Geometric Brownian motion problem Here's the question:
Let $S(t)$, $t \geq 0$ be a Geometric Brownian motion process with drift parameter $\mu = 0.1$ and volatility parameter $\sigma = 0.2$. Find $P(S(3) < S(1) > S(0)).$
Is there something wrong with the following reasoning:
$P(S(3) < S(1) > S(0))=P(S(1)>S(3) \geq S(0)) + P(S(1) > S(0) \geq S(3))$, where 
$P(S(1)>S(3) \geq S(0))=P(S(1) > S(3))-P(S(3) \leq S(0)$ and
$P(S(1) > S(0) \geq S(3))=P(S(1) > S(0)) - P(S(0) \leq S(3))$.

Tiny point: You are missing a "$)$".
Major point: $P(S(1)\gt S(3) \geq S(0))=P(S(1) \gt S(3))-P(S(3) \leq S(0))$ is wrong and similarly with the following equality.  You should have something like $P(S(1)\gt S(3) \geq S(0))  =P(S(1) \gt S(3))-P(S(3) \leq S(0)) + P(S(0) \gt S(1) \geq S(3))$
Critical point:  $S(0)$, $S(1)$ and $S(3)$ are not independent, so this is the wrong approach.  But $S(1)-S(0)$ and $S(3)-S(1)$ are independent, so their probabilities multiply, and you will probably find it easier to deal with the rather more obvious and more easily handled $$P(S(3) \lt S(1) \gt S(0)) = P(S(3) \lt S(1)) \times P(S(1) \gt S(0)).$$ 
