Suppose that 16% of the CD's that are distributed are defective. If you choose 5 CD's randomly, what is the probability that at least 1 of them is defective?

prob(defect) = .16

prob(not defect) = .84

prob(at least 1 defective)
= 1 - prob(none defective)
= 1 - C(5,0) (.16)^0 (.84)^5
= 1 - .84^5
= appr.582

90

To find the probability that at least one CD is defective when choosing 5 CDs randomly, we can use the concept of complementary probability.

First, we find the probability that none of the CDs are defective. Since the probability of choosing a defective CD is 16%, the probability of choosing a non-defective CD would be 100% - 16% = 84%, or 0.84 as a decimal.

To calculate the probability that none of the CDs are defective, we multiply the probabilities of choosing a non-defective CD for each of the 5 CDs:

0.84 * 0.84 * 0.84 * 0.84 * 0.84 = 0.32768

Next, we subtract this probability from 1 to find the complementary probability of at least one CD being defective:

1 - 0.32768 = 0.67232

Therefore, the probability that at least one CD is defective when choosing 5 CDs randomly is approximately 0.67232, or 67.232%.