Question 633353
<pre>
n=4, p= 0.05

 x   p(x) = C(5,x)(.05)<sup>x</sup>(.95)<sup>5-x</sup>
-------------------------------------------------------
 0   p(0) = C(5,0)(.05)<sup>0</sup>(.95)<sup>5-0</sup> = .774
 1   p(1) = C(5,1)(.05)<sup>1</sup>(.95)<sup>5-1</sup> = .204
 2   p(2) = C(5,2)(.05)<sup>2</sup>(.95)<sup>5-2</sup> = .021
 3   p(3) = C(5,3)(.05)<sup>3</sup>(.95)<sup>5-3</sup> = .001
 4   p(4) = C(5,4)(.05)<sup>4</sup>(.95)<sup>5-4</sup> = .000  
 5   p(5) = C(5,5)(.05)<sup>5</sup>(.95)<sup>5-5</sup> = .000

Shorten the table to just two columns:

 x   p(x) 
---------
 0   .774
 1   .204
 2   .021
 3   .001
 4   .000  
 5   .000
---------
sum=1.000

Edwin</pre>