Answer: The probability that two of the three CDs will be defective is approximately 0.01044 or 1.044%.
Step-by-step explanation: We can use the binomial distribution to solve this problem. Let X be the number of defective CDs in a sample of three. Then X follows a binomial distribution with parameters n = 3 and p = 6/100, where n is the sample size and p is the probability of a CD being defective.
The probability of getting exactly two defective CDs in a sample of three can be calculated using the binomial probability formula:
P(X = 2) = (3 choose 2) * (6/100)^2 * (94/100)^1
where (3 choose 2) = 3 is the number of ways to choose 2 defective CDs out of 3.
Simplifying this expression, we get:
P(X = 2) = 3 * (6/100)^2 * (94/100)
P(X = 2) = 0.01044
Therefore, the probability that two of the three CDs will be defective is approximately 0.01044 or 1.044%.