A dealership sells an average of 0.5 cars per day. The company needs to pay a debt that matures in two days. Must sell at least three cars in two days, what is the likelihood that the company can pay the debt?
Well the company must sell at LEAST 3 cars. Not exactly 3 cars. The probability that they sell exactly three cars is 0.0613. How would you go about finding the probability that they sell at LEAST 3 cars; we are still using Poisson here, but you're going to have to take the compliment of something.
Well if they can sell more than 3 cars, then theoretically they can sell 100 cars or 1000 cars or "n" number of cars. So to find out the Poisson Distribution of "at least 3 cars. . ." you would have to calculate Poisson(3)+Poisson(4)+Poisson(5)+. . .Poisson(n). Of course this isn't possible (or how you're supposed to do this). Thus you need another alternative.
As I said, you can use the idea of the compliment of an event. If they don't sell at LEAST 3 cars in two days, then how many cars at MOST did they sell. Find the Poisson Distribution of the AT MOST statement, and see if something clicks.
Apprentice, you know the Poisson formula. Add them up from 0 to 2 and subtract from 1.
They say at least 3 must be sold. That can be 3 up to however many.
But, since there is no given bound, find the probabilties from 0 to 2 and subtract from 1.
$\displaystyle 1-\sum_{k=0}^{2}\frac{e^{-1}}{k!}$
Because $\displaystyle {\lambda}=.5(2)=1$