Answer:
P(X=2) = 0.302
Step-by-step explanation:
With the conditions mentioned in the question, we can model this variable as a binomial random variable, with parameters n=9 and p=0.2.
The probability of having k defective items in the sample of nine chips is:
[tex]P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{9}{k} 0.2^{k} 0.8^{9-k}\\\\\\[/tex]
Then, the probability of having 2 defective chips in the sample is:
[tex]P(x=2) = \dbinom{9}{2} p^{2}(1-p)^{7}=36*0.04*0.2097=0.302\\\\\\[/tex]
