/*! 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 163 A Wired Equivalent Privacy (WEP)... [FREE SOLUTION] | 91Ó°ÊÓ

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A Wired Equivalent Privacy (WEP) key is a security code that one must enter in order to access a secure WiFi network. The characters in the key are used from the numbers 0 to 9 and letters from \(\mathrm{A}\) to \(\mathrm{F}\), which gives 16 possibilities for each character of the key. Note that repeats are allowed, that is, the same letter or number can be used more than once in a key. A WEP key for a WiFi network with 64 -bit security is 10 characters long. a. How many different 64 -bit WEP keys can be made by using the given numbers and letters? b. A specific 64 -bit network has a WEP key in which the 2 nd, 5 th, 8 th, and 9 th characters are numbers and the other 6 characters are letters. How many different WEP keys are possible for this network? c. A hacker has determined that the WiFi network mentioned in part b will lock him out if he makes 20,000 unsuccessful attempts to break into the network. What is the probability that the hacker will be locked out of the network?

Short Answer

Expert verified
a) There are \(16^{10}\) different 64-bit WEP keys. b) There are \(10^4 * 6^6\) different WEP keys for the specific network. c) The probability that the hacker will be locked out is \(20000 / (10^4 * 6^6)\).

Step by step solution

01

Determine total combinations for a 64-bit WEP key

A WEP key for a WiFi network with 64-bit security is 10 characters long. The characters in the key consist of the numbers 0 to 9 and letters from A to F, which gives 16 possible characters for each character of the key. Each character in the key is chosen from these 16 possibilities, and each character is independent of the others. So the total number of possible keys is the product of the number of possibilities for each character. Therefore, the total number of keys is \(16^{10}\).
02

Calculate specific WEP key combinations

A specific 64-bit network has a WEP key in which the 2nd, 5th, 8th, and 9th characters are numbers and the other 6 characters are letters. This means that for the 2nd, 5th, 8th, and 9th characters we have 10 choices (0-9), and for the remaining 6 characters we have 6 choices (A-F). Applying the multiplication principle again, the number of different WEP keys for this network is \(10^4 * 6^6\).
03

Calculate the probability that a hacker will be locked out

A hacker is locked out after 20,000 unsuccessful attempts. This means we can divide the number of unsuccessful attempts by the total number of possible keys calculated in Step 2 to find the probability of being locked out. The probability that the hacker will be locked out of the network is \(20000 / (10^4 * 6^6)\).

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Key Concepts

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

Probability
Probability is the measure of the likelihood of an event occurring. It's expressed as a fraction or percentage denoting how often one outcome will happen compared to the possible outcomes. In simple terms, if there are 10 possible events and only 1 is favorable, the probability of that event is 1/10 or 10%. In the context of network security, probability can describe how likely a hacker is to guess a correct WEP key by brute force. The more possible keys, the lower the probability of guessing correctly. The formula used is:
  • The number of successful outcomes
  • Divided by the total number of possible outcomes
By calculating probabilities, network administrators can understand how secure a network is against random guessing attempts.
Network Security
Network security is essential in protecting data and resources from unauthorized access and attacks. A secure network uses various protocols and encryption methods to keep information safe. WEP, or Wired Equivalent Privacy, is a network security protocol for wireless networks. It was one of the initial attempts to secure wireless networks. However, over time, vulnerabilities were found, and more advanced protocols like WPA or WPA2 are used nowadays for better security.
In ensuring network security, each access point employs security keys such as the WEP key, which uses a combination of letters and numbers to restrict access. Regular updating and on-time patching of security practices are vital to safeguarding network integrity.
Permutations
Permutations refer to different arrangements or orderings of a set of elements. For example, if you have the letters A, B, and C, different permutations would include ABC, ACB, BAC, etc. In the context of a WEP key, the characters can be arranged in many different ways. Each arrangement represents a unique key. Since each position in the key is independent, the number of total permutations is determined by raising the number of choices for each character to the power of the length of the key.
Problems involving permutations are fundamental in understanding combinatorics concepts crucial for creating and evaluating secure access codes.
Cryptography
Cryptography is the science of encrypting and decrypting information to protect data. It transforms readable data into an unreadable format, making it accessible only to authorized individuals. WEP keys use cryptography to guard wireless communications. Encryption methods like WEP masks data with a code, requiring someone attempting to access the network to input the correct key.
However, cryptography is ever-evolving. As vulnerabilities in systems like WEP are discovered, new encryption methods are developed. Keeping up-to-date with cryptographic advancements is crucial for maintaining effective security measures against potential breaches.

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Most popular questions from this chapter

Given that \(A, B\), and \(C\) are three independent events, find their joint probability for the following. a. \(P(A)=.81, \quad P(B)=.49\), and \(P(C)=.36\) b. \(P(A)=.02, \quad P(B)=.03\), and \(P(C)=.05\)

Many states have a lottery game, usually called a Pick-4, in which you pick a four-digit number such as 7359 . During the lottery drawing, there are four bins, each containing balls numbered 0 through 9\. One ball is drawn from each bin to form the four-digit winning number. a. You purchase one ticket with one four-digit number. What is the probability that you will win this lottery game? b. There are many variations of this game. The primary variation allows you to win if the four digits in your number are selected in any order as long as they are the same four digits as obtained by the lottery agency. For example, if you pick four digits making the number 1265 , then you will win if \(1265,2615,5216,6521\), and so forth, are drawn. The variations of the lottery game depend on how many unique digits are in your number. Consider the following four different versions of this game. i. All four digits are unique (e.g., 1234 ) ii. Exactly one of the digits appears twice (e.g., 1223 or 9095 ) iii. Two digits each appear twice (e.g., 2121 or 5588 ) iv. One digit appears three times (e.g., 3335 or 2722 ) Find the probability that you will win this lottery in each of these four situations.

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Given that \(P(A \mid B)=.44\) and \(P(A\) and \(B)=.33\), find \(P(B)\).

A gambler has four cards - two diamonds and two clubs. The gambler proposes the following game to you: You will leave the room and the gambler will put the cards face down on a table. When you return to the room, you will pick two cards at random. You will win $$\$ 10$$ if both cards are diamonds, you will win $$\$ 10$$ if both are clubs, and for any other outcome you will lose $$\$ 10$$. Assuming that there is no cheating, should you accept this proposition? Support your answer by calculating your probability of winning $$\$ 10$$.

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