computer system uses passwords that contain exactly eight

Applied Statistics and Probability for Engineers | 6th Edition | ISBN: 9781118539712 | Authors: Douglas C. Montgomery, George C. Runger

Problem 141E Chapter 2.5

Applied Statistics and Probability for Engineers | 6th Edition

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Applied Statistics and Probability for Engineers | 6th Edition | ISBN: 9781118539712 | Authors: Douglas C. Montgomery, George C. Runger

Applied Statistics and Probability for Engineers | 6th Edition

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Problem 141E

computer system uses passwords that contain exactly eight characters, and each character is one of the 26 lowercase letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9). Let ? denote the set of all possible password, and let A and B denote the events that consist of passwords with only letters or only integers, respectively. Suppose that all passwords in ? are equally likely. Determine the following robabilities:

Step-by-Step Solution:
Step 1 of 3

Solution 141E

Step1 of 4:

From the given problem we have:

The number of characteristics contain a password = 8,

Lower case letters = 26,

Upper case letters = 26, and

Integers = 10.

Here our goal is:

a). We need to find

b). We need to find

c). We need to find the probability of password contains exactly 2 integers given that it contains at least 1 integer.

Step2 of 4:

a).

Let us define events are:

A = the event that password consists only letters.

B = the event that password consists only integers.

Now,

The total number of characteristics = 26 + 26 + 10

                                                         = 62.

The possible number of 8 digit password =   [Because digits are independent to each other]

The number of characteristics in letters = 26 + 26

                                                                = 52.

The possible number of 8 digit letters password =

The number of characteristics in Integers = 10

The possible number of 8 digit Integer password =

Conside,

                                     

                                                          …..(1)

Where,

                                           

                                                   

                                                   

Hence,

Similarly,

                                         [Because A and B are mutually Exclusive events ]  

Finally,

                                           

                                                   

                                                   

                                     

                                                            0

Hence,

Substitute these values in equation (1), then we get:

                                 

               ...

Step 2 of 3

Chapter 2.5, Problem 141E is Solved
Step 3 of 3

Textbook: Applied Statistics and Probability for Engineers
Edition: 6th
Author: Douglas C. Montgomery, George C. Runger
ISBN: 9781118539712

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