Consider a random walk $S_n=S_0+\sum^b_{i=1}X_i$ with i.i.d steps $X_i$ taking value $4$ and $-7$ with probabilities $\frac{7}{11}$ and $\frac{4}{11}$ respectively.
Suppose the walk starts at $S_0=3$, and terminates whenever the boundaries at $1600$ or $-4900$ are reached or exceeded for the first time. Determine the probability that the walk terminates at or beyond the right boundary $1600$ using the optional stopping theorem.
My attempt to solve this:
The random walk $S_n$ terminates at time $T$ such that $S_T=1600$ or $S_T=-4900$
Assuming validity of $E[S_{T}]=E[S_0]$, we can express $E[S_{T}]$ as $1600P+(-4900)(1-P)$ using the fact that $S_0=3$ with probability $1$ hence $$1600P-4900(1-P)=3\Leftrightarrow 6500P=4903\Leftrightarrow P=0.75$$
So the probability the walk terminates at or beyond $1600$ is $0.75$
Is this correct?