+0

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

0
627
5

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

Jul 10, 2015

#4
+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
+102763
+5

that would  be 1

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

Jul 10, 2015
#2
+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
+102763
0

Jul 25, 2015
#4
+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:

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

.

Alan Jul 25, 2015
#5
+102763
0

Ok Alan, thank you :)

Jul 25, 2015