Calculate probability using brute-force method I have to calculate a probability using the following probabilities from a Bayesian network:
$$P(+a)=...$$
$$P(+a|+b)=..., P(+a|¬b)=...$$
$$P(+b|+a)=..., P(+b|¬a)=...$$
$$P(+d|+b)=..., P(+d|¬b)=...$$
$$P(+f|+b,+c)=..., P(+f|¬b,+c)=..., P(+f|¬b,¬c)=...$$
$$P(+g|+c)=..., P(+g|¬c)=...$$
I have to calculate $P(a|+d,+f,¬g)$.
I have no idea about how to do it, but I guess it is something like that:
$$P(a|+d,+f,¬g)=\frac{P(a,+d,+f,¬g)}{P(+d,+f,¬g)}$$
And then, I have to calculate $P(a,+d,+f,¬g)$ and $P(+d,+f,¬g)$ using the previous probabilities. Is that correct?
But the $P(a|$ is confusing me because it hasn't a + or a ¬.
NOTE: you don't have to elaborate each formula, only point me to the right direction.
 A: The Bayesian network tells you that the joint distribution factorizes according to
$$P(a,b,c,d,f,g) = P(a)P(b|a)P(c|a) P(d|b)P(f|b,c) P(g|c),$$
So you can compute any truth assignments using the conditional distributions you have presented.
For instance
$$P(+a, \neg b, +c, \neg d, +f, +g) = P(+a)P(\neg b|+ a)P(+ c| + a) P(\neg d| \neg b)P(+ f|\neg b,+c) P(+g|+ c).$$
Note that you have all these probabilities: for those that you do not have explicit, consider the normalization constraint that tells you that $P(\neg E) = 1 - P(E)$.
You have correctly written that 
$$P(a| +d, +f, \neg g) = \frac{P(a,+d,+f,\neg g)}{P(+d,+f,\neg g)}.$$
Note that this is a short-hand that represents two probabilities: $P(+a|+d,+f,\neg g)$ and $P(\neg a|+d,+f,\neg g)$. By normalization, you only need to compute one of them, and the other one follows by $P(+a|+d,+f,\neg g) = 1 - P(\neg a|+d,+f,\neg g)$; so we only look at  $P(+a| +d, +f, \neg g)$.
Now we need observe that the numerator and denominator probability involve a subset of the variables. This means that we must marginalize. We need to remove the variables that do not appear using the sum-rule of probability theory:
$P(A) = \sum_B P(A, B).$
In your case, we have that
$$P(+a, +d, +f, \neg g) = \sum_{b,c} P(+a, b, c +d, +f, \neg g),$$
where the summation sign means that we sum over all possible outcomes of $b$, and $c$:
$$P(+a, +d, +f, \neg g) = 
P(+a, +b, +c +d, +f, \neg g) +
P(+a, +b, \neg c +d, +f, \neg g) +
P(+a, \neg b, +c +d, +f, \neg g) +
P(+a, \neg b, \neg c +d, +f, \neg g).
$$
To compute each one of these probabilities, you use the definition of the joint at the beginning of the answer.
Similarly, you can compute $P(\neg a, +d, +f, \neg g)$. Then, you use the same rule to compute
$$P(+d, +f, \neg g) = P(+a, +d, +f, \neg g) + P(\neg a, +d, +f, \neg g).$$
It's a lot of products and sums :) good luck!
