Conditional probability given two variables from conditional probabilities of one variable

I have a question which comes from an example of a textbook, but all I am concerned with is how we go from having probabilities $P(X|A)$ and $P(X|B)$ to having $P(X|A,B)$. In the example events $A$ and $B$ are independent, each influencing $X$. The example comes from Pattern Classification by Duda et. al, edition 2.

The example gives us a table defining $P(x_i|a_j)$ and $P(x_i|b_j)$. Shortly thereafter the authors have a summation over $P(x_1|a_i,b_j)$:

$$\sum\limits_{i,j}P(x_1|a_i, b_j)$$

Later, they expand this sum into the particular values, which appear to me to be the values of $P(x_i|a_j)$. The numbers end up adding up to 4 which makes sense since there are four rows in $P(x_i|a_j)$ (each which must sum to 1).

The probability $P(x_1|a_i, b_j)$ makes a second appearance in the example, but unfortunately the authors have left a magic step and I am unsure of the values they used.

My question is why is this OK to do? What is the relationship between $P(X|A)$, $P(X|B)$, and $P(X|A,B)$? Do we need more information to get $P(X|A,B)$ and are they doing some sort of approximation?

The section is dealing with Bayesian belief networks. If you happen to have the book, it is Example 4 in Section 2.11.

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I may have figured something out something unrelated to this question that may help me, though this question is still on the table and I'd really be interested to know the answer. – OEP Mar 22 '13 at 4:49
In case anyone is struggling with the same problem, I'm pretty certain now it is an error with the book. There is a revised section 2.11 with an example that's been almost completely rewritten. – OEP Mar 22 '13 at 13:55
I can see this Example 4 with Google Books but I don't see this summation. – Stéphane Laurent Mar 22 '13 at 14:18
You are right. I think I have an old version of the book which has errors in it. I'm currently separated from my newer version so that's probably the problem. – OEP Mar 22 '13 at 16:48