

    P(both draws are R ) withreplacement withoutreplacement= (3/5)(3/5) = (3/5)(2/4)

    Independence doesnt determinewhetheryou multiply; thatsdetermined byboth events have to happen.

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 5 / 6

    http://find/
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    Independence and multiplication

    Two draws at random from R R R G G

    P(both draws are R ) withreplacement withoutreplacement= (3/5)(3/5) = (3/5)(2/4)

    Independence doesnt determinewhetheryou multiply; thatsdetermined byboth events have to happen.

    Independence affectswhatyou multiply.

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 5 / 6

    http://find/
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    What to multiply

    Example 1. A die is rolled twice. Can you find the chance that the firstroll shows an even number of spots and the second roll shows one spot?

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 6 / 6

    http://find/
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    What to multiply

    Example 1. A die is rolled twice. Can you find the chance that the firstroll shows an even number of spots and the second roll shows one spot?Answer. Yes;(3/6)(1/6)

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 6 / 6

    Wh l i l

    http://find/
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    What to multiply

    Example 1. A die is rolled twice. Can you find the chance that the firstroll shows an even number of spots and the second roll shows one spot?Answer. Yes;(3/6)(1/6)

    Example 2. A person is picked at random from a population. 50% of the population is male 10% of the population is left-handedCan you find the chance that the person is a left-handed male?

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 6 / 6

    Wh l i l

    http://find/
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    What to multiply

    Example 1. A die is rolled twice. Can you find the chance that the firstroll shows an even number of spots and the second roll shows one spot?Answer. Yes;(3/6)(1/6)

    Example 2. A person is picked at random from a population. 50% of the population is male 10% of the population is left-handedCan you find the chance that the person is a left-handed male?

    Answer. No. You get stuck at

    P(male)P(left handed|male) = 0.5P(left handed| male) = 0.5??

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 6 / 6

    Wh t t lti l

    http://find/
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    What to multiply

    Example 1. A die is rolled twice. Can you find the chance that the firstroll shows an even number of spots and the second roll shows one spot?Answer. Yes;(3/6)(1/6)

    Example 2. A person is picked at random from a population. 50% of the population is male 10% of the population is left-handedCan you find the chance that the person is a left-handed male?

    Answer. No. You get stuck at

    P(male)P(left handed|male) = 0.5P(left handed| male) = 0.5??

    The proportion of left-handers among males is not necessarily the same asthe overall proportion, 0.1.

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 6 / 6

    Wh t t lti l

    http://find/
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    What to multiply

    Example 1. A die is rolled twice. Can you find the chance that the firstroll shows an even number of spots and the second roll shows one spot?Answer. Yes;(3/6)(1/6)

    Example 2. A person is picked at random from a population. 50% of the population is male 10% of the population is left-handedCan you find the chance that the person is a left-handed male?

    Answer. No. You get stuck at

    P(male)P(left handed|male) = 0.5P(left handed| male) = 0.5??

    The proportion of left-handers among males is not necessarily the same asthe overall proportion, 0.1. Gender and handedness are dependent.

    Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 6 / 6

    http://find/