Explanation Given Information: The Agriculture Department of U.S. in the plan that there are many tests that can detect the mad cow disease, and plan to test approximately 243 , 000 cows per year for the disease in March, 2004 and the disease is present only in 5 cows out of 45 million. Formula Used: If the outcomes of an event are equally likely to occur, then the probability of an event E, is given by P ( E ) = n ( E ) n ( S ) Where n ( E ) is the number of favorable outcomes and n ( S ) is the total number of outcomes. If n is the number of trials and X is the number of successes, then the probability for particular value of X is given by P ( X = x ) = C ( n , x ) p x q n − x : Where p is the probability of success and q is the probability of failure. The unordered set of r items chosen from n taken r at a time is given by rule of combination C ( n , r ) = n ! r ! ( n − r ) ! Calculation: Consider the total number of cows, n ( S ) = 45 , 000 , 000 Evaluate the number of favorable outcomes, n ( E ) = 5 Evaluate the probability that the cow selected had the mad cow disease. As, P ( E ) = n ( E ) n ( S ) Substitute n ( E ) = 5 and n ( S ) = 45 , 000 , 000 in P ( E ) = n ( E ) n ( S ) : P ( E ) = 5 45 , 000 , 000 P ( E ) = 1.1111 × 10 − 7 Consider the total number of cows tested, n = 243 , 000 Assume cows with positive result to be success. The probability of success is p = 1

Finite Mathematics and Applied Calculus (MindTap Course List)

7th Edition

ISBN: 9781337274203

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Estimation

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Chapter 9.2, Problem 45E

To determine

To calculate: The probability that at least 1 cow out of 45 million cows would test positive in one year for the mad cow disease, if the disease is present only in 5 cows. It was stated by the Agriculture Department of U.S. in the plan that there are many tests that can detect the mad cow disease, and plan to test approximately 243,000 cows per year for the disease in March, 2004.

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Chapter 9.2, Problem 44E

Chapter 9.2, Problem 46E

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Prob. 15E Ch. 9.5 - Prob. 16E Ch. 9.5 - Prob. 17E Ch. 9.5 - Prob. 18E Ch. 9.5 - Prob. 19E Ch. 9.5 - Prob. 20E Ch. 9.5 - Prob. 21E Ch. 9.5 - Prob. 22E Ch. 9.5 - Prob. 23E Ch. 9.5 - Prob. 24E Ch. 9.5 - Prob. 25E Ch. 9.5 - Prob. 26E Ch. 9.5 - SAT Scores SAT test scores are normally...Ch. 9.5 - Prob. 28E Ch. 9.5 - Prob. 29E Ch. 9.5 - Prob. 30E Ch. 9.5 - IQ Scores IQ scores (as measured by the...Ch. 9.5 - Prob. 32E Ch. 9.5 - Prob. 33E Ch. 9.5 - Prob. 34E Ch. 9.5 - Prob. 35E Ch. 9.5 - Prob. 36E Ch. 9.5 - Baseball The mean batting average in major league...Ch. 9.5 - Prob. 38E Ch. 9.5 - Marketing Your pickle company rates its pickles on...Ch. 9.5 - Prob. 40E Ch. 9.5 - Prob. 41E Ch. 9.5 - Prob. 42E Ch. 9.5 - Prob. 43E Ch. 9.5 - Prob. 44E Ch. 9.5 - Prob. 45E Ch. 9.5 - Prob. 46E Ch. 9.5 - Prob. 47E Ch. 9.5 - Prob. 48E Ch. 9.5 - Prob. 49E Ch. 9.5 - Prob. 50E Ch. 9.5 - Prob. 51E Ch. 9.5 - Prob. 52E Ch. 9.5 - Prob. 53E Ch. 9.5 - Prob. 54E Ch. 9.5 - Prob. 55E Ch. 9.5 - Prob. 56E Ch. 9.5 - Prob. 57E Ch. 9.5 - 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Solution Summary: The author calculates the probability that at least 1 cow out of 45 million cows would test positive in one year for the mad cow disease.

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