Question

Assume that there are 5 different issues of sports illustrated, 6 different issues of newsweek, and 3 different issues of time, including the december 1st issue, on the rack. you choose 4 of them at random. (1) what is the probability that you choose 1 issue of sports illustrated and 3 issues of newsweek? (2) what is the probability that you choose at least 2 of the newsweek magazines?

Answer 1

Answer:

a. P(1 Sports illustrated and 3 Newsweek) =  $\frac{100}{1001}$

b. P(at least 2 Newsweek) = $\frac{595}{1001}$

We follow these steps to arrive at the answer:

The total number of magazines is $5+6+3 =14$

Of these we need to choose 4 books randomly. Since order doesn’t matter here, we use combinations.

There are $_{14}C_{4}$ ways to choose 4 books.  

$_{14}C_{4}=\frac{14!}{10!4!} = 1001 ways$

1. 1. We need to choose 1 Sports illustrated and 3 newsweeks.

We can choose 1 Sports illustrated from 5 in $_{5}C_{1}$ ways.

$_{5}C_{1}=\frac{5!}{4!1!} = 5 ways$

We can choose 3 issues of Newsweek in $_{6}C_{3}$ ways.  

$_{6}C_{3}=\frac{6!}{3!3!} = 20 ways$

So,

$P(1 Sports Illustrated and 3 Newsweek) = \frac{_{6}C_{3} * _{5}C_{1}}{_{14}C_{4}}$

$P(1 Sports Illustrated and 3 Newsweek) = \frac{20 * 5}{1001}$

$P(1 Sports Illustrated and 3 Newsweek) = \frac{100}{1001}$

2. P(at least 2 Newsweek)

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).

$P(at least 2 Newsweek) = 1 - [P(no Newsweek) + P(one newsweek)]$

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:

${ _{8}C_{4} = \frac{8!}{4!4!} = 70 ways$

So,

$P(no newsweek) = \frac{70}{1001}$

We can compute P(1 Newsweek) as follows:

We can choose 1 Newsweek in $_{6}C_{1} =\frac{6!}{5!1!} = 6 ways$

We can choose 3 other books in $_{8}C_{3} =\frac{8!}{5!3!} = 56 ways So, P(at least 1 Newsweek) = [tex]\frac{6*56}{1001} or \frac{336}{1001}$

Since

$P(at least 2 Newsweek) = 1 - [P(no Newsweek) + P(one newsweek)]$,

$P(at least 2 Newsweek) = 1 - [\frac{70}{1001} + \frac{336}{1001}]$

$P(at least 2 Newsweek) = 1 - [\frac{406}{1001}]$

$P(at least 2 Newsweek) = \frac{595}{1001}$