Question

On average 83.1% of welds performed by a particular welder are defective. if, in one day, the welder creates three welds. what is the probability all of the samples are defective?

Answer 1

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.
$P(X = 3) = \binom{3}{3}(p)^{3}{q}^{3 - 3}$
$P(X = 3) = \binom{3}{3}(0.831)^{3}(1)$
$P(X = 3) = (0.831)^{3} \approx 0.573856191$