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sum from 0 to N of (N choose k)*(P^k)(1-P)^k

 Jul 10, 2015

Best Answer 

 #4
avatar+28125 
+5

heureka has given the result for p equals any value, not necessarily between 0 and 1.  The sum isn't 1 when 0<p<1

The probabilistic expression (which is probably what was meant) is:

 

$$\sum_{k=0}^N(\frac{N}{k})p^{N-k}(1-p)^k$$

.

 Jul 25, 2015
 #1
avatar+102763 
+5

that would  be 1

P is of course between 0 and 1 since it represents a probability.

 Jul 10, 2015
 #2
avatar+22881 
+5

$$\sum \limits_{k=0}^N {N \choose k} \cdot (P^k)(1-P)^k
=[1+p\cdot(1-p)]^N$$

 

 Jul 25, 2015
 #3
avatar+102763 
0

I do not understand your answer Heureka :/

 Jul 25, 2015
 #4
avatar+28125 
+5
Best Answer

heureka has given the result for p equals any value, not necessarily between 0 and 1.  The sum isn't 1 when 0<p<1

The probabilistic expression (which is probably what was meant) is:

 

$$\sum_{k=0}^N(\frac{N}{k})p^{N-k}(1-p)^k$$

.

Alan Jul 25, 2015
 #5
avatar+102763 
0

Ok Alan, thank you :)

 Jul 25, 2015

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