Determine fair price of a digital option 
A digital option pays one dollar at time $t = T$ if the asset price is
  above a fixed level (strike) $K$ and is worthless otherwise. 
Consider the following
  model, with $r = 0$:
\begin{array}{|c|c|c|}
\hline
\omega& S(0) & S(1) & S(2) \\ \hline
 \omega_1&6 & 10&12 \\ \hline
  \omega_2&6 &10 & 7\\ \hline
  \omega_3&6 &4 & 7\\ \hline
 \omega_4&6 &4 & 3\\ \hline
\end{array}
Evaluate $\Bbb E_\Bbb Q[X]$ and determine the fair price of the digital option struck at $4$.

What I have done for a part that preceded this was find the risk neutral probabilities $\Bbb Q = (p, \frac{1}{3}-p,\frac{5}{12}-\frac{5}{4}p,\frac{1}{4}+\frac{5}{4}p)$ with a restriction on $0<p<\frac{1}{3}$ using systems of equations. 
I am not sure how I can find the expectation based on this and what the random variable $X$ represents exactly. I guess the expected value that I am supposed to find will be the fair price?
 A: The elementary events $\{\omega_j\}$ correspond to paths on a binomial lattice with transition probabilities $p_1, \, p_2,$ and $p_3$ as shown:
$$\omega_1: \quad S(0) = 6 \underbrace{\to}_{p_1} S_u = 10\underbrace{\to}_{p_2} S_{uu}=12\\\omega_2: \quad S(0) = 6 \underbrace{\to}_{p_1} S_u = 10\underbrace{\to}_{1-p_2} S_{ud}=7\\ \omega_3: \quad S(0) = 6 \underbrace{\to}_{1-p_1} S_d = 4\underbrace{\to}_{p_3} S_{du}=7\\ \omega_4: \quad S(0) = 6 \underbrace{\to}_{1-p_1} S_d = 4\underbrace{\to}_{1-p_3} S_{dd}=3\\$$
The risk-neutral probabilities are found by enforcing expected future prices to be equal to forward prices, which coincide with spot prices as the interest rate is assumed to be $r = 0$.  Under the risk-neutral measure the asset price process is a martingale.
Thus,
$$S(0) = \mathbb{E}[S(1)] = p_1S_u + (1-p_1) S_d\\ \implies 6 = 10p_1 + 4(1-p_1) \implies p_1 = \frac{1}{3}$$
$$S_u = \mathbb{E}[S(2)|S(1) = S_u] = p_2S_{uu} + (1-p_2) S_{ud}\\ \implies 10 = 12p_2 + 7(1-p_2) \implies p_2 = \frac{3}{5}$$
$$S_d = \mathbb{E}[S(2)|S(1) = S_d] = p_3S_{du} + (1-p_3) S_{dd}\\ \implies 4 = 7p_3 + 3(1-p_3) \implies p_3 = \frac{1}{4}$$
We can now compute risk-neutral path probabilities as,
$$P(\omega_1) = \frac{1}{3}\frac{3}{5} = \frac{1}{5}, \,\,P(\omega_2) = \frac{1}{3}\frac{2}{5} = \frac{2}{15}, \,\,P(\omega_3) = \frac{2}{3}\frac{1}{4} = \frac{2}{12}, \,\,P(\omega_4) = \frac{2}{3}\frac{3}{4} = \frac{1}{2} \,\,$$
The digital option expiring at time $T=2$ with strike $K= 4$ pays $1$ on paths $1,2,3$ and pays $0$ on path $4$.  The fair price is the risk-neutral expected payoff which is
$$\mathbb{E}[X]= P(\omega_1) \cdot 1 + P(\omega_2) \cdot 1 + P(\omega_3) \cdot 1 + P(\omega_4) \cdot 0 = \frac{1}{2}$$
