+118703 Yes it would be very difficult. Maybe you select the book off the shelf but then read it off a kindle. That way the selected book would never leave the shelf and you could select 4 of the same book. You could almost recite it after reading it 4 times. 
Then your probability would be correct. :)
+118703 Since you have told me that you are 12 you will not understand this but I will show you.
The number of ways that 4 books can be selected from 15 is
15C4=15!4!11! on the calc this is (15!4!×(15−4)!)=1365
The number of ways that 2 Lewis Carroll books can be chosen from the 5 is 5C2 = (5!2!×(5−2)!)=10
The number of ways the other 2 can be chosen from the other 10 books is 10C2 = (10!2!×(10−2)!)=45
so the number of ways that the 2 Lewis Carroll books can be chosen is 45*10 = 450
The probability of choosing exactly 2 Lewis Carroll books is 4501365=3091=0.3296703296703297
so the answer is 3091
+1038 Probability of selecting one Lewis Carroll book (1/3)
Probability of NOT selecting one Lewis Carroll book (1-(1/3))
Probability of choosing exactly 2 Lewis Carroll books from 4 books
\Text {Solution }\\\\ \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}\; \hspace{15pt}| \hspace{15pt} \Text {n=4 \; k=2\\
(42)∗(13)2∗(23)2=827≈0.29629
+118703 Hi Nauseated,
I do not believe your answer is correct.
Binomial expansion are only used when the trials are independent. :)
+1038 Oh, that’s right. The book would have to be replaced after it was selected.
It softly and suddenly vanishes away after each selection like the baker, after the boojum variety of snark selects him.
Makes it kind of difficult to select 4 books for reading, doesn’t it?
+118703 Yes it would be very difficult. Maybe you select the book off the shelf but then read it off a kindle. That way the selected book would never leave the shelf and you could select 4 of the same book. You could almost recite it after reading it 4 times.
Then your probability would be correct. :)
