I'm reading through the Koeller and Friedman PGM book and there's an example PGM in Chapter 3. The student example. On page 54 the authors calculate $P(i^1|g^3) \approx 0.079$. The text states a priori $P(i^1)=0.3$ and the measurement is $g=3$.
The PGM looks like this:
First of all it should be obvious, that the value of $P(g^3)=1.22$ must be wrong, since it is greater than $1$. We use the law of total probability to obtain
$$P(g^3)=P(g^3|i^0,d^0)\cdot P(i^0)\cdot P(d^0)+P(g^3|i^0,d^1)\cdot P(i^0)\cdot P(d^1)$$
$$+P(g^3|i^1,d^0)\cdot P(i^1)\cdot p(d^0)+P(g^3|i^1,d^0)\cdot P(i^1)\cdot P(d^0)$$
$$=0.3\cdot 0.7\cdot 0.6+0.7\cdot 0.7\cdot 0.4+0.02\cdot 0.3\cdot 0.6+0.2\cdot 0.3\cdot 0.4$$
$P(g^3|i^l,d^k)$ are the values of the coloured column. It is worth to mention, that $P(i^l|d^k)=P(i^l)$.
To calculate $P(i^1, g^3)$ we just look for the summands of $P(g^3)$ which contains the event $i^1$: $$P(i^1, g^3)=0.02\cdot 0.3\cdot 0.6+0.2\cdot 0.3\cdot 0.4$$ Thus in total we have
$P(i^1|g^3)=\frac{0.02\cdot 0.3\cdot 0.6+0.2\cdot 0.3\cdot 0.4}{0.3\cdot 0.7\cdot 0.6+0.7\cdot 0.7\cdot 0.4+0.02\cdot 0.3\cdot 0.6+0.2\cdot 0.3\cdot 0.4}=0.07894...\approx 0.079=7.9\%$