MathRevolution wrote:

In a certain factory, the production line which produces the bags has the probability that a bag selected at random is defective is 0.02. If 6 bags are selected at random, what is the probability that at least one bag is defective?

A. \(0.02^6\)

B. \((0.02)(0.98)^6\)

C. \(1-(0.02)^6\)

D. \(0.98^6\)

E. \(1-(0.98)^6\)

We are given that the probability that a selected bag is defective is 0.02, so we can determine that the probability of a non-defective bag is 1 - 0.02 = 0.98. We need to determine the probability that when 6 bags are selected at least one bag is defective.

We know the following:

1 = P(at least one defective bag is selected) + P(no defective bags are selected)

Thus,

P(at least one defective bag is selected) = 1 - P(no defective bags are selected)

Therefore, we need to determine P(no defective bags are among the 6 selected bags).

P(no defective bags are among the 6 selected bags) = 0.98^6

Thus, P(at least one defective bag is among the 6 selected bags) = 1 - 0.98^6

Answer: E

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