Explanation Calculation: The following information provided in this case. There are n cards numbered 1 through n and these are to be turned one a time. The problem is to guess the card in such way that the strategy that maximizes the expected number of correct guesses fixes permutation of n cards. So, that is continually guesses 1 until it is correct then move to the next number that is 2 and choose the number until it is correct or all cards have been turned over, and continually guesses the remaining cards. The object is to find P ( G = K ) where G is the number of correct guesses. To get only one correct guess, which means the guessing of the card with number one must be picked in the last trial or the card with number one must ne picked after the card with number two. Therefore, leaving the cards with numbers one and two out of the pool and the number of ways to ordered is ( n − 2 ) ! ways. The two cards can be ordered in 2 ! ways. The correct order to get the exact one correct guess should be in 2 , 1 . Therefore, the total number of ways to order the cards is 2 ! ( n − 2 ) ! ways. The favorable number of ways to get one exact guess is 1 ! ( n − 2 ) ! ways. The probability of one correct guess is, P ( G = K ) = 1 ( n − 2 ) ! 2 ! ( n − 2 ) ! = 1 2 ! To get only 3 correct guesses, the card with number 4 must be before three, the card with number must ne before 2 and the card with number 2 must be before one. Therefore, leaving the cards with number 1 , 2 , 3 and 4 out of the pool and the number of ways to ordered is ( n − 4 ) ! ways. The four cards can be ordered in 4 ! ways. Among them the correct guesses are possible in 3 ways. That is ( 4 , 1 , 2 , 3 ) or ( 1 , 4 , 2 , 3 ) or ( 1 , 2 , 4 , 3 ) . Therefore, the total number of ways to order the cards is 4 ! ( n − 4 ) ! ways. The favorable number of ways to get exactly three correct guesses is 3 ( n − 4 ) ! ways. The probability of three correct guesses is, P ( G = 3 ) = 3 ( n − 4 ) ! 4 ! ( n − 4 ) ! = 3 ( 3 + 1 ) ! Proceeding similarity to get k − 1 correct guesses the favorable number of ways of ordered them is k ( n − k ) ! ways

A First Course in Probability (10th Edition)

10th Edition

ISBN: 9780134753119

Author: Sheldon Ross

Publisher: PEARSON

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Chapter 4 Solutions

A First Course in Probability (10th Edition)

Chapter 4, Problem 4.1P Chapter 4, Problem 4.2P Chapter 4, Problem 4.3P Chapter 4, Problem 4.4P Chapter 4, Problem 4.5P Chapter 4, Problem 4.6P Chapter 4, Problem 4.7P Chapter 4, Problem 4.8P Chapter 4, Problem 4.9P Chapter 4, Problem 4.10P

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Chapter 4, Problem 4.16P

Solution Summary: The author calculates the probability of P(G=K), where G is the number of correct guesses.

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