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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