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4. In a batch of 100 CDs, 6 are defective. A sample of three CDs is to be selected at random. What is the probability that two of the three CDs will be defective?​

User Svartalf
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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%.

User Tamura
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