The table lists recommended amounts of food to order for 19 party guests. Nathan and Sydney are hosting a graduation party for 40 guests. They know there will also be guests stopping by who may have come from other parties. For ordering purposes, they will count each of these "drop-in" guests as half a guest. How much of each food item should Nathan and Sydney order for a graduation party with 5 drop-in guests? Fried Chicken: 26 pieces

Deli Meats: 5 2/3 pounds

Lasagna: 10 1/4 pounds

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the answer   is 200

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this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

the fist one is not equalvalent but i think th second one isand the third isnt definetly or the last one is i think. hope this !

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