There are several ways to solve this question. I will try to explain it in simple terms:

We have 4 cases to analyze:

1) The probability of picking only fiction books.
2) The probability of picking at least a fiction book and a biography.
3) The probability of picking at least a fiction book and non-biography books.
4) The probability of picking a fiction book, a biography book, and a non-biography book.

With those cases we have covered all the possibilities. Lets calculate them:

1) The probability of picking only fiction books.
\(\frac{4}{10}\frac{3}{9}\frac{2}{8} = \frac{1}{30}\)
2) The probability of picking at least a fiction book and a biography.
We have 3 combinations in this case (\(B\) stands for Biography, and \(F\) for fiction):
\({B,F,F}\)
\({F,B,F}\)
\({F,F,B}\)
\((\frac{4}{10}\frac{3}{9}\frac{3}{8}) *3 = \frac{3}{20}\)
3) The probability of picking at least a fiction book and non-byography books.
We have 6 cases (\(N_{b}\) stands for non-biography):
\({N_{b},F,F}\)
\({F,N_{b},F}\)
\({F,F,N_{b}}\)
whose probability could be written as:
\((\frac{4}{10}\frac{3}{9}\frac{3}{8}) *3 = \frac{3}{20}\)

and also, we have:
\({N_{b},N_{b},F}\)
\({N_{b},F,N_{b}}\)
\({F,N_{b},N_{b}}\)
whose probability could be written as:
\((\frac{3}{10}\frac{2}{9}\frac{4}{8}) *3 = \frac{1}{10}\)
4) The probability of picking a fiction book, a biography book, and a non-biography book.
In this case, we have 6 combinations (\(3!\) because we have 3 different items that have to be mixed):
\({F,N_{b},B}\)
\({F,B,N_{b}}\)
\({N_{b},F,B}\)
\({B,N_{b},F}\)
\({B,F,N_{b}}\)
\({N_{b},B,F}\)
and the probability is:
\((\frac{4}{10}\frac{3}{9}\frac{3}{8}) *6 = \frac{6}{20}\)

Now, we have to sum each probability in order to consider all the cases:
\(\frac{1}{30} + \frac{3}{20} + \frac{3}{30} + \frac{1}{10} + \frac{6}{20} = \frac{44}{60} = \frac{11}{15}\)