Let p be 0.831 denote the percentage of defective welds and q be 0.169 denote the percentage of non-defective welds.
Using the binomial distribution, we want all three to be defective.
[tex]P(X = 3) = \binom{3}{3}(p)^{3}{q}^{3 - 3}[/tex]
[tex]P(X = 3) = \binom{3}{3}(0.831)^{3}(1)[/tex]
[tex]P(X = 3) = (0.831)^{3} \approx 0.573856191[/tex]
