Chapter 13.2, Problem 69SR

a)

To determine

To find : Probability of ( $4$ blue, $3$ red, $3$ yellow in that) order

Answer to Problem 69SR

  $\frac{\text{1}}{\text{4200}}$

Explanation of Solution

Given information :

An administrative assistant has $4$ blue file folders, $3$ red folders and $3$ yellow folders.

Each file contains different information. She put files randomly in a box.

Calculation:

P( $4$ blue, $3$ red, $3$ yellow in that order)

  $\begin{array}{l}\text{Total files = (}4+3+3\text{) = 10 files}\\ n(s)=10!(\because \text{ in 10! ways file can be kept})\\ Now,\end{array}$

  

  $\begin{array}{l}n(\epsilon )=4!3!3!\\ \left[\begin{array}{l}\because \text{4 blue files can be kept by 4! ways}\\ 3\text{ red files can be kept by 3! ways}\\ 3\text{ yellow files can be kept by 3! ways}\end{array}\right]\\ P\text{(4 Red, 3 Blue, 3 Yellow)}=\frac{n(\epsilon )}{n(s)}=\frac{4!3!3!}{10!}=\frac{1}{4200}\end{array}$

b)

To determine

To find : Probability of (first $2$ blue, last $2$ blue)

Answer to Problem 69SR

  $\frac{\text{1}}{\text{1360}}$

Explanation of Solution

Given Information :

We know that, $3$ Red file folders, $4$ Blue files, $3$ Yellow files are kept randomly.

Calculation:

P(first $2$ blue, last $2$ blue)

  $\begin{array}{l}\text{n(S) = 10!}\text{[}\because \text{ total no}\text{. of files are 10]}\end{array}$

  

  $\begin{array}{l}n(\epsilon )=2!6!2!\\ \left[\begin{array}{l}\because \text{2 Blue files at first can be kept by 2!}\text{.}\\ \text{6 files in between blue files can be kept by 6!}\text{.}\\ \text{Last 2 blue files can be kept by 2!}\end{array}\right]\\ P(\epsilon )=\frac{2!6!2!}{10!}=\frac{1}{1360}\end{array}$