A random walk in a random environment starts at $0$. There are two types of sites: green and blue. At a green site, the probability of stepping to the right is $1/4$, and the probability of stepping to the left is $3/4$. At a blue site, the probability of stepping to the right is $2/3$, and the probability of stepping to the left is $1/3$. Each site (and in particular, site $0$) is blue with probability p and green with probability $1 − p$. For what value of p will the first step be to the right with probability exactly $1/2$.
Here is what I have so far: Let $\left\{B_R\right\}$ be the event where $x=0$ is a blue spot and we move right. Let ${G_R}$ be the event where $x=0$ is a green spot and we move right. Their probabilities are $$\Bbb P(B_R)=p(2/3) \text{ and } \Bbb P(G_R)=(1-p)(1/4)$$ Would I solve for $p$ by solving $P(B_R\cup B_L\cup G_R\cup G_L)=1$?
For the question asked, you can ignore the fact that it's a random walk on a line. Just worry about what's happening at the origin.
You're asking for $P(R)$, the probability of moving right. Let $B, G$ be the events that the origin is blue/green respectively. Then $$ \begin{align*} P(R) &= P(R \cap (B \cup G)) \text{ since } P(B \cup G) = 1 \\ &= P(R \cap B) + P(R \cap G) \text{ since } B \cap G = \emptyset \\ &= P(R | B)P(B) + P(R|G)P(G) \text{ by def. of conditional probability} \end{align*} $$
You know all the probabilities in the last line of the equation. Now solve $P(R) = 1/2$ for $p$.