# Simple Bayes network

Given the following Bayes network:

with

• $p(k=t)=.2$
• $p(o=t)=.1$
• $p(s=t|k=f,o=f)=.0$
• $p(s=t|k=f,o=t)=.2$
• $p(s=t|k=t,o=f)=.5$
• $p(s=t|k=t,o=t)=.95$

how would I calculate $p(s=t|o=t)$ and $p(o=t|s=t)$?

I tried the following way:

$p(s=t|o=t) = \dfrac{p(s=t \land o=t)}{p(o=t)} = \dfrac{p(s=t)p( o=t | s=t)}{p(o=t)}$, which obviously doesn't make sense, since I have to caluclate $p( o=t | s=t)$ with Bayes rule.

From the graph you know $P(s|k,o)$, so to get $P(s|o)$ you need to sum over all possible values of $k$, it will be then:
$$P(s=t|o=t) = \sum_{i \in {t,f}} P(s=t|o=t,k=i)$$
$$P(s=t|o=t) = P(s=t|o=t,k=f) + P(s=t|o=t,k=t)$$ In your case it doesn't sum to one your distribution, I think you should change the values.