Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
1334
5
avatar

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

 Jul 10, 2015

Best Answer 

 #4
avatar+33654 
+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:

 

Nk=0(Nk)pNk(1p)k

.

 Jul 25, 2015
 #1
avatar+118696 
+5

that would  be 1

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

 Jul 10, 2015
 #2
avatar+26396 
+5

Nk=0(Nk)(Pk)(1P)k=[1+p(1p)]N

 

.
 Jul 25, 2015
 #3
avatar+118696 
0

I do not understand your answer Heureka :/

 Jul 25, 2015
 #4
avatar+33654 
+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:

 

Nk=0(Nk)pNk(1p)k

.

Alan Jul 25, 2015
 #5
avatar+118696 
0

Ok Alan, thank you :)

 Jul 25, 2015

0 Online Users