a. To determine To find: The probability that vote of the 5 th county commissioner counts. Explanation Given info: There are 5 county commissioners voting for an issue, each of whom have equal probability to vote for or against. The vote of the 5 th county commissioner counts only if the other votes are split. Justification: Let X be the number of commissioners voting in favor of the issue. The total number of commissioners is ( n ) 4. The probability of voting in favor of the issue is p 0.5. Thus, the random variable X follows binomial distribution. Binomial Distribution: A random variable X is said to follow Binomial distribution if X has only two outcomes “success” and “failure” and the mass function of X is, p ( x ) = ( n x ) p x q n − x , x = 0 , 1 , 2 , ... , n , Where n is the number of trials and p is the probability of success, q = 1 − p . The mean is μ = n p and the standard deviation is σ = ( n p q ) . Thus, X has a binomial distribution. Here, number of commissioners voting for the issue is 4. Thus, number of trials, n = 4 . As each commissioner has equal probability of voting for and against, thus, probability of success is p = 0.5 . Thus, mass function of X is: p ( x ) = ( 4 x ) ( 0 b. To determine To find: The probability that vote of the 5 th county commissioner counts, under the changed scenario.

13th Edition

ISBN: 9780134080215

Author: James T. McClave, Terry T Sincich

Publisher: PEARSON

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Chapter 4.4, Problem 77ACI

To determine

To find: The probability that vote of the 5^th county commissioner counts.

To determine

To find: The probability that vote of the 5^th county commissioner counts, under the changed scenario.

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Chapter 4.4, Problem 76ACI

Chapter 4.4, Problem 78ACI

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The probability that vote of the 5 th county commissioner counts.