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LectureNotes5 - Probability Part II Cyr Emile MLAN Ph.D...

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Probability: Part II Cyr Emile M’LAN, Ph.D. [email protected] Probability: Part II – p. 1/33
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Counting Rules Text Reference : Introduction to Probability and Statistics for Engineers and Scientists, Chapter 3. Reading Assignment : Section 3.5, September 22-September 24 When the various outcomes of an experiment are equally likely, the task of computing probabilities reduces to counting. However, when the number of simple events is large, this task of counting the number of simple events in the sample space and the event of interest becomes tedious or even impossible. Probability: Part II – p. 2/33
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Counting Techniques Product Rule for Ordered Pairs Our first counting rule applies to any situation in which an event consists of ordered pairs of objects and we wish to count the number of such pairs. Proposition 3.1a : If the first element of an ordered pair can be selected in n 1 ways, and for each of these n 1 ways the second element of the pair can be selected in n 2 ways, then the number of pairs is n 1 n 2 . Example 3.6 : A homeowner doing some remodeling requires the services of both a plumbing contractor and an electrical contractor. If there are 12 plumbing contractors and 9 electrical contractors available in the area, in how many ways can the contractors be chosen? Probability: Part II – p. 3/33
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Counting Techniques Solution : If we denote the plumbers by P 1 , . . . , P 12 and the electricians by Q 1 , . . . , Q 9 , then we wish the number of pairs of the form ( P i , Q j ) . With n 1 = 12 and n 2 = 9 , the product rule yields N = (12)(9) = 108 possible ways of choosing the two types of contractors. Product Rule for Ordered Collections of k objects Proposition 3.1b : Suppose an event consists of ordered collections of k el- ements ( k -tuples) and that there are n 1 possible choices for the first element; for each choice of the first element, there are n 2 possible choices of the second element;...; for each possible choice of the first k - 1 elements, there are n k choices of the k th element. Then there are n 1 n 2 . . . n k possible k -tuples. Probability: Part II – p. 4/33
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Counting Techniques Example 3.7 : The fixed-price dinner at a local restaurant provides the following choices: Appetizer : seafood soup or salad or egg rolls or crab rangoon Entrée : baked chicken or sweet and sour chicken or diced lemon chicken or broiled beef patty or teriyaki beef or beef with snow pea or beef boneless rib Desert : ice cream or cheesecake Solution : There are four choices of an appetizer. For each choice of appetizer, there 7 choices of entrée, and that for each of the 4 · 7 = 28 choices of appetizer and entrée, there are 2 choices for desert. Hence, a total of 4 · 7 · 2 = 56 different meals can be ordered. Probability: Part II – p. 5/33
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Counting Techniques Example 3.8 : Mr. Jones has 10 books that he is going to put on his bookshelf. Of these, 4 are mathematics books, 3 are chemistry books, 2 are history books, and 1 is a language book. Jones wants to arrange his books so that all the books dealing with the same subjects are together on the shelf. How many different arrangement are possible?
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