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 / 8 , what are the odds of it burning out during the period from 6 months to 1 year following its purchase ?
"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 ) = 1 / 8 - - - > p ( of not burning out in 1 st 6 mnth ) = 1 - 1 / 8 = 7 / 8 - - - - > p ( of not burning out in a six mnth period ) = 1 / 2 * 7 / 8 = 4 / 9 - - - > p ( of burning out in a six mnth period ) = 1 - 4 / 9 = 4 / 7 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 ) = 7 / 8 * 4 / 7 = 1 / 2 ans c"
multiply(subtract(1, divide(1, 8)), subtract(1, divide(subtract(1, divide(const_2.0, 8)), 1)))