probability over 3 values with dependency At the exercise, there is no information that B and C are independent, but with logical reasoning, there must be a pendency. 
The problem is, I can not create a connection with depency of B and C, is that even possible and if yes, how and I calculate this?
My current approach:
$$
A=rain; B=wet; C=sun \\
$$
Given values:
$$
P(A)=0.44; P(B|A)=0.99, P(B|\bar A)=0.1 \\
P(C|A)=0.05; P(C|\bar A)=0.7
$$
P(B) can be calculated with the law of total probability:
$$
P(B)= P(B|A) \cdot P(A)+P(B|\bar A)\cdot P(\bar A) \\
P(B)= 0.99 \cdot 0.44+0.1\cdot 0.56 \approx 0.49 \\
P(\bar B)= 1-P(B) \approx 0.51 \\
$$
Bayes' theorem:
$$
P(A|B)= \frac {P(B|A)P(A)}{P(B)} \\
= \frac{0.99*0.44}{0.49} \approx 0.89 \\ 
P(C)= P(C|A) \cdot P(A)+P(C|\bar A) \cdot P(\bar A) \\
P(C)= 0.05 \cdot 0.44+0.7 \cdot 0.56 \approx 0.41 \\
P(\bar B \cap C) = P(\bar B) \cdot P(C)=0.51\cdot 0.41 \approx 0.21 \\
$$
$P(\bar B \cap C)$ can not be calculated as above if depency exist
target: 
$$
P(A|\bar B \cap C)=?
$$
I hope someone can help me with that problem, thanks.

Assuming this is a Bayesian network problem, events $B,C$ are dependent, but given $A$, are conditionally independent. So you can find $P(A\mid \bar B\cap C)$ as follows:
\begin{eqnarray*}
P(A\mid \bar B\cap C) &=& \dfrac{P(A\cap\bar B\cap C)}{P(\bar B\cap C)} \\
&=& \dfrac{P(A\cap\bar B\cap C)}{P(A\cap\bar B\cap C) + P(\bar A\cap\bar B\cap C)} \\
&& \\
P(A\cap\bar B\cap C) &=& P(\bar B\mid A)P(C\mid A)P(A)\qquad\qquad\text{(using Chain Rule)} \\
&=& 0.01\cdot 0.05\cdot 0.44 \;=\; 0.00022 \\
&& \\
P(\bar A\cap\bar B\cap C) &=& P(\bar B\mid \bar A)P(C\mid \bar A)P(\bar A) \\
&=& 0.9\cdot 0.7\cdot 0.56 \;=\;0.3528 \\
&& \\
\therefore\quad P(A\mid \bar B\cap C) &=& 0.00022 / 0.35302 \approx 0.000623.
\end{eqnarray*}
