15. A particle $M$ starting from the origin moves according to the vector $\boldsymbol{a}=(0,1)$ with a probability of $\frac{2}{3}$. It moves according to the vector $\boldsymbol{b}=(0,2)$ with a probability of $\frac{1}{3}$. Let $P_{n}$ be the probability that $M$ reaches the point $(0, n)$. Find $P_{n}$.
15. $M$ reaches the point $(0, n)$ in two scenarios:
(1) Moves from point $(0, n-1)$ to point $(0, n)$ along the vector $a=(0,1)$, with a probability of $\frac{2}{3} P_{n-1}$;
(2) Moves from point $(0, n-2)$ to point $(0, n)$ along the vector $\boldsymbol{b}=(0,2)$, with a probability of $\frac{1}{3} P_{n-2}$.
Since these two scenarios are mutually exclusive, we have: $P_{n}=\frac{2}{3} P_{n-1}+\frac{1}{3} P_{n-2}(n \geqslant 3)$, which simplifies to $P_{n}-P_{n-1}=-\frac{1}{3}\left(P_{n-1}-\right.$ $\left.P_{n-2}\right)(n \geqslant 3)$.

Given $P_{1}=\frac{2}{3}, P_{2}=\frac{7}{9}$. Therefore, $\left\{P_{n}-P_{n-1}\right\}$ is a geometric sequence with the first term $P_{2}-P_{1}=\frac{1}{9}$ and common ratio $-\frac{1}{3}$, so $P_{n}-P_{n-1}=\frac{1}{9}\left(-\frac{1}{3}\right)^{n-2}=\left(-\frac{1}{3}\right)^{n}$.

Thus, $P_{n}=P_{1}+\left(P_{2}-P_{1}\right)+\cdots+\left(P_{n}-P_{n-1}\right)=\frac{2}{3}+\left(-\frac{1}{3}\right)^{2}+\left(-\frac{1}{3}\right)^{2}+\cdots+\left(-\frac{1}{3}\right)^{n}$. $=\frac{2}{3}+\frac{\left(-\frac{1}{3}\right)^{2}\left[1-\left(-\frac{1}{3}\right)^{n-1}\right]}{1-\left(-\frac{1}{3}\right)}=\frac{2}{3}+\frac{1}{12}\left[1-\left(-\frac{1}{3}\right)^{n-1}\right]$.