How can I find the conditional entropy? I was given the joint pmf of 3 variables :
$$P_{X,Y,Z}(x,y,z) = 
\begin{cases}
       {\frac{3}{16}} &\quad\text{if }(x,y,z)\in \{001,111\}\\
       {\frac{1}{8}} &\quad\text{if }(x,y,z)\in  \{000,010,100,110\}\\
       {\frac{1}{16}} &\quad\text{if } (x,y,z)\in  \{011,101\}
        \\ 
     \end{cases}$$
and the pmf of each variable:
 $$P_{X}(x) = 
\begin{cases}
       {\frac{1}{2}} &\quad\text{if }x\in \{0\}\\
       {\frac{1}{2}} &\quad\text{if } x\in  \{1\}
        \\ 
     \end{cases}$$
$$P_{Y}(y) = 
\begin{cases}
       {\frac{1}{2}} &\quad\text{if }y\in \{0\}\\
       {\frac{1}{2}} &\quad\text{if } y\in  \{1\}
        \\ 
     \end{cases}$$
$$P_{Z}(z) = 
\begin{cases}
       {\frac{1}{2}} &\quad\text{if }z\in \{0\}\\
       {\frac{1}{2}} &\quad\text{if } z\in  \{1\}
        \\ 
     \end{cases}$$
How can I find the conditional entropy $H(X,Y|Z=0)$?

Note that simply $P_{X,Y|Z}(x,y|z=0)=\frac{P_{X,Y,Z}(x,y,z=0)}{p(z=0)}$, which is equal to $\frac{1}{4}$, in the given distribution. Hence
$$H(X,Y|Z=0)=2.$$
