Question 879528
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 = 3/5 and q = 2/5,  n = 10
{{{nCx = (n!)/x!(n - x)!)}}} 
P(x = 9) = 10C9(3/5)^9(2/5)^1  = 10(3/5)^9(2/5)^1 = .0403 0r 4.03% chance