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mojorising800
Joined: 11 Dec 2009
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Schools: Tepper '14 Goizueta '14 ISB Tippie Kelley '14 Georgia Tech Mays Katz Kenan-Flagler '14 Schulich Smith '15
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01 Feb 2010, 04:16
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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0% (00:00) correct
100%
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Not Attempted Yet
Plz explain
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Hi there,
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01 Feb 2010, 05:29
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mojorising800
I think there should be "probability" instead of "odds".
We want light bulb NOT to bun out during the first 6 month and to burn out during the next 6 months period:
Probability of not burning out during the first 6 months \(1-\frac{1}{3}=\frac{2}{3}\);
Probability of not burning out during the next 6 months \(\frac{\frac{2}{3}}{2}=\frac{1}{3}\), hence probability of burning out is \(1-\frac{1}{3}=\frac{2}{3}\).
Probability of burning out during the period from 6 months to 1 year = Probability of not burning out in first 6 months * Probability of burning out in next 6 months = \(\frac{2}{3}* \frac{2}{3} =\frac{4}{9}\)
Answer: D.
