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For each 6-month period during a light bulb's life span, the odds of

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For each 6-month period during a light bulb's life span, the odds of [#permalink]

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New post Updated on: 20 Apr 2015, 08:11
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For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month 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

Originally posted by donisback on 17 Dec 2009, 05:42.
Last edited by Bunuel on 20 Apr 2015, 08:11, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: For each 6-month period during a light bulb's life span, the odds of [#permalink]

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New post 17 Dec 2009, 06:34
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donisback wrote:
For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month 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

Spoiler: :: OA
D


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 1-1/3=2/3
Probability of not burning out during the next 6 months 2/3/2=1/3, hence probability of burning out 1-1/3=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 6-month period during a light bulb's life span, the odds of [#permalink]

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New post 17 Dec 2009, 07:47
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)=1-1/3=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)=1-1/3=2/3
now
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 6-month period during a light bulb's life span, the odds of [#permalink]

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New post 20 Apr 2015, 07:58
donisback wrote:
For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month 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

Spoiler: :: OA
D


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 6-month period during a light bulb's life span, the odds of [#permalink]

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New post 13 Dec 2017, 14:15
Lucky2783 wrote:
donisback wrote:
For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month 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

Spoiler: :: OA
D


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 6-month period during a light bulb's life span, the odds of [#permalink]

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New post 13 Dec 2017, 14:39
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Madhavi1990 wrote:
Lucky2783 wrote:
donisback wrote:
For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month 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

Spoiler: :: OA
D


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 over-use are half what they were in the previous 6-month 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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Re: For each 6-month period during a light bulb's life span, the odds of   [#permalink] 13 Dec 2017, 14:39
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