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donisback
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


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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donisback
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


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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donisback
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


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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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?

Probability of light bulb burning out during the first 6-month period following its purchase = 1/3
Probability of light bulb NOT burning out during the first 6-month period following its purchase = 1-1/3= 2/3

Probability of light bulb NOT burning out during next 6-month period = 1/3
Probability of light bulb burning out during next 6-month period = 1-1/3 = 2/3

Probability of light bulb burning out during the period from 6months to 1 year following its purchase = Probability of light bulb NOT burning out during the first 6-month period following its purchase*Probability of light bulb burning out during next 6-month period = 2/3*2/3 = 4/9

IMO D
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