Chapt 1e - 1.10 Basic Combinatorial Analysis 41 Solution...

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Unformatted text preview: 1.10 Basic Combinatorial Analysis 41 Solution Let N denote the number of ways of indiscriminately selecting 10 com- ponents from a batch of 100 components. Then N is given by N = C ( 100 , 10 ) = 100 ! 90 ! × 10 ! Let E denote the event “the batch containing 10 defective components is ac- cepted by the inspector.” The number of ways that E can occur is the number of ways of selecting 10 components from the 90 nondefective components and no component from the 10 defective components. This number, N ( E ) , is given by N ( E ) = C ( 90 , 10 ) × C ( 10 , ) = C ( 90 , 10 ) = 90 ! 80 ! × 10 ! Because the components are selected at random, the combinations are equiprob- able. Thus, the probability of event E is given by P ( E ) = N ( E ) N = 90 ! 80 ! × 10 ! × 90 ! × 10 ! 100 ! = 90 ! × 90 ! 100 ! × 80 ! = 90 × 89 × ··· × 81 100 × 99 × ··· × 91 = . 3305 trianglesolid Example 1.34 The Applied Probability professor gave the class a set of 12 re- view problems and told them that the midterm exam would consist of 6 of the 12 problems selected at random. If Lidya memorized the solutions to 8 of the 12 problems but could not solve any of the other 4 problems, what is the probability that she got 4 or more problems correct in the exam? Solution By choosing to memorize only a subset of the review problems Lidya partitioned the 12 problems into two sets: a set consisting of the 8 problems she memorized and a set consisting of the 4 problems she could not solve. If she got k problems correct in the exam, then the k problems came from the first set and the 6- k problems she failed came from the second set, where k = , 1 , 2 , . . . , 6. The number of ways of choosing 6 problems from 12 problems is C ( 12 , 6 ) . The number of ways of choosing k problems from the 8 problems that she memorized is C ( 8 , k ) , and the number of ways of choosing 6- k problems from the four she did not memorize is C ( 4 , 6- k ) , where 6- k ≤ 4 or 2 ≤ k ≤ 6. Because the problems have been partitioned, the number of ways in which the 8 problems can be chosen so that Lidya could get 4 or more of them correct in the exam is C ( 8 , 4 ) C ( 4 , 2 ) + C ( 8 , 5 ) C ( 4 , 1 ) + C ( 8 , 6 ) C ( 4 , ) = 420 + 224 + 28 = 672 42 Chapter 1 Basic Probability Concepts Thus, the probability p that she got 4 or more problems correct in the exam is given by p = C ( 8 , 4 ) C ( 4 , 2 ) + C ( 8 , 5 ) C ( 4 , 1 ) + C ( 8 , 6 ) C ( 4 , ) C ( 12 , 6 ) = 672 924 = 8 11 trianglesolid 1.11 Reliability Applications As discussed earlier in the chapter, reliability theory is concerned with the du- ration of the useful life of components and systems of components. That is, it is concerned with determining the probability that a system with possibly many components will be functioning at time t . The components of a system can be arranged in two basic configurations: series configuration and parallel configura- tion. A real system consists of a mixture of series and parallel components, whichtion....
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This note was uploaded on 01/05/2010 for the course STAT 350 taught by Professor Carlton during the Fall '07 term at Cal Poly.

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Chapt 1e - 1.10 Basic Combinatorial Analysis 41 Solution...

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