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:
$\frac{1}{30} + \frac{3}{20} + \frac{3}{30} + \frac{1}{10} + \frac{6}{20} = \frac{44}{60} = \frac{11}{15}$