12) Two dice are rolled; the probability that the score on the first die is strictly less than that on the second is:
(A) $\frac{1}{6}$
(B) $\frac{1}{3}$
(C) $\frac{5}{12}$
(D) $\frac{1}{2}$
(E) $\frac{7}{12}$.
12) The answer is $(\mathbf{C})$. Indeed, if the score of one of the two dice is $n$, the favorable cases are $n-1$; In total, the favorable cases are thus $5+4+3+2+1+0=15$, while the possible cases are $6^{2}=36$. The probability is therefore $\frac{15}{36}=\frac{5}{12}$.