Problem: if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 6 ?
Rationale: i get 5 / 8 as well 1 to 96 inclusive means we have 48 odd and 48 even integers e o e / 6 = integer , therefore we have 48 / 96 numbers divisible by 6 o e o / 6 = not integer we can not forget multiples of 6 from 1 to 96 we have 16 numbers that are multiple of 8 therefore , 48 / 96 + 16 / 96 = 64 / 96 = 2 / 3 answer : a
annotated_formula: divide(add(divide(96, 2), divide(96, 6)), 96)