I was solving an exercise which asks the reader to calculate the probability that a Brownian particle $B(t) = (B_1(t),...,B_n(t))$ starting at the origin in $\mathbb{R}^n$ will strike the surface of a cube $B = [-1, 1]^n$ within time $t$. I understand how to do this, because each coordinate $B_i(t)$ is an independent Brownian motion whose distribution is known. But I am wondering about the following: is this probability the same as being outside of the cube within time $t$? In other words, is $$\mathbb{P}_0(B(s) \in \partial B : \text{ for some } s \leq t) = \mathbb{P}_0(B(s) \in \mathbb{R}^n \setminus \overline{B} : \text{ for some } s \leq t)?$$
Thanks for any help!