8) A perfectly balanced die is rolled 3 times. What is the probability that neither the score of the first roll, nor the sum of the scores of the first and second rolls, nor the sum of the scores of the first three rolls is divisible by 7?
(A) $\frac{1}{3}$
(B) $\frac{1}{2}$
(C) $\frac{25}{36}$
(D) $\frac{3}{4}$
(E) $\frac{5}{6}$
8) The answer is (C). The score obtained on the first roll is not necessarily divisible by 7, but whatever it is, there is a unique possible second score such that their sum is divisible by 7. Therefore, the probability that the sum of the first two scores is divisible by 7 is $\frac{1}{6}$. Similarly, the sum of the three scores will be divisible by 7 only if the sum of the first two is not (which occurs with a probability of $\frac{5}{6}$) and if the third score is exactly that one number that makes the total a multiple of 7; so this event will have a probability of $\frac{5}{6} \times \frac{1}{6}=\frac{5}{36}$. In conclusion, the sought probability is given by $1-\frac{1}{6}-\frac{5}{36}=\frac{25}{36}$.