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Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 73e
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 73e

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# Twenty pairs of individuals playing in a bridge tournament ISBN: 9780321629111 32

## Solution for problem 73E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 73E

Twenty pairs of individuals playing in a bridge tournament have been seeded 1, . . . , 20. In the first part of the tournament, the 20 are randomly divided into 10 east–west pairs and 10 north–south pairs. a.? ?What is the probability that x of the top 10 pairs end up playing east–west? b.? ?What is the probability that all of the top five pairs end up playing the same direction? c. ?If there are 2n pairs, what is the pmf of X = the number among the top n pairs who end up playing east–west? What are E(X) and V(X)?

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Answer: Step1: Given, twenty pairs of individuals playing in a bridge tournament have been seeded 1, . . . , 20. In the first part of the tournament, the 20 are randomly divided into 10 east–west pairs and 10 north–south pairs. Here, M = 10, n = 10, N = 20 P(X = x) = h(x;n,M,N) ( x)( n x = N (n) Step2: a). The aim is what is the probability that x of the top 10 pairs end up playing east–west. 20 pairs of individuals were randomly divided into 10 east–west pairs and 10 north–south pairs. Let ‘x’ be the number of top 10 pairs from east-west group. Probability that ‘x’ of the top 10 pairs end up playing east-west is. 10 2010 h(x;10,10,20) = (x )( 10 x (10 Step 3: b). To find the probability that all of the top five pairs end up playing the same direction. ( )( )15 P( all top five pairs may east-west) = 5 205 ( 10 1×3003 = 184756 = 0.0163 Therefore, the probability that all of the top five pairs end up playing the same direction is 0.0163. Step 4: c). If there are 2n pairs, to find the pmf of X = the number among the top n pairs who end up playing east–west. The pmf of ‘x’ is ( x )( NM ) P(X = x) = Nn x ( n 10 2010 ( x)( 2n x h(x;2n,10,20) = ( 2n Then, Mean E(x) = n.( M ) N = 2n.( 10) 20 = n var(x) = = 2 nM(Nn)(Nm) N (N1) 2n(10)(202n)(2010) = 2 20 (201) 20n(202n)(10) = 400(19) 2 = (400n 40n )10 7600 2 4000n 400n = 7600 If there are 2n pairs, the mean and variance is E(x) = n 2 4000n 400n2 var(x) = = 7600

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##### ISBN: 9780321629111

Since the solution to 73E from 3 chapter was answered, more than 1536 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. The full step-by-step solution to problem: 73E from chapter: 3 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM. This full solution covers the following key subjects: pairs, playing, top, west, end. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. The answer to “Twenty pairs of individuals playing in a bridge tournament have been seeded 1, . . . , 20. In the first part of the tournament, the 20 are randomly divided into 10 east–west pairs and 10 north–south pairs. a.? ?What is the probability that x of the top 10 pairs end up playing east–west? b.? ?What is the probability that all of the top five pairs end up playing the same direction? c. ?If there are 2n pairs, what is the pmf of X = the number among the top n pairs who end up playing east–west? What are E(X) and V(X)?” is broken down into a number of easy to follow steps, and 102 words.

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