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Probabilities of boolean combinations

As in the nonindependent case, we may utilize the minterm expansion and the minterm probabilities to calculate the probabilities of Boolean combinations of events. However,it is frequently more efficient to manipulate the expressions for the Boolean combination to be a disjoint union of intersections.

A simple boolean combination

Suppose the class { A , B , C } is independent, with respective probabilities 0.4, 0.6, 0.8. Determine P ( A B C ) . The minterm expansion is

A B C = M ( 3 , 4 , 5 , 6 , 7 ) , so that P ( A B C ) = p ( 3 , 4 , 5 , 6 , 7 )

It is not difficult to use the product rule and the replacement theorem to calculate the needed minterm probabilities. Thus p ( 3 ) = P ( A c ) P ( B ) P ( C ) = 0 . 6 0 . 6 0 . 8 = 0 . 2280 . Similarly p ( 4 ) = 0 . 0320 , p ( 5 ) = 0 . 1280 , p ( 6 ) = 0 . 0480 , p ( 7 ) = 0 . 1920 . The desired probability is the sum of these, 0.6880.

As an alternate approach, we write

A B C = A A c B C , so that P ( A B C ) = 0 . 4 + 0 . 6 0 . 6 0 . 8 = 0 . 6880

Considerbly fewer arithmetic operations are required in this calculation.

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In larger problems, or in situations where probabilities of several Boolean combinations are to be determined, it may be desirable to calculate all minterm probabilities then use theminterm vector techniques introduced earlier to calculate probabilities for various Boolean combinations. As a larger example for which computational aid is highly desirable, consideragain the class and the probabilities utilized in [link] , above.

Consider again the independent class { E 1 , E 2 , , E 10 } with respective probabilities { 0 . 13 0 . 37 0 . 12 0 . 56 0 . 33 0 . 71 0 . 22 0 . 43 0 . 57 0 . 31 } . We wish to calculate

P ( F ) = P E 1 E 3 ( E 4 E 7 c ) E 2 ( E 5 c E 6 E 8 ) E 9 E 10 c

There are 2 10 = 1024 minterm probabilities to be calculated. Each requires the multiplication of ten numbers. The solution with MATLAB is easy, as follows:

>>P = 0.01*[13 37 12 56 33 71 22 43 57 31];>>minvec10 Vectors are A1 thru A10 and A1c thru A10cThey may be renamed, if desired.>>F = (A1|(A3&(A4|A7c)))|(A2&(A5c|(A6&A8)))|(A9&A10c);>>pm = minprob(P);>>PF = F*pm' PF = 0.6636

Writing out the expression for F is tedious and error prone. We could simplify as follows:

>>A = A1|(A3&(A4|A7c));>>B = A2&(A5c|(A6&A8));>>C = A9&A10c;>>F = A|B|C; % This minterm vector is the same as for F above

This decomposition of the problem indicates that it may be solved as a series of smaller problems. First, we need some central facts about independence ofBoolean combinations.

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Independent boolean combinations

Suppose we have a Boolean combination of the events in the class { A i : 1 i n } and a second combination the events in the class { B j : 1 j m } . If the combined class { A i , B j : 1 i n , 1 j m } is independent, we would expect the combinations of the subclasses to be independent. It is importantto see that this is in fact a consequence of the product rule, for it is further evidence that the product rule has captured the essence of the intuitive notion of independence.In the following discussion, we exhibit the essential structure which provides the basis for the following general proposition.

Proposition . Consider n distinct subclasses of an independent class of events. If for each i the event A i is a Boolean (logical) combination of members of the i th subclass, then the class { A 1 , A 2 , , A n } is an independent class.

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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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