<s>[INST] If an integer $n$ is chosen at random from $1$ to $96$ inclusive ,what is the probability that $n(n+1)(n+2)$ is divisible by 8? In this one If I consider that n is even then probability that the number $n(n+1)(n+2)$ will be divisible by 8 will be 1/2 ,Now if n is odd then for n(n+1)(n+2) to be divisible by 8 ,n+1 should be a multiple of 8 ,Now how to find the probability among 96 numbers such that it is a multiple of 8 ?
 [/INST] the total numbers are $96$. for an odd number to be multiple of 8 using expression $n(n+1)(n+2)$ the number $n$ should be a number preceding to a multiple of $8$ starting from $8$ itself so first number is $7$ and such is an AP whose last term is $95$. So total terms which are $1$ less than a multiple of $8$ ie odd are $12$ thus the probability is $\frac{12}{96}=\frac{1}{8}$. Numbers with $4m-2=\frac{1}{4}$ and then $4m=\frac{1}{4}$ so addition is $\frac{5}{8}$
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