We have a Bayesian Network that A to D is Boolean variable. we want to calculate the probability which C and D be True and A be false. my answer sheet calculate the last result and is 0.0424.
any expert would help me in a teach manner, how this probability is calculated?
The usual method to solve such questions for Bayesian networks is to use the Chain Rule for probability and use the rules regarding independence and conditional independence implied by the lines joining the nodes.
Firstly, $P(\lnot A, C, D) = P(\lnot A, C, D, B) + P(\lnot A, C, D, \lnot B)$, since either $B$ must occur or not occur.
We can calculate each of the probabilities on the RHS as follows. The Chain Rule gives us:
\begin{eqnarray*} P(D, C, B, \lnot A) &=& P(D \mid C, B, \lnot A) \times P(C \mid B, \lnot A) \times P(B \mid \lnot A) \times P(\lnot A) \\ &=& P(D \mid C, B) \times P(C \mid B, \lnot A) \times P(B) \times P(\lnot A) \\ &=& 0.1 \times 0.5 \times 0.8 \times 0.7 \\ &=& 0.028 \end{eqnarray*}
\begin{eqnarray*} P(D, C, \lnot B, \lnot A) &=& P(D \mid C, \lnot B, \lnot A) \times P(C \mid \lnot B, \lnot A) \times P(\lnot B \mid \lnot A) \times P(\lnot A) \\ &=& P(D \mid C, \lnot B) \times P(C \mid \lnot B, \lnot A) \times P(\lnot B) \times P(\lnot A) \\ &=& 0.6 \times 0.1 \times 0.8 \times 0.3 \\ &=& 0.0144 \end{eqnarray*}
The second step in each of these calculations uses the Bayesian network independence rules. E.g. $D$ and $\lnot A$ are conditionally independent given $C$, and also $\lnot A$ and $B$ are independent events.
Summing these two results, we get the answer:
$$P(\lnot A, C, D) = 0.028 + 0.0144 = 0.0424$$