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 2 / 3, what are the odds of it burning out during the period from 6 months to 1 year following its purchase? Choose one:	A) 5 / 27	B) 2 / 9	C) 1 / 3	D) 4 / 9	E) 2 / 7
<gadget id="calculator">2 / 3</gadget>
<output>2/3 = around 0.666667</output>

<gadget id="calculator">1 - (2/3)</gadget>
<output>1/3 = around 0.333333</output>

<gadget id="calculator">(1/3) / 2</gadget>
<output>1/6 = around 0.166667</output>

<gadget id="calculator">1 - (1/6)</gadget>
<output>5/6 = around 0.833333</output>

<gadget id="calculator">(1/3) * (5/6)</gadget>
<output>5/18 = around 0.277778</output>

<result>E</result>
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 2 / 3, what are the odds of it burning out during the period from 6 months to 1 year following its purchase?
multiply(subtract(1, divide(2, 3)), subtract(1, divide(subtract(1, divide(2, 3)), 2)))
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 1 st 6 mnth ) = 2 / 3 - - - > p ( of not burning out in 1 st 6 mnth ) = 1 - 2 / 3 = 1 / 3 - - - - > p ( of not burning out in a six mnth period ) = 1 / 2 * 1 / 3 = 1 / 6 - - - > p ( of burning out in a six mnth period ) = 1 - 1 / 3 = 2 / 3 now p ( of burning out in 2 nd six mnth period ) = p ( of not burning out in 1 st six mnth ) * p ( of burning out in a six mnth ) = 2 / 3 * 1 / 6 = 2 / 7 ans e