The U.S. Department of transportation maintains statistics for mishandled bags. Assume​ that, in one​ year, an airline had 0.74 mishandled bags per month. Complete parts​ (a) through​ (c) below. Question content area bottom Part 1 Bold a. nbsp What is the probability that in the next​ month, the airline will have no mishandled​ bags? While a table of Poisson​ probabilities, technology, or the equation shown below can be used to find the probability that Xequalsx events in an area of opportunity given lambda​, where lambda is the expected number of​ events, e is the mathematical​ constant, and x is the number of events ​(xequals​0, ​1, 2,...), use technology. ​P(Xequalsx ​| lambda​)equalsStartFraction e Superscript negative lambda Baseline lambda Superscript x Over x exclamation mark EndFraction Part 2 The probability that in the next​ month, there will be no mishandled bags is the same as finding Upper P left parenthesis Upper X equals 0 right parenthesis for lambda equals 0.74 . Use technology to find Upper P left parenthesis Upper X equals 0 right parenthesis​, rounding to four decimal places. ​P(Xequals​0)equals0.4771 Part 3 ​Therefore, the probability that there will be no mishandled bags in the next month is approximately 0.4771 . Part 4 Bold b. nbsp What is the probability that in the next​ month, the airline will have at least one mishandled​ bag? The probability that in the next​ month, there will be at least one mishandled bag is the same as finding Upper P left parenthesis Upper X greater than or equals 1 right parenthesis. Part 5 To find the probability that Xgreater than or equals​1, remember that the Poisson distribution is a discrete probability distribution and the sum of the probabilities for every value of X must always equal 1. First complete the equation below. Upper P left parenthesis Upper X greater than or equals 1 right parenthesis equals 1 minus Upper P left parenthesis Upper X less than 1 right parenthesis equals 1 minus Upper P left parenthesis Upper X equals 0 right parenthesis Part 6 Substitute the value found in part​ (a) for Upper P left parenthesis Upper X equals 0 right parenthesis. Upper P left parenthesis Upper X greater than or equals 1 right parenthesis equals 1minus​P(Xequals​0) equals 1minus0.4771 equals 0.5229 Part 7 ​Therefore, the probability that at least one mishandled bag in the next month is approximately 0.5229. Part 8 Bold c. nbsp What is the probability that in the next​ month, the airline will have at least two mishandled​ bags? The probability that in the next​ month, there will be at least two mishandled bags is the same as finding Upper P left parenthesis Upper X greater than or equals 2 right parenthesis. Part 9 Use the fact that the sum of the probabilities for every value of X must always equal 1 to restate the probability. Upper P left parenthesis Upper X greater than or equals 2 right parenthesis equals 1 minus Upper P left parenthesis Upper X less than 2 right parenthesis equals 1 minus left bracket Upper P left parenthesis Upper X equals 0 right parenthesis plus Upper P left parenthesis Upper X equals 1 right parenthesis right bracket Part 10 Use technology to find Upper P left parenthesis Upper X equals 1 right parenthesis​, rounding to four decimal places. Upper P left parenthesis Upper X equals 1 right parenthesisequals0.3531 Part 11 Substitute ​P(Xequals​1) and the value found in part​ (a) for Upper P left parenthesis Upper X equals 0 right parenthesis. Upper P left parenthesis Upper X greater than or equals 2 right parenthesis equals 1 minus left bracket Upper P left parenthesis Upper X equals 0 right parenthesis plus Upper P left parenthesis Upper X equals 1 right parenthesis right bracket equals 1minus​[0.4771plus0.3531​] equals 0.1698 (explain part c)