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?

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.