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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.
 Fall '07
 Carlton

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