BinomialDistribution
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The Binomial distribution can be described as the sum of a specific number of independent trials, each of which results in either a zero or a one with constant probability. It provides a reasonable description of coin tossing experiments, among others.
To get X heads in a sequence of N tosses, several things have to happen. If the probability of a head on a single trial is p and the probability of a tail is q=(1-p), then each sequence with X heads and N-X tails has a probability calculated by finding the product of multiplying p times itself X times (p^X), and q times itself (N-X) times (q^(N-X)), that is (p^X q^(N-X)). However, there are many sequences which match this description. By the methods of CombinaTorics, we can find that there are N!/(X!*(N-X)!) different combinations with X heads and N-X tails. So, the probability of X heads in N trials is
X (N-X)
N! p q
X!* (N-X)!
DickBeldin----
If you go to previous versions and look at the first one, 02/15/2001, which is yours?, you will see :
1). q (1-p), maybe a typo?
2). And the formula for the numbers of ways of picking X items out of N items was:
N!/X!/(N-X)!. This is plain wrong. Yes, after requesting a change for a week, I changed it.
3).There were also wording problems. RoseParks.
I see now the problem. (1-p) was intended as a parenthetical definition. I guess N1/X!/(N-X)! worked in my programming codes so I couldn't see the ambiguity. How would you calculate N!/X!/(N-X)!? From right to left? On the other hand, Today is 02/20/2001, so I think your "requesting a change for a week" is a bit off. Today is only the 20th by my calendar. In any case, the criticism has led to something better.
DickBeldin----
In answer to your question on how you evaluate, I guess, X!/N!/(N-X)!, this is ambiguous. In any easy example.
2/4/12 is ambiguous since
- (2/4)/12= 2/48=1/24 while
- 2/(4/12)= 24/4= 6.
Multiplication is associative over the reals. If you look at division as the inverse operation of multplication, i.e. 2/4/12=2*4^1*12^1=1/24 you are okay. If you look at division in the ordinary sense, you must specify the order of operations.RoseParks