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For each 6month period during a light bulb's life span, the odds of
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Updated on: 20 Apr 2015, 07:11
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For each 6month period during a light bulb's life span, the odds of it not burning out from overuse are half what they were in the previous 6month period. If the odds of a light bulb burning out during the first 6month period following its purchase are 1/3, what are the odds of it burning out during the period from 6months to 1 year following its purchase? A. 5/27 B. 2/9 C. 1/3 D. 4/9 E. 2/3
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Originally posted by donisback on 17 Dec 2009, 04:42.
Last edited by Bunuel on 20 Apr 2015, 07:11, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




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Re: For each 6month period during a light bulb's life span, the odds of
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17 Dec 2009, 05:34
donisback wrote: For each 6month period during a light bulb's life span, the odds of it not burning out from overuse are half what they were in the previous 6month period. If the odds of a light bulb burning out during the first 6month period following its purchase are 1/3, what are the odds of it burning out during the period from 6months to 1 year following its purchase? A. 5/27 B. 2/9 C. 1/3 D. 4/9 E. 2/3 PLEASE EXPlain. I think there should be "probability" instead of "odds" (these two things are not the same and GMAT always asks about probability, not odds). Probability of not burning out during the first 6 months 11/3=2/3 Probability of not burning out during the next 6 months 2/3/2=1/3, hence probability of burning out 11/3=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 = 2/3 * 2/3 =4/9 Answer: D.
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Re: For each 6month period during a light bulb's life span, the odds of
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17 Dec 2009, 06:47
P(of not burning out in a six mnth period)=1/2 of P(of not burning out in prev 6 mnth period) P(of burning out in 1st 6 mnth)= 1/3 > P( of not burning out in 1st 6 mnth)=11/3=2/3>P(of not burning out in a six mnth period)=1/2 *2/3=1/3> P(of burning out in a six mnth period)=11/3=2/3now P( of burning out in 2nd six mnth period)= P( of not burning out in 1st six mnth)* P(of burning out in a six mnth)=2/3 * 2/3=4/9
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Re: For each 6month period during a light bulb's life span, the odds of
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20 Apr 2015, 06:58
donisback wrote: For each 6month period during a light bulb's life span, the odds of it not burning out from overuse are half what they were in the previous 6month period. If the odds of a light bulb burning out during the first 6month period following its purchase are 1/3, what are the odds of it burning out during the period from 6months to 1 year following its purchase? A. 5/27 B. 2/9 C. 1/3 D. 4/9 E. 2/3 PLEASE EXPlain. P(Not Burning out in first 6months) = 2/3 P(Not Burning out in 6months 12months ) = \(\frac{1}{2} * \frac{2}{3}\) = \(\frac{1}{3}\) hence , P(Burning out in 6months 12months ) = \(\frac{2}{3}\) P(Not Burning out in first 6months) * P(Burning out in 6months 12months ) = \(\frac{2}{3}* \frac{2}{3} = \frac{4}{9}\) Ans: D



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Re: For each 6month period during a light bulb's life span, the odds of
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13 Dec 2017, 13:15
Lucky2783 wrote: donisback wrote: For each 6month period during a light bulb's life span, the odds of it not burning out from overuse are half what they were in the previous 6month period. If the odds of a light bulb burning out during the first 6month period following its purchase are 1/3, what are the odds of it burning out during the period from 6months to 1 year following its purchase? A. 5/27 B. 2/9 C. 1/3 D. 4/9 E. 2/3 PLEASE EXPlain. P(Not Burning out in first 6months) = 2/3 P(Not Burning out in 6months 12months ) = \(\frac{1}{2} * \frac{2}{3}\) = \(\frac{1}{3}\) hence , P(Burning out in 6months 12months ) = \(\frac{2}{3}\) P(Not Burning out in first 6months) * P(Burning out in 6months 12months ) = \(\frac{2}{3}* \frac{2}{3} = \frac{4}{9}\) Ans: D Could you explain why 1/2 *2/3 > where did that derive from?



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Re: For each 6month period during a light bulb's life span, the odds of
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13 Dec 2017, 13:39
Madhavi1990 wrote: Lucky2783 wrote: donisback wrote: For each 6month period during a light bulb's life span, the odds of it not burning out from overuse are half what they were in the previous 6month period. If the odds of a light bulb burning out during the first 6month period following its purchase are 1/3, what are the odds of it burning out during the period from 6months to 1 year following its purchase? A. 5/27 B. 2/9 C. 1/3 D. 4/9 E. 2/3 PLEASE EXPlain. P(Not Burning out in first 6months) = 2/3 P(Not Burning out in 6months 12months ) = \(\frac{1}{2} * \frac{2}{3}\) = \(\frac{1}{3}\) hence , P(Burning out in 6months 12months ) = \(\frac{2}{3}\) P(Not Burning out in first 6months) * P(Burning out in 6months 12months ) = \(\frac{2}{3}* \frac{2}{3} = \frac{4}{9}\) Ans: D Could you explain why 1/2 *2/3 > where did that derive from? The odds of not burning out in the first 6 months are 2/3. The odds of not burning out over the next 6 month period are half of that. (The problem reads "the odds of it not burning out from overuse are half what they were in the previous 6month period"). Half of 2/3 is 1/2*2/3 = 1/3. Likewise, the odds of not burning out over the next 6 month period would be 1/2*1/3 = 1/6, and so on.
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