3 bottles per case are underfilled on average. What is the chance that at least 4 underfilled bottles will be contained in a random case?
Using the formula: $1-\left( \dfrac{3^0 e^{-3} }{ 0!} + \dfrac{3^1 e^{-3}}{1!} + \dfrac{3^2 e^{-3}}{2!} + \dfrac{3^3 e^{-3}}{3!} \right)$, I obtained $0.3528$
However, my answer was marked as incorrect. How do you solve this correctly?
My guess is you were told something about the number of bottles in each case and you should have used the binomial distribution rather than the Poisson distribution.