/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A computer system uses passwords... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 7

A computer system uses passwords that are six characters, and each character is one of the 26 letters \((a-z)\) or 10 integers \((0-9)\). Uppercase letters are not used. Let \(A\) denote the event that a password begins with a vowel (either \(a, e, i, o,\) or \(u\) ), and let \(B\) denote the event that a password ends with an even number (either \(0,2,4,6,\) or 8 ). Suppose a hacker selects a password at random. Determine the following probabilities: a. \(P(A)\) b. \(P(B)\) c. \(P(A \cap B)\) d. \(P(A \cup B)\)

Short Answer

Expert verified
a. \(P(A) = \frac{5}{36}\); b. \(P(B) = \frac{5}{36}\); c. \(P(A \cap B) = \frac{25}{1296}\); d. \(P(A \cup B) = \frac{335}{1296}\).

Step by step solution

01

Determine the Total Number of Passwords

Each of the six characters in the password can be either one of 26 letters \((a-z)\) or one of 10 digits \((0-9)\). So, there are a total of \(26 + 10 = 36\) possible characters for each position in the password. The total number of possible passwords is then \(36^6\).
02

Calculate Probability of Event A

Event \(A\) occurs if a password begins with one of the 5 vowels \((a, e, i, o, u)\). There are 5 choices for the first character and 36 choices for each of the remaining 5 characters. Thus, the number of favorable outcomes for \(A\) is \(5 \cdot 36^5\). The probability \(P(A)\) is given by \(\frac{5 \cdot 36^5}{36^6}\).
03

Calculate Probability of Event B

Event \(B\) occurs if a password ends with an even digit \((0, 2, 4, 6, 8)\). There are 5 choices for the last character and 36 choices for each of the first 5 characters. Hence, the number of favorable outcomes for \(B\) is \(36^5 \cdot 5\). The probability \(P(B)\) is \(\frac{36^5 \cdot 5}{36^6}\).
04

Calculate Probability of Event A and B (Intersection)

\(P(A \cap B)\) represents passwords that begin with a vowel and end with an even number. There are 5 choices for the first character and 5 choices for the last character, with \(36\) choices for each of the 4 middle characters. The number of favorable outcomes for \(A \cap B\) is \(5 \cdot 36^4 \cdot 5\). Thus, \(P(A \cap B) = \frac{5 \cdot 36^4 \cdot 5}{36^6}\).
05

Calculate Probability of Event A or B (Union)

Using the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), we substitute the values from the previous steps. Calculate \(P(A)\), \(P(B)\), and \(P(A \cap B)\), then substitute these into the formula to get \(P(A \cup B)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Password Security
Password security is crucial for maintaining the confidentiality and integrity of information systems. A strong password is critical in safeguarding against unauthorized access. In today's digital environment, this means choosing passwords that are both complex and unique. For the system in this exercise, passwords are made up of six characters chosen from lowercase letters or numbers. This setup provides a significant number of combinations, making it difficult for a hacker to guess the correct password. However, even with these combinations, it's important to regularly update passwords and employ additional security measures, such as two-factor authentication, to further enhance security.
Event Probability Calculation
In probability, an event is a set of outcomes to which a probability is assigned. Calculating the probability of an event involves determining how likely it is to occur when compared to the total number of possible outcomes. For example, to compute the probability that a password begins with a vowel, we first consider the total number of possible passwords. In this scenario, the password can have any of 36 characters (26 letters + 10 digits) in each of its six positions, totaling to 366 possible combinations. The event in question occurs when one of these positions meets a specific criteria, such as starting with a vowel. By evaluating the number of ways this can happen, we can find the probability of this event.
Union and Intersection of Events
The concepts of union and intersection help us understand the probability of combined events. The intersection of two events, denoted as \( A \cap B \), refers to the probability of both events occurring simultaneously. For instance, in this exercise, it is the event where a password starts with a vowel and ends with an even number. The union of two events, denoted as \( A \cup B \), represents the probability of either event occurring, or both. To calculate \( P(A \cup B) \), we add the probabilities of the individual events and then subtract the probability of their intersection to avoid overcounting. This formula provides a useful tool in combining probabilities into a cohesive solution.
Combinatorics in Probability
Combinatorics plays a vital role in calculating probabilities, especially when dealing with password security. It involves counting and arranging distinct objects into sets, a method we use extensively in determining the likelihood of specific events. For the example of our six-character password, combinatorics helps compute the total number of configurations a password can have. When determining the probability of a password beginning with a vowel or ending with an even digit, we use combinatorial methods to count the number of ways these specific conditions can be fulfilled. This approach allows us to understand the complexity and vastness of possibilities in password configuration, which is crucial for ensuring robust security measures.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A robotic insertion tool contains 10 primary components. The probability that any component fails during the warranty period is \(0.01 .\) Assume that the components fail independently and that the tool fails if any component fails. What is the probability that the tool fails during the warranty period?

Incoming calls to a customer service center are classified as complaints (75\% of calls) or requests for information ( \(25 \%\) of calls). Of the complaints, \(40 \%\) deal with computer equipment that does not respond and \(57 \%\) deal with incomplete software installation; in the remaining \(3 \%\) of complaints, the user has improperly followed the installation instructions. The requests for information are evenly divided on technical questions ( \(50 \%\) ) and requests to purchase more products \((50 \%)\) a. What is the probability that an incoming call to the customer service center will be from a customer who has not followed installation instructions properly? b. Find the probability that an incoming call is a request for purchasing more products.

The edge roughness of slit paper products increases as knife blades wear. Only \(1 \%\) of products slit with new blades have rough edges, \(3 \%\) of products slit with blades of average sharpness exhibit roughness, and \(5 \%\) of products slit with worn blades exhibit roughness. If \(25 \%\) of the blades in manufacturing are new, \(60 \%\) are of average sharpness, and \(15 \%\) are worn, what is the proportion of products that exhibit edge roughness?

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in \(60 \%\) of the spam messages and only \(4 \%\) of the valid messages. Also, \(20 \%\) of the messages are spam. Determine the following probabilities: a. The message contains free. b. The message is spam given that it contains free. c. The message is valid given that it does not contain free.

A maintenance firm has gathered the following information regarding the failure mechanisms for air conditioning systems: $$ \begin{array}{lccc} & & \text { Evidence of Gas Leaks } \\ & & \text { Yes } & \text { No } \\ \text { Evidence of } & \text { Yes } & 55 & 17 \\ \text { electrical failure } & \text { No } & 32 & 3 \end{array} $$ The units without evidence of gas leaks or electrical failure showed other types of failure. If this is a representative sample of AC failure, find the probability a. That failure involves a gas leak b. That there is evidence of electrical failure given that there was a gas leak c. That there is evidence of a gas leak given that there is evidence of electrical failure
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.