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Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 2 - Problem 110e
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 2 - Problem 110e

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# A particular airline has 10 A.M. flights from Chicago to ISBN: 9780321629111 32

## Solution for problem 110E Chapter 2

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

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Problem 110E

A particular airline has 10 A.M. flights from Chicago to New York, Atlanta, and Los Angeles. Let ?A ?denote the event that the New York flight is full and define events ?B ?and ?C ?analogously for the other two flights. Suppose P(A) = .6, P(B) = .5, P(C) = .4 and the three events are independent. What is the probability that a. All three flights are full? That at least one flight is not full? b. Only the New York flight is full? That exactly one of the three flights is full?

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Problem 110E Answer: Step1: We have A particular airline has 10 A.M. flights from Chicago to New York, Atlanta, and Los Angeles. Let A denote the event that the New York flight is full and define events B and C analogously for the other two flights. Suppose P(A) = 0.6, P(B) = 0.5, P(C) = 0.4 and the three events are independent. We need to find the probability that, a. All three flights are full That at least one flight is not full b. Only the New York flight is full That exactly one of the three flights is full Step2: a). P(All three flights are full) = P(ABC) = P(A)×P(B)×P(C) = 0.6×0.5×0.4 = 0.12 Therefore,The probability that the all three flights are full is 0.12. P(at least one flight is not full) = 1 - P(All three flights are full) = 1 - 0.12 = 0.88 Therefore,At least one flight is not full is 0.88. Step3: b). P( nly the New York flight is ful = P(A)×P(B )×P(C ) = P(A)×[1-P(B)]×[1-P(C)] = 0.6×[1 - 0.5]×[1 - 0.4] = 0.6×0.5×0.6 = 0.18 Therefore,Only the New York flight is full is 0.18. P(exactly one of the three flights is full) = P(AB C ) + P(A BC ) + P(A B C) 1 1 = {P(A)×[1 - P(B)]×[1 - P(C)]}+{[1 - P(A)]×P(B)×[1 - P(C)]}+{[1 - P(A)]×[1 - P(B)]× P(C)} = {0.6×[1 - 0.5]×[1 - .4]} + {[1 - 0.6]×0.5×[1 - 0.4]} + {[1 - 0.6]×[1 - 0.5]×0.4} = {0.6×0.5×0.6} + {0.4×0.5×0.6} + {0.4×0.5×0.4} = 0.18 + 0.12 + 0.08 = 0.38 Therefore,The probability that exactly one of the three flights is full is 0.38.

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##### ISBN: 9780321629111

Since the solution to 110E from 2 chapter was answered, more than 470 students have viewed the full step-by-step answer. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. The answer to “A particular airline has 10 A.M. flights from Chicago to New York, Atlanta, and Los Angeles. Let ?A ?denote the event that the New York flight is full and define events ?B ?and ?C ?analogously for the other two flights. Suppose P(A) = .6, P(B) = .5, P(C) = .4 and the three events are independent. What is the probability that a. All three flights are full? That at least one flight is not full? b. Only the New York flight is full? That exactly one of the three flights is full?” is broken down into a number of easy to follow steps, and 92 words. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. This full solution covers the following key subjects: full, flights, new, Flight, York. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. The full step-by-step solution to problem: 110E from chapter: 2 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM.

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