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Suppose identical tags are placed on both the left ear and

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 89E Chapter 2

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

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

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

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Problem 89E

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events C1={left ear tag is lost}and C2={left ear tag is lost}. Let ? = P(C1)= P(C2), and assume ?C?1 and ?C?2 are independent events. Derive an expression (involving _) for the probability that exactly one tag is lost, given that at most one is lost (“Ear Tag Loss in Red Foxes,” ?J. Wildlife Mgmt., ?1976: 164–167). [?Hint: ?Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

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Problem 89E Answer: Step1: We have Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events C1={left ear tag is lost} and C2={left ear tag is lost}. Let = P(C1)= P(C2), and assume C1 and C2 are independent events. Our goal is to Derive an expression (involving ) for the probability that exactly one tag is lost, given that at most one is lost. Step2: 1).Let L be the event “left ear tag is lost ” and Let R be the event “right ear is tag lost” Then, P(L) = P(R) = 2).Let L be the event “left ear tag is not lost” and 1 Let R be the event “right ear is not lost” Then, 1 1 P(L ) = P(R ) = 1 - 3).The event that only one is lost is “LR or L R” 1 1 1 1 1 P(L or L R) = P(L)P(R ) + P(L )P(R) = (1 - ) + (1 - ) + 2 (1 - ) 4).The probability that at most 1 is lost = (1 - )×(1 - ) + 2 (1 - ) = (1 - ) + 2 (1 - ) 5).The probability that exactly one is lost given that at most one is lost = 2(1 ) (1 )(1 ) + 2(1 ) = 2(1 2) 1

Step 2 of 3

Chapter 2, Problem 89E is Solved
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Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

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Suppose identical tags are placed on both the left ear and