Question 913000
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 = .032 and q = .968, n = 4
{{{nCx = (n!)/x!(n - x)!)}}} 
P(x ≤ 1) = P(0) + P(1) = (.968)^4 + 4C1 (.032)^1(.968)^2
Using TI
P(x ≤ 1) = binomcdf(4, .032, 1)