7. Let $n$ be a positive integer. From the set $\{1,2, \cdots, 2015\}$, the probability that a randomly chosen positive integer $n$ is a solution to the equation
$$
\left[\frac{n}{2}\right]=\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]
$$

is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
7. $\frac{1007}{2015}$.

When $n=6 k\left(k \in \mathbf{Z}_{+}\right)$,
$$
\begin{array}{l}
{\left[\frac{n}{2}\right]=\left[\frac{6 k}{2}\right]=3 k,} \\
{\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\left[\frac{6 k}{3}\right]+\left[\frac{6 k}{6}\right]=2 k+k=3 k,}
\end{array}
$$

The $n$ that satisfies the equation in the problem is $6,12, \cdots, 2010$, a total of 335;
When $n=6 k-5\left(k \in \mathbf{Z}_{+}\right)$,
$$
\begin{array}{l}
{\left[\frac{n}{2}\right]=\left[\frac{6 k-5}{2}\right]=3 k-3,} \\
{\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\left[\frac{6 k-5}{3}\right]+\left[\frac{6 k-5}{6}\right]} \\
=2 k-2+k-1=3 k-3,
\end{array}
$$

The $n$ that satisfies the equation in the problem is $1,7,13, \cdots, 2011$, a total of 336;
When $n=6 k-4\left(k \in \mathbf{Z}_{+}\right)$,
$$
\begin{array}{l}
{\left[\frac{n}{2}\right]=\left[\frac{6 k-4}{2}\right]=3 k-2,} \\
{\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\left[\frac{6 k-4}{3}\right]+\left[\frac{6 k-4}{6}\right]} \\
=2 k-2+k-1=3 k-3,
\end{array}
$$

There is no $n$ that satisfies the equation in the problem;
$$
\begin{array}{l}
\text { When } n=6 k-3\left(k \in \mathbf{Z}_{+}\right) \text {, } \\
{\left[\frac{n}{2}\right]=\left[\frac{6 k-3}{2}\right]=3 k-2,} \\
{\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\left[\frac{6 k-3}{3}\right]+\left[\frac{6 k-3}{6}\right]} \\
=2 k-1+k-1=3 k-2,
\end{array}
$$

The $n$ that satisfies the equation in the problem is $3,9,15, \cdots, 2013$, a total of 336;
When $n=6 k-2\left(k \in \mathbf{Z}_{+}\right)$,
$$
\begin{array}{l}
{\left[\frac{n}{2}\right]=\left[\frac{6 k-2}{2}\right]=3 k-1,} \\
{\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\left[\frac{6 k-2}{3}\right]+\left[\frac{6 k-2}{6}\right]} \\
=2 k-1+k-1=3 k-2,
\end{array}
$$

There is no $n$ that satisfies the equation in the problem;
When $n=6 k-1\left(k \in \mathbf{Z}_{+}\right)$,
$$
\begin{array}{l}
{\left[\frac{n}{2}\right]=\left[\frac{6 k-1}{2}\right]=3 k-1,} \\
{\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\left[\frac{6 k-1}{3}\right]+\left[\frac{6 k-1}{6}\right]} \\
=2 k-1+k-1=3 k-2,
\end{array}
$$

There is no $n$ that satisfies the equation in the problem.
Therefore, the probability that a positive integer $n$ chosen from the set $\{1,2, \cdots, 2015\}$ is a solution to the equation in the problem is
$$
\frac{335+336+336}{2015}=\frac{1007}{2015} \text {. }
$$