# Bayesian Network

I am preparing for midterm exam and need to know what is the step by step solution to this question? Answer is shown in red. Also any external related link is very much appreciated.

First, there is a standard factorization of probability that always holds:

$$P(\sim A , \sim B , \sim C) = P(\sim A) P(\sim B | \sim A)P(\sim C | \sim B, \sim A)$$

Given the shape of the Bayesian network, we conclude that $C$ is independent of $A$ given $B$. That means:

$P(\sim C | \sim B, \sim A) = P(\sim C | \sim B)$

Since we know $P(A) = 0.3$, we also know $P(\sim A) = 0.7$. Furthermore $P(\sim B | \sim A) = 1 - P(B | \sim A) = 1 - 0.5 = 0.5$. Finally $P(\sim C | \sim B) = 1 - P(C | \sim B) = 1 - 0.1 = 0.9$

Putting it all together we get $$P(\sim A , \sim B \sim C) = 0.7 \cdot 0.5 \cdot 0.9 = 0.315$$

• Where is this rule coming from? P(∼A,∼B,∼C)=P(∼A)P(∼B|∼A)P(∼C|∼B,∼A) – Mona Jalal Oct 19 '14 at 22:21
• Also how did you get to this conclusion? can you please explain further? Given the shape of the Bayesian network, we conclude that C is independent of A given B. – Mona Jalal Oct 19 '14 at 22:21
• The rule is called the chain rule (en.wikipedia.org/wiki/Chain_rule_%28probability%29). The independence comes from the Markov condition: a node is independence of its non-descendants given its parents. – George Oct 19 '14 at 22:47
• Thanks. Helped me understand the steps to attack Bayesian networks. The A->B->C implies C & is d-separated by B was a killer. – KarthikS Jun 7 '15 at 6:11