Question 865884
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Hi
p(def) = .14,  n = 100
P(x = 11) = binompdf(100, .14, 11) = .0849  0r  8.49%
p(def) = .14,  n = 15
P(x = 4) = binompdf(15, .14, 4) = .0998   0r 9.98%
Or for singular probability:
Using {{{P (x)= highlight_green(nCx)(p^x)(q)^(n-x) }}} 
p and q are the probabilities of success and failure respectively. 
In this case p= .14 and q = .86
{{{nCx = (n!)/x!(n - x)!)}}}