Consider a random walk $(X_n)_{n≥0}$ with $p = 0.7$, starting from $X_0 = 3$. Find the probability that $X_{10} = 5$, but $X_n ≥ 1$ for $n = 0, . . . , 10.$.
Essentially what I got from the reflection principle is this. $P(x_0=1,\text{staying} \geq 1, X_{10}=5)=P(\text{going from (0,3) to (10,5)})-P(\text{going from (0,3) to (10,-3))}={10\choose 6}(.7)^6(.3)^4-{10\choose 2}(.7)^2(.3)^8=.1519332$.
But I am not sure if this is correct or not. Thanks
I think you're reflecting in the wrong line. You want to exclude paths from $(0,3)$ to $(10,5)$ that touch/cross the $X_n=0$ line, not the $X_n=1$ line.
The number of such paths equals the number of all paths from $(0,3)$ t0 $(10,-5)$, by the reflection principle (or equivalently from $(0,-3)$ to $(10,5)$, depending on which way you reflect).
Also, you don't want to use the probability of each reflected path but of each original path.
So you should have
$$\left[\binom{10}{6} - \binom{10}{9}\right]p^6q^4.$$