kondicionalna vjerovatnoća
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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
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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
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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
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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
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7/24/2019 kondicionalna vjerovatnoa
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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
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7/24/2019 kondicionalna vjerovatnoa
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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
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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
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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
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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 )
Ani Adhikari and Philip Stark Statistics 2.2X Lecture 2.1 2 / 6
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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/
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