(New page: Start out by replacing the value of Y by N-X. You then get P(N-X) = (N N-X) * [P^(N-X)] * [(1-P)^(N-(N-X))] = (N N-X) * [(1-P)^X] * [P^(N-X) Then just expand the com...)
 
Line 1: Line 1:
 
Start out by replacing the value of Y by N-X.
 
Start out by replacing the value of Y by N-X.
  
You then get P(N-X) = (N N-X) * [P^(N-X)] * [(1-P)^(N-(N-X))]
+
You then get  
 +
            P(N-X) = (N N-X) * [P^(N-X)] * [(1-P)^(N-(N-X))]
  
                    = (N N-X) * [(1-P)^X] * [P^(N-X)
+
            P(N-X) = (N N-X) * [(1-P)^X] * [P^(N-X)
  
 
Then just expand the combination and prove that it is equal to (N X).
 
Then just expand the combination and prove that it is equal to (N X).
  
 
The last step is to define a new variable P' = 1-P, which is the probability parameter for Y.
 
The last step is to define a new variable P' = 1-P, which is the probability parameter for Y.

Revision as of 15:01, 23 September 2008

Start out by replacing the value of Y by N-X.

You then get

            P(N-X) = (N N-X) * [P^(N-X)] * [(1-P)^(N-(N-X))]
            P(N-X) = (N N-X) * [(1-P)^X] * [P^(N-X)

Then just expand the combination and prove that it is equal to (N X).

The last step is to define a new variable P' = 1-P, which is the probability parameter for Y.

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett