# Conditional Probability in Bayesian Network

I am working on exercise 13.5 from Nong Ye's Data Mining - Theories, Algorithms, and Examples. The problem gives the following Bayesian network

as well as conditional probability tables showing the probabilities of each variable given its parents. All the variables are binary. The problem is to compute $$P(y=1|x_6=1)$$. I have broken up the probability a sum over conditional probabilities of each variable given its parents:

$$P(y=1|x_6=1)=\frac{P(y=1,x_6=1)}{P(x_6=1)}$$ $$=\frac{1}{P(x_6=1)}\sum_{x_1,x_2,x_3,x_4,x_5,x_7,x_8,x_9}P(x_1)P(x_2)P(x_3)P(x_4|x_2,x_3)P(x_5|x_1)P(x_6=1|x_3)P(x_7|x_5,x_6=1)P(x_8|x_4)P(x_9|x_5)P(y=1|x_7,x_8,x_9)$$

In this, I need to compute $$P(y=1|x_7, x_8, x_9)$$. Conditional probability tables are given for $$P(y|x_7)$$, $$P(y|x_8)$$, and $$P(y|x_9)$$, but not for $$P(y|x_7, x_8, x_9)$$. How can $$P(y=1|x_7, x_8, x_9)$$ be broken up into the probabilities given each of it parents individually?

Using Bayes theorem and the definition of conditional probability we can write

$$P(y=1|x_7, x_8, x_9)=\frac{P(y=1,x_7, x_8, x_9)}{P(x_7, x_8, x_9)}=\frac{P(x_7, x_8, x_9|y=1)P(y=1)}{P(x_7, x_8, x_9)}$$

but I'm not sure how to proceed from there.

Although I think that tables in a binary bayesian network should contain $$2^{par(n)}$$ rows where $$par(n)$$ is the number of parents for each node, the book wanted you to solve this problem like other examples in chapter 13. instead of marginalizing on all of the other parameters, you can do this: $$\mathbb{P}(y = 1, x_6 = 1) = \sum_{x_1x_3x_5x_7} \mathbb{P}(y = 1, x_6 = 1, x_1, x_3, x_5, x_7) = \sum_{x_1x_3x_5x_7} \mathbb{P}(x_1)\mathbb{P}(x_3)\mathbb{P}(x_5|x_1)\mathbb{P}(x_6 = 1 | x_3)\mathbb{P}(x_7|x_5, x_6=1)\mathbb{P}(y = 1|x_7)$$ And likewise for the other one: $$\mathbb{P}(x_6 = 1) = \sum_{x_3} \mathbb{P}(x_6 = 1, x_3) = \sum_{x_3} \mathbb{P}(x_6 = 1| x_3)\mathbb{P}(x_3)$$