Problem: 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 ?
Rationale: 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
annotated_formula: multiply(subtract(1, divide(2, 3)), subtract(1, divide(subtract(1, divide(2, 3)), 2)))