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26 May 2017, 04:44
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In a certain factory, the production line that produces the toys has the probability of 0.001 that a randomly selected toy is damaged. If 100 toys are selected at random, what is the probability that at least one toy is damaged?
A. \(0.001^{100}\)
B. \((0.001)(0.999)^{100}\)
C. \(1-(0.001)^{100}\)
D. \(0.999^{100}\)
E. \(1-(0.999)^{100}\)
A. \(0.001^{100}\)
B. \((0.001)(0.999)^{100}\)
C. \(1-(0.001)^{100}\)
D. \(0.999^{100}\)
E. \(1-(0.999)^{100}\)
26 May 2017, 04:46
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Official Solution:
In a certain factory, the production line that produces the toys has the probability of 0.001 that a randomly selected toy is damaged. If 100 toys are selected at random, what is the probability that at least one toy is damaged?
A. \(0.001^{100}\)
B. \((0.001)(0.999)^{100}\)
C. \(1-(0.001)^{100}\)
D. \(0.999^{100}\)
E. \(1-(0.999)^{100}\)
When randomly selecting a toy, the probability of selecting a damaged toy is 0.0001. Thus, when selecting 100 toys, the probability of selecting at least one damaged toy is equal to the probability that 1-100 selected toys are not damaged. Hence, \(1-(1-0.001)(1-0.001)\ldots \ldots \ldots ..(1-0.001) = 1 - (0.999)^{100}\). Therefore, the answer is E.
Answer: E
In a certain factory, the production line that produces the toys has the probability of 0.001 that a randomly selected toy is damaged. If 100 toys are selected at random, what is the probability that at least one toy is damaged?
A. \(0.001^{100}\)
B. \((0.001)(0.999)^{100}\)
C. \(1-(0.001)^{100}\)
D. \(0.999^{100}\)
E. \(1-(0.999)^{100}\)
When randomly selecting a toy, the probability of selecting a damaged toy is 0.0001. Thus, when selecting 100 toys, the probability of selecting at least one damaged toy is equal to the probability that 1-100 selected toys are not damaged. Hence, \(1-(1-0.001)(1-0.001)\ldots \ldots \ldots ..(1-0.001) = 1 - (0.999)^{100}\). Therefore, the answer is E.
Answer: E
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