Three fair six-faced dice are rolled together. The probability that the product of the three numbers shown is divisible by 12 is m/n, where m and n…

$12=4\cdot3$, so the product must be divisible by $3$ and by $4$. Let $A$ = no face is $3$ or $6$ (product not divisible by 3), $B$ = product not divisible by $4$. Required count $=216-|A|-|B|+|A\cap B|$. $|A|=4^3=64$. For $B$ the power of $2$ in the product is at most $1$: faces $\{1,3,5\}$ contribute $2^0$, face $\{2,6\}$ contribute $2^1$, face $4$ contributes $2^2$. So either all three faces are odd ($3^3=27$ ways) or exactly one face is from $\{2,6\}$ and the other two odd ($3\cdot2\cdot3^2=54$ ways): $|B|=81$. For $A\cap B$ use faces without $3,6$: odd faces $\{1,5\}$, even faces $\{2,4\}$. All odd: $2^3=8$; exactly one face equal to $2$, others odd: $3\cdot1\cdot2^2=12$; so $|A\cap B|=20$. Count $=216-64-81+20=91$. Probability $=\dfrac{91}{216}$, already in lowest terms ($91=7\cdot13$), so $m+n=307$. A direct enumeration of all $216$ outcomes confirms $91$.