$W_t$ is a brownian motion, I have this exponential value: $$v(t)= e^{0.00025 + 0.3W_t}$$ what's the probability that $v(1)<0.5$?
By taking natural log on both size, I got $0.00025 + 0.3W_1 < -ln(2), W_1 = -2.3$ Then the question becomes $$P(W_1<-2.3) =?$$
You can check from the definition of a Brownian motion that $W_1$ has the normal distribution $N(0,1)$. So the problem really amounts to figuring out what
$$P(W_1<-2.3)$$
is for a standard normal random variable $W_1$. You can look this up in a table. You have that
$$P(W_1<-2.3) = \Phi(-2.3),$$ where $\Phi$ is the cumulative distribution function of the standard normal distribution.