Answer:
a. P(1 Sports illustrated and 3 Newsweek) = 
b. P(at least 2 Newsweek) = 
We follow these steps to arrive at the answer:
The total number of magazines is 
Of these we need to choose 4 books randomly. Since order doesn’t matter here, we use combinations.
There are
ways to choose 4 books.

1. 1. We need to choose 1 Sports illustrated and 3 newsweeks.
We can choose 1 Sports illustrated from 5 in
ways.

We can choose 3 issues of Newsweek in
ways.

So,



2.<u> P(at least 2 Newsweek)</u>
Here we divide the number of books into two categories. One is the Newsweek Magazines (6), the other group consists of all other books (8).
![P(at least 2 Newsweek) = 1 - [P(no Newsweek) + P(one newsweek)]](https://tex.z-dn.net/?f=P%28at%20least%202%20Newsweek%29%20%3D%201%20-%20%5BP%28no%20Newsweek%29%20%2B%20P%28one%20newsweek%29%5D)
Now, if we have to find P(no newsweek), we first need to find the number of ways we can select 4 books from the group that does NOT contain Newsweek magazines.
We can calculate that by:

So,

We can compute P(1 Newsweek) as follows:
We can choose 1 Newsweek in 
We can choose 3 other books in ![_{8}C_{3} =\frac{8!}{5!3!} = 56 ways So, P(at least 1 Newsweek) = [tex]\frac{6*56}{1001} or \frac{336}{1001}](https://tex.z-dn.net/?f=_%7B8%7DC_%7B3%7D%20%3D%5Cfrac%7B8%21%7D%7B5%213%21%7D%20%3D%2056%20ways%20%3C%2Fstrong%3E%20%3C%2Fp%3E%3Cp%3ESo%2C%3Cstrong%3E%20P%28at%20least%201%20Newsweek%29%20%3D%20%5Btex%5D%5Cfrac%7B6%2A56%7D%7B1001%7D%20or%20%5Cfrac%7B336%7D%7B1001%7D)
Since
,
