# Why $H(A)=H(C)$ where $C$ is $A$ with an additional parity bit?

Let $$A=\{00,01,10,11\}$$ with equal probabilities for each symbol, and $$B=\{0, 1\}$$ be a parity generator such that $$b=\begin{cases} 0, & \text{if} \,\, a=00 \quad \text{or} \quad a=11 \\ 1, & \text{if} \,\, a=01 \quad \text{or} \quad a=10 \end{cases}$$ Now assume we transmit $$(a_i,bj)$$ where $$a_i$$ is a symbol in $$A$$ and $$b_j$$ is the parity bit associated with it. I calculated the entropy for $$A$$ as follows: $$H(A)=4 [4 \log_2(4)]=2$$ and to calculate the entropy of the new symbols, call it alphabet $$C=\{000,011,101,110\}$$, we need the probabilities: $$p_c(0)=\mathbb{P}(a=00 \, \text{and} \, b=0)=\mathbb{P}(a=00)\mathbb{P}(b=0 \mid a=00)=\mathbb{P}(a=00)=1/4$$ Similarly, $$p_c(1)=p_c(2)=p_c(3)=p_c(0)=1/4$$, the entropy of $$C$$ is $$H(C)=4 [4 \log_2(4)]=2=H(A)$$ How is this the case? We have more bits per symbols in $$C$$.

The source $$A$$ contains $$4$$ equiprobable symbols hence it is obvious that we need $$\log_2(4) = 2$$ bits per symbol to represent the source.
The source $$C$$ is simply $$A$$ with its parity bits hence no new information is added. We have more bits per symbols, but the number of information bits is unchanged.