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27 Oct 2024, 07:56
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
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Question Stats:
77% (01:04) correct
23%
(01:26)
wrong
based on 527
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When a certain manufacturer produces a unit of its product, the probability that the unit will be defective is 0.01. If the manufacturer is to produce 10 units of its product, what is the probability that at least 1 of the 10 units produced will be defective?
A. \((0.01)^{10}\)
B. \((0.01)^9(0.99)\)
C. \((0.01)(0.99)^9\)
D. \(1-(0.01)^{10}\)
E. \(1-(0.99)^{10}\)
GMAT-Club-Forum-5x64e1hi.png [ 43.16 KiB | Viewed 2849 times ]
A. \((0.01)^{10}\)
B. \((0.01)^9(0.99)\)
C. \((0.01)(0.99)^9\)
D. \(1-(0.01)^{10}\)
E. \(1-(0.99)^{10}\)
Attachment:
GMAT-Club-Forum-5x64e1hi.png [ 43.16 KiB | Viewed 2849 times ]
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29 Oct 2024, 18:04
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When a certain manufacturer produces a unit of its product, the probability that the unit will be defective is 0.01. If the manufacturer is to produce 10 units of its product, what is the probability that at least 1 of the 10 units produced will be defective?
The question asks for the probability that "at least 1" of the 10 units produced will be defective.
One way to answer such a question would be to determine the probability that 1 unit will be defective, the probability that 2 units will be defective, the probability that 3 units will be defective, and so on up to the probability that 10 units will be defective and add those probabilities to determine the total probability that at least 1 will be defective.
However, that method would take a lot of work, and there's a much more efficient way to calculate the probability that at least 1 will be defective.
We can simply find the probability that 0 units will be defective and subtract that probability from 1.
The reason that approach works is that the probability that 0 will be defective and the probabilities that 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 will be defective add to 1.
Since the probability that a given unit will be defective is 0.01, the probability that a given unit will not be defective is 0.99.
So, the probability that 0 units will be defective is \(0.99^{10}\).
Thus, the probability that at least 1 will be defective is \(1 - 0.99^{10}\).
A. \((0.01)^{10}\)
B. \((0.01)^9(0.99)\)
C. \((0.01)(0.99)^9\)
D. \(1-(0.01)^{10}\)
E. \(1-(0.99)^{10}\)
Correct answer: E
The question asks for the probability that "at least 1" of the 10 units produced will be defective.
One way to answer such a question would be to determine the probability that 1 unit will be defective, the probability that 2 units will be defective, the probability that 3 units will be defective, and so on up to the probability that 10 units will be defective and add those probabilities to determine the total probability that at least 1 will be defective.
However, that method would take a lot of work, and there's a much more efficient way to calculate the probability that at least 1 will be defective.
We can simply find the probability that 0 units will be defective and subtract that probability from 1.
The reason that approach works is that the probability that 0 will be defective and the probabilities that 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 will be defective add to 1.
Since the probability that a given unit will be defective is 0.01, the probability that a given unit will not be defective is 0.99.
So, the probability that 0 units will be defective is \(0.99^{10}\).
Thus, the probability that at least 1 will be defective is \(1 - 0.99^{10}\).
A. \((0.01)^{10}\)
B. \((0.01)^9(0.99)\)
C. \((0.01)(0.99)^9\)
D. \(1-(0.01)^{10}\)
E. \(1-(0.99)^{10}\)
Correct answer: E
General Discussion
27 Oct 2024, 07:59
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SergejK
Given that the probability that the unit will be defective is 0.01, the probability that the unit will NOT be defective is 1 - 0.01 = 0.99.
P(at least one defective) = 1 - P(0 defective) =
\(=1-(0.99)^{10}\)
Answer: E.
