<s>[INST] joint and conditional probability This is table of conditional probability with $X$ is for red coin, and $Y$ for blue coin.  
$$\begin{array}{rc} 
               & X_{\text{red}} \\
Y_{\text{blue}} & \begin{aligned} 
\  P (X \mid Y)&& X=\text{Tail} && X=\text{Head} 
\\Y=\text{Head} &&\frac{6}{10} && \frac{4}{10} 
\\ Y=\text{Tail} &&\frac{2}{10}&&\frac{8}{10}
\end{aligned}
\end{array}$$
It is known that $P(Y_{\text{blue}} = \text{Head} )=0.45$ and $P(X_{\text{red}} = \text{Tail})=0.55$
I'm trying to find the $p(x,y)$ joint probability of $x$ and $y$
such as 
$$P(X=\text{Head},Y=\text{Tail})=P(Y=\text{Tail})\cdot P(X = H \mid Y= T)=0.55 \cdot 0.2 = 0.11$$
and 
$$P(X= H,\, Y=H)=P(Y=H)\cdot P(X=H \mid Y=H)=0.6 \cdot 0.45=0.27$$
$$\begin{array}{rc} 
               & X_{\text{red}} \\
Y_{\text{blue}} & \begin{aligned} 
\  P (X ,\, Y)&& X=\text{Tail} && X=\text{Head} 
\\Y=\text{Head} &&\frac{27}{100} && \frac{18}{100} 
\\ Y=\text{Tail} &&\frac{11}{100}&&\frac{44}{100}
\end{aligned}
\end{array}$$
is this right?
Also I want to know why for $p(x \mid y)$ , probability of $ Y(\text{Head}) =\frac{6}{10} + \frac{4}{10} = 1 $ and $Y(\text{Tail}) = 1$ but for $X( \text{Head})= \frac{4}{10} + \frac{8}{10}$ not equal to $1$? 
 [/INST] \begin{align}
P(X=H, Y=H) &= P(Y=H) P(X=H|Y=H)\\
&= 0.45 \cdot \color{red}{0.4}
\end{align}
For the table of conditional probability.
Summing up the rows give you $1$, because given that $Y$ has occured, and the probability of possible outcome of $X$ must sum to $1$.
$$\sum_xP(X=x|Y=y)=1$$
However, for the columns, 
$$\sum_y P(X=x|Y=y),$$
there is no reason for it to sum to $1$.
</s>