7. Randomly throw three dice. The probability that the sum of the numbers on two of the dice is 7 is $\qquad$ .
7. $\frac{5}{12}$.

Rolling three dice has a total of $6^{3}=216$ possible outcomes.
Notice that, $7=1+6=2+5=3+4$.
When rolling three dice, the number of outcomes where two dice show 1 and 6 is $6 \times 6-6=30$, which can be divided into
$$
\begin{array}{l}
(1,6, \times),(1, \times, 6),(6,1, \times), \\
(6, \times, 1),(\times, 1,6),(\times, 6,1)
\end{array}
$$

These six possibilities, each with 6 scenarios, include
$$
\begin{array}{l}
(1,6,1),(1,6,6),(1,1,6), \\
(6,1,1),(6,1,6),(6,6,1)
\end{array}
$$

which are repeated.
Similarly, the number of outcomes where two dice show 2 and 5, and the number of outcomes where two dice show 3 and 4, are both 30.

Thus, the number of outcomes where two of the three dice show a sum of 7 is $3 \times 30=90$.
Therefore, the required probability is $\frac{90}{216}=\frac{5}{12}$.