Question 1160050
This is a Poisson with prob proportional to time, count data, and theoretically infinite number.
lambda=4.06
a. e^(-4.06)*(4.06)^)/0!=0.0172
b. At least 1 is 1-P(0)=0.9828
c. At least 2 mishandled bags is 1-p(0) -p(1), where p(1)=e^(-4.06)*4.06=0.0700
This is 1-0.0172-0.0700=0.9128