Question 1181130: So the question I had is this
If the pizza baker scatters n pepperoni slices at random before cutting the pie, what's the probability that the pizza can be divided in half through its diameter, with all the pepperoni slices on one of the halves?
My thinking is that the answer is 1/2, but that cant be it? It seems too easy? Please help. Thank you.
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i'm not exactly sure, but i think this is what happens.
the probability that one of the pepperoni slices ends up on one side of the half of the pizza would be 1/2.
the probability that 2 slices end up on one side of the half of the pizza is 1/2 * 1/2 = 1/4.
this assumes, of course, that all of the pepperoni slices end up in the pizza and not on the table or the floor.
the probability that all of the pepperoni slices end up on one half of the pizza is therefore (1/2) ^ n, where n is the number of pepperoni slices thrown on the pizza.
the more pepperoni slices you randomly throw on the pizza, the lower the probability that they will all end up on one half of the pizza.
this is similar to flipping a coin where the proability of getting heads is 1/2.
the probability with one toss all heads is 1/2.
the probability with two tosses all heads is 1/2 * 1/2 = 1/4
the probability with n tosses all heads is (1/2) ^ n
bottom line:
it's definitely not 1/2.
that's only if you toss one slice of pepperoni randomly on the pizza.
this is really a moot point, because the pizza person always aims the slices so that they are, more or less, evenly spaced on the pizza.
if you order a pizza with half pepperoni and half plain, the pizza person will aim the pepperoni slices to one half of the pizza only, so random doesn't apply.
Answer by ikleyn(52887)  (Show Source):
You can put this solution on YOUR website! .
As formulated in the post, the problem is posed INCORRECTLY.
The solution by @Theo is NOT adequate to the problem, as it is posed in the post.
The solution by @Theo is the solution to another problem.
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