How do you find the probability of a brownian motion?

Question

If $B(t)$ is a brownian motion what do these two questions mean?
1. What is the probability of $B(2)$
2. What is the probability of $B(2) \gt B(1)$

I know this is also called a Wiener Process and I think a brownian motion has a normal distribution with mean $0$ and variance $t$, but I don't understand what the probability of $B(t)$ means?

Answer 1

1. It should be distribution of $B(2)$ most likely, and thus is $N(0,2)$.
2. $1/2$, since this is $P(B(2)-B(1)>0)$ and $B(2)-B(1)$ is a normal distribution with zero mean and is thus >0 with probability $1/2$