(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...) |
m (4.2a Emir Kavurmacioglu moved to 4.1a Emir Kavurmacioglu: Improperly named originally) |
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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))] | ||
| − | + | 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. | ||
Latest revision as of 06:40, 15 October 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.
