An Application of the Neyman-Pearson Lemma. 

Problem: Using the Neyman-Pearson Lemma, determine the most powerful test of size $ 5 \% $.

I know the Neyman-Pearson Lemma says that the test with the critical region
$$
\left\{
x \in \{ 1,2,3,4 \} ~ \Bigg| ~ \frac{L(x \mid \theta = 0)}{L(x \mid \theta = 1)} \leq A
\right\},
$$
where $ A $ satisfies
$$
\mathbf{Pr} \left(
            \frac{L(X \mid \theta = 0)}{L(X \mid \theta = 1)} \leq A ~ \Bigg| ~ H_{0}
            \right)
= 0.05,
$$
is the most powerful test of size $ 5 \% $. However, I’m not sure what the likelihood functions are, as I’m more used to seeing them given by a formula.
Thanks.
 A: According to the data, we have
$$
L(x \mid \theta = 0) =
\begin{cases}
0.02 & \text{if $ x = 1 $}; \\
0.02 & \text{if $ x = 2 $}; \\
0.03 & \text{if $ x = 3 $}; \\
0.93 & \text{if $ x = 4 $},
\end{cases} \\
L(x \mid \theta = 1) =
\begin{cases}
0.10 & \text{if $ x = 1 $}; \\
0.20 & \text{if $ x = 2 $}; \\
0.30 & \text{if $ x = 3 $}; \\
0.40 & \text{if $ x = 4 $}.
\end{cases}
$$
Therefore,
$$
\Lambda(x) \stackrel{\text{def}}{=} \frac{L(x \mid \theta = 0)}{L(x \mid \theta = 1)}
=
\begin{cases}
0.2   & \text{if $ x = 1 $}; \\
0.1   & \text{if $ x = 2 $}; \\
0.1   & \text{if $ x = 3 $}; \\
2.325 & \text{if $ x = 4 $}.
\end{cases}
$$
The chosen size is $ 0.05 $; the Neyman-Pearson Lemma says that in order to find the most powerful test of this size, we need to find an $ \eta \in \mathbb{R} $ such that
$$
\mathbf{Pr}(\Lambda(X) \leq \eta \mid \theta = 0) = 0.05.
$$
The choice of $ \eta = 0.1 $ works, so the most powerful test of size $ 0.05 $ is:

Reject $ H_{0} $ in favor of $ H_{1} $ when the observed value of $ \Lambda(X) $ is $ \leq 0.1 $, or equivalently, when the observed value of $ X $ is either $ 2 $ or $ 3 $.

