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The minterms generated by the class { A , B , C } have minterm probabilities

p m = [ 0 . 15 0 . 05 0 . 02 0 . 18 0 . 25 0 . 05 0 . 18 0 . 12 ]

Show that the product rule holds for all three, but the class is not independent.

pm = [0.15 0.05 0.02 0.18 0.25 0.05 0.18 0.12];y = imintest(pm) The class is NOT independentMinterms for which the product rule fails y =1 1 1 0 1 1 1 0 % The product rule hold for M7 = ABC
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The class { A , B , C , D } is independent, with respective probabilities 0.65, 0.37, 0.48, 0.63. Use the m-function minprob to obtain the minterm probabilities. Usethe m-function minmap to put them in a 4 by 4 table corresponding to the minterm map convention we use.

P = [0.65 0.37 0.48 0.63];p = minmap(minprob(P)) p =0.0424 0.0249 0.0788 0.0463 0.0722 0.0424 0.1342 0.07880.0392 0.0230 0.0727 0.0427 0.0667 0.0392 0.1238 0.0727
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The minterm probabilities for the software survey in Example 2 from "Minterms" are

p m = [ 0 0 . 05 0 . 10 0 . 05 0 . 20 0 . 10 0 . 40 0 . 10 ]

Show whether or not the class { A , B , C } is independent: (1) by hand calculation, and (2) by use of the m-function imintest.

pm = [0 0.05 0.10 0.05 0.20 0.10 0.40 0.10];y = imintest(pm) The class is NOT independentMinterms for which the product rule fails y =1 1 1 1 % By hand check product rule for any minterm 1 1 1 1
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The minterm probabilities for the computer survey in Example 3 from "Minterms" are

p m = [ 0 . 032 0 . 016 0 . 376 0 . 011 0 . 364 0 . 073 0 . 077 0 . 051 ]

Show whether or not the class { A , B , C } is independent: (1) by hand calculation, and (2) by use of the m-function imintest.

npr04_04 Minterm probabilities for [link] are in pm y = imintest(pm)The class is NOT independent Minterms for which the product rule failsy = 1 1 1 11 1 1 1
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Minterm probabilities p ( 0 ) through p ( 15 ) for the class { A , B , C , D } are, in order,

p m = [ 0 . 084 0 . 196 0 . 036 0 . 084 0 . 085 0 . 196 0 . 035 0 . 084 0 . 021 0 . 049 0 . 009 0 . 021 0 . 020 0 . 049 0 . 010 0 . 021 ]

Use the m-function imintest to show whether or not the class { A , B , C , D } is independent.

npr04_05 Minterm probabilities for [link] are in pm imintest(pm)The class is NOT independent Minterms for which the product rule failsans = 0 1 0 10 0 0 0 0 1 0 10 0 0 0
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Minterm probabilities p ( 0 ) through p ( 15 ) for the opinion survey in Example 4 from "Minterms" are

p m = [ 0 . 085 0 . 195 0 . 035 0 . 085 0 . 080 0 . 200 0 . 035 0 . 085 0 . 020 0 . 050 0 . 010 0 . 020 0 . 020 0 . 050 0 . 015 0 . 015 ]

Show whether or not the class { A , B , C , D } is independent.

npr04_06 Minterm probabilities for [link] are in pm y = imintest(pm)The class is NOT independent Minterms for which the product rule failsy = 1 1 1 11 1 1 1 1 1 1 11 1 1 1
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The class { A , B , C } is independent, with P ( A ) = 0 . 30 , P ( B c C ) = 0 . 32 , and P ( A C ) = 0 . 12 . Determine the minterm probabilities.

P ( C ) = P ( A C ) / P ( A ) = 0 . 40 and P ( B ) = 1 - P ( B c C ) / P ( C ) = 0 . 20 .

pm = minprob([0.3 0.2 0.4])pm = 0.3360 0.2240 0.0840 0.0560 0.1440 0.0960 0.0360 0.0240
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The class { A , B , C } is independent, with P ( A B ) = 0 . 6 , P ( A C ) = 0 . 7 , and P ( C ) = 0 . 4 . Determine the probability of each minterm.

P ( A c C c ) = P ( A c ) P ( C c ) = 0 . 3 implies P ( A c ) = 0 . 3 / 0 . 6 = 0 . 5 = P ( A ) .

P ( A c B c ) = P ( A c ) P ( B c ) = 0 . 4 implies P ( B c ) = 0 . 4 / 0 . 5 = 0 . 8 implies P ( B ) = 0 . 2

P = [0.5 0.2 0.4];pm = minprob(P) pm = 0.2400 0.1600 0.0600 0.0400 0.2400 0.1600 0.0600 0.0400
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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