Given that a manufacturing machine has a 8% defect rate, thus p = 0.08.
The probability of a binomial distribution of n trials with p as the probability of success is given by:
[tex]P(x)={ ^nC_x}p^r(1-p)^{n-x}[/tex]
If 4 items are chosen at random, the probability that at least one will have a defect is given by the probablity that it is not the case that no one will have defect.
Thus,
[tex]P(x\geq1)=1-P(0) \\ \\ =1-{ ^4C_0}(0.08)^0(1-0.08)^{4-0} \\ \\ =1-1\times1\times0.92^4=1-0.7164 \\ \\ =0.2836.[/tex]
