Question

5. The probability that a defect will occur over the surface of a semiconductor chip is 0.2. Assuming the occurrences of defects are independent, what is the probability that two out of nine chips selected with replacement will be defective

Answer 1

P(X=2) = 0.302

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:

`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)\\\\\\`

Then, the probability of having 2 defective chips in the sample is:

`P(x=2) = \dbinom{9}{2} p^(2)(1-p)^(7)=36*0.04*0.2097=0.302\\\\\\`