Does shift cipher with non-uniform plain text probability have perfect secrecy? I am having a hard time grasping the perfect secrecy concept. (I have already tried solving a couple of similar problems, so please don't think I am trying to get you guys to do my work, I really am stumped on this).

Consider a shift cipher with plaintext, key and cipher-text space is $[0..7]$ i.e. 

$$
P = \{0, 1, 2, 3, 4, 5, 6, 7, 8\}\\
K = \{0, 1, 2, 3, 4, 5, 6, 7, 8\}\\
C = \{0, 1, 2, 3, 4, 5, 6, 7, 8\}
$$

plain-text's even alphabets have twice the probability of odd alphabet i.e.

$$
\forall p \in P: 
\Pr_P(p) = \begin{cases} \frac{2}{12} &amp; \textrm{ if p = 2(b) for some b $\in$ Z } \\
\frac{1}{12} &amp; \textrm{ if p = 2(b) + 1 for some b $\in$ Z }
\end{cases}\\
\\
\forall k \in K: \Pr_K(k) = \frac{1}{8}
$$

From here I have calculated that $\Pr_C(c) = \frac{1}{8}$.


How do I calculate $\Pr(y|x)$
How do I use Baye's theorem to calculate $\Pr(x|y)$
How do I show whether $\Pr(x|y) = \Pr(x)$ or $\Pr(x|y) \neq \Pr(x)$ (thereby determining whether this cipher has perfect secrecy or not).
For any ciphertext $c \in C$ and message $p \in P$ we have
$$P(c|p) = \frac{P(p\cap c)}{P(p)} = \frac{P(p\cap k)}{P(p)},$$
where $k = c-p$. This is because if the plaintext is $p$, the ciphertext is $c$ if and only if the key is $k$. Now, if the key is chosen independently of the plaintext we have $P(p\cap k) = P(p)\cdot P(k)$, and so finally $P(c|p) = P(k) = 1/8$ if the key is chosen uniformly.