Question

Lindsay is checking out the books at the library, and she's probably interested in Mysteries and nonfiction. She knows her selection down to 11 Mysteries and 6 non-fiction books. Is she randomly chooses four books from her selection, what's the probability that they will all be Mysteries?As a fraction or round your answer to four decimal places if necessary

Answer 1

We have a group of books that is conformed by 11 books about mysteries and 6 non-fiction books.

If she chooses 4 books, we have to calculate the probaiblity that all of them are mystery books.

We can calculate this probability as the quotient between the combinations that include 4 mystery books and all the possible combinations.

We can start calculating all the possible combinations of 4 books from a group of 17 books. It can be calculated as:

$C(17;4)=\frac{17!}{4!(17-4)!}=\frac{17!}{4!13!}=2380$

Then, we can calculate the combinations that only include mystery books by excluding the non-fiction books from the options. Now, we have only 11 options fo fill the selection of 4 books, so the possible combinations are:

$C(11;4)=\frac{11!}{4!(11-4)!}=\frac{11!}{4!7!}=330$

Then, we can calculate the probability as the quotient between the combinations we have just calculated:

$P=\frac{C(11;4)}{C(17;4)}=\frac{330}{2380}=\frac{33}{238}\approx0.1387$

Answer: the probability that the 4 books are mystery books is P = 33/238 or approximately 0.1387.