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`