Binomial Probability
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If the probability of occurrences of an event E (success) in any single trial
is p where p lies between 0 and 1
so that the probability of not occurring of E(failure) =q=1-p
then the probability of exactly r occurrances and n-r nonoccurences of E that
is  exactly s=r successes (and f= n-r failures)
in some order in n independent trials is given by

Prob{s=r}[n,p]= n!/(r!*(n-r)!) *p^r*(1-p)^(n-r)=( n over r) *p^r*(1-p)^(n-r)

Probability of at least r occurrences ( r or more to n) is 
 sum {s=r to n)  (n over s)*p^s*(1-p)^(n-s)
p= probability of failure in a single event
r= 0 ,1,2.3...n

A sample of 10 items taken at random from infinite population
Assume 0.08  are defective in population
What is probability sample has exactly  2 defectives?
p=0.08  s=2
answer 0.1478071

What is probability of at least two defectives   
prob s>=2 = 0.1878825

what is chance of not more than 2 defectives? n=10  r=3    p=.08  s>=3  
0.0400754

what is probability of less than 2 defectives?  s<2     1- prob(s>2)
= 1-0.1878825 =  0.8121175

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TI-89
n= total number of  items    r=number of  items selected     
p= probability an item is bad

Define pp(n,r,p)=n!/(r!*(n-r)!) *p^r*q^(n-r)
 = probability of a single case

Define ss(n,r,p)=Sigma( n!/(s!*(n-s)!) *p^s*(1-p)^(n-s),s,r,n) 
= probability of sum of single cases

pp(10, 2, 0.08) =  0.14807

ss(10, 2, 0.08) = 0.18788

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On Home screen

Permutations
5 things 3 at a time
nPr(5,3) = 60 
nPr(n,p)=n!/(n-r)!

Combinations
5 things 3 at a time
nCr(5,3) = 100
nPr(n,p)=n!/(r!*(n-r)!)

Note for typical use.
If you select nPr and nCr out of Statistics menu, it uses list1 and list2 as 
arguments and list3=nPr(list1, list2) and list4 = nCr(list1, list2)

Any list number can be used as input and output lists.