A prize wheel has 12 prizes listed. The wheel will be spun a total of
72 times. What's the probability of each slot coming up more than 6
times (72/12=6)?
How many of each prize should be on hand to give away?

Hello pann!
We can easily solve this problem thinking of it as involving a
binomial distribution.

The binomial distribution answers the follwing question. Given an
experiment that is repeated 'n' times, and an event that has a
probability 'p' of happening in each of the trials, what is the
probability that the event will happen 'x' times after the 'n' trials
are finished?

In order to apply this to your problem, let's assume we're interested
only in finding what's the probability of a single prize (let's say,
prize number 1) coming up more than 6 times. In your problem, the
experiment is repeated n=72 times. The probability of prize number 1
coming up is p=1/12. The probability function for a binomial
distribution is the follwing:

P(x) =       n!
       -------------- * p^x * (1-p)^(n-x)
         x! (n-x)!

Binomial Distribution -- MathWorld
http://mathworld.wolfram.com/BinomialDistribution.html

In your problem, you're interested in the probability of x being
greater than 6. That is you want to know:

P(x>6) = P(7) + P(8) + P(9) + ...

which is an infite sum. Fortunately, we can still calculate it in this way:

P(x>6) = 1 - ( P(6)+P(5)+P(4)+P(3)+P(2)+P(1)+P(0) )

which involves calculating only 7 terms using the previous equation.
This can be calculated by hand, or you can use the binomial
probability calculator at the following link:

Binomial Calculator
http://www.stat.sc.edu/~west/applets/binomialdemo.html

If you use the calculator, fill n=72, p=0.0833 (which is 1/12) and
choose "Prob. X is at least 7". You will find that this value is
0.3932.

So we've found that the probability of prize number 1 coming up more
than 6 times is 0.3932, or 39.32%. But clearly, prize number one is
equivalent to any of the other 11 prizes. Therefore, we conlcude that
the probability of any prize coming up more than 6 times is 0.3932.

The second question, how many of each prize should be on hand to give
away, it's a subjective question. If you want to have a zero
probability of running out of prizes, you should keep 72 of each, as
it's possible (although extremely unlikely) that the same prize will
come up 72 times. So it much depends on the probability of running out
of prizes you want to set. For example, we've just found that the
probability of a any prize coming up more than 6 times is 0.3932. This
means that if you keep 6 of each prize, the probability of running out
of a prize is 0.3932. More prizes would clearly reduce this
probability, but would also increase the cost of buying prizes, so the
answer to this question involves a trade-off for which there is no
clear answer without further information.

If what youe meant with the second question is to find the expected
value of the number of times a prize can come up, then it can again be
answered using the binomial distribution. The expected value of a
variable following this distribution is n*p. In your case, it is
72*(1/12) = 6.


Google search terms
"binomial distribution"
://www.google.com/search?hl=en&q=%22binomial+distribution%22
"binomial distribution" calculator
://www.google.com/search?hl=en&lr=&q=%22binomial+distribution%22+calculator


I hope this helps! If you have any further questions regarding my
answer, please don't hesitate to request clarification.

Best wishes!
elmarto

homework questions are deleted.

Just a correction - not very important.

Elmarto wrote 
"In your problem, you're interested in the probability of x being
greater than 6. That is you want to know:
P(x>6) = P(7) + P(8) + P(9) + ... which is an infite sum."

He has proceeded correctly on calculating the probability, but he's
not 100% right on saying that it's an infinite, since the probability
of a single prize coming up more than 72 times is 0 (because the wheel
will be spun only 72 times),
hence P(73) + P(74) + P(75) + ... = 0, so you could convert it into
the finite sum P(x>6) = P(7) + P(8) + ... + P(71) + P(72),
but you'll get the same result with a bit more difficult calculation.

Anyway, I thought it might interest you.