Explanation Given: 10 % of the raccoons are carriers of rabies. Approach: For a binomial experiment with two outcomes with outcome of success in each trial equal to p . The probability of r successes in n trials is equal to P ( r successes in n trials) = C ( n , r ) p r ( 1 − p ) n − r (1) Calculation: Interpret this problem as a binomial experiment with two outcomes: a dog is bitten by a raccoon carrier or not bitten by a raccoon carrier. “Success” is the event that dog is bitten by a raccoon who is a rabies carrier. Probability of success, p = 0.1 Number of trials, n = 4 Problem asks probability of being bitten by at least one rabies carrier. This gives: r ≥ 1 Substitute: n , p , r in equation (1) to get P ( r ≥ 1 ) = 1 − P ( r = 0

4th Edition

ISBN: 9781305719781

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: PEARSON

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Chapter 14.3, Problem 24E

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The probability that dog was bitten by at least one rabies carrier.

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Chapter 14.3, Problem 23E

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Chapter 14 Solutions

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Ch. 14.1 - CONCEPTS The Fundamental Counting Principle says...Ch. 14.1 - CONCEPTS The number of ways of arranging r objects...Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 9E Ch. 14.1 - Prob. 10E

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The probability that dog was bitten by at least one rabies carrier.

Solution Summary: The author explains that the probability of being bitten by at least one rabies carrier is equal to 0.3439.

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Chapter 14 Solutions

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The probability that dog was bitten by at least one rabies carrier.