A subset of the binomial distribution... Message #6 Posted by Karl Schneider on 5 Oct 2004, 11:52 p.m., in response to message #1 by Jim
The answers that others have given are correct; this is a simple problem. A tougher question is, what are the chances of getting, say, 8 out of 10 right?
The binomial distribution is as follows:
P(n,k,p) = C(n,k)* pk * (1-p)(n-k)
P is the overall probability
n is the number of trials
k is the specified number of successes
p is the probability of success of each trial
C(n,k) is the statistical combination function
Now, if there are n successes in n trials (get 'em all correct), this reduces to
P(n,n,p) = pn
If you have a 65% chance of picking each of 10 teams correctly, your chances of getting 8 correct are:
P(0.65, 10, 8) = C(10,8) * 0.658 * (1-0.65)(10-8) = 0.175652953
-- KS
|