I am self-studying graphical models and I have come across the, what seems to be famous, "Student Network". I have included a picture of this graph below. There is a question that asks to compute $$p(\text{Intelligence}=1| \text{Letter}=1, \text{SAT}=1)$$. My question is - do I have to start from the joint probability that can be read from the model namely that (i use abbreviations, first letter, for the naming) $$p(I, D, G, S, L) = p(D)p(I)p(G|I, D)p(S|T)p(L|G)$$ And rewrite this probability until i get the necessary parts to compute what I want, which I have attempted but gotten stuck with, or is there something I am missing from the graph or tables that makes the computation simpler?
You have, using the same subscript notation as your table, that
$$\begin{align}p(\imath^I, d^D, g^G, s^S, \ell^L) &= p(d^D)~p(\imath^I)~p(g^G\mid \imath^I, d^D)~p(s^S\mid \imath^I)~p(\ell^L\mid g^G)\\[1ex]&= p(\imath^I)~p(s^S\mid \imath^I)~p(\ell^L\mid g^G)~p(d^D)~p(g^G\mid \imath^I, d^D)\end{align}$$
So by marginalising
$$\begin{align}p(\imath^I, s^1, \ell^1) &=p(\imath^I)~p(s^1\mid \imath^I)\sum_{G} p(\ell^1\mid g^G)\sum_D p(d^D)~p(g^G\mid \imath^I, d^D)\\&= p(\imath^I)~p(s^1\mid \imath^I)~p(\ell^1\mid \imath^I)\\[2ex]p(i^1\mid s^1, \ell^1)&=\dfrac{p(\imath^1)~p(s^1\mid \imath^1)~p(\ell^1\mid \imath^1)}{p(\imath^0)~p(s^1\mid \imath^0)~p(\ell^1\mid\imath^0)+p(\imath^1)~p(s^1\mid \imath^1)~p(\ell^1\mid\imath^1)}\\[2ex]\text{where:}\\[2ex]p(\ell^1\mid\imath^I)&=\sum_G p(\ell^1\mid g^G)\sum_D p(d^D)~p(g^G\mid\imath^I, d^D)\\[1ex]&={p(\ell^1\mid g^0)~\big(p(d^0)~p(g^0\mid\imath^I, d^0)+p(d^1)~p(g^0\mid\imath^I, d^1)\big)+p(\ell^1\mid g^1)~\big(p(d^0)~p(g^1\mid\imath^I, d^0)+p(d^1)~p(g^1\mid\imath^I, d^1)\big)}\end{align}$$
Everything else is reading the tables and crunching the numbers.