Consider a simple (symmetric) random walk $p=q=\frac{1}{2}$ and $(X_n)_{n\geq 0}$ with $X_0 = 0$. Using the reflection principle, find the probability that $X_{12} = -4$ and $X_1 < 2$, $X_2 < 2$, etc. $X_{11} < 2$.
What I did was reflect $X_{12}=-4$about the $y=2$ which we end up at $X_{12}=8$. By the reflection the total number of paths that pass $y=2$ from $(0,0)-->(12,8)$ is equal to the number of paths intersecting $y=2$ from $(0,0)--->(12,-4)$. The total number of paths from $(0,0)-->(12,8)$ is ${12\choose 10}=66.$ Thus it follows that $P(x_{12}=-4,x_n<2)=({12\choose 4}-66)*2^{-12}$. Would this be right thanks.