#1**+5 **

that would be 1

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

Melody Jul 10, 2015

#2**+5 **

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

=[1+p\cdot(1-p)]^N$$

heureka Jul 25, 2015