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A 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 \(\Omega\) denote the set of all possible passwords. Suppose that all passwords in \(\Omega\) are equally likely. Determine the probability for each of the following: (a) Password contains all lowercase letters given that it contains only letters (b) Password contains at least 1 uppercase letter given that it contains only letters (c) Password contains only even numbers given that is contains all numbers

Short Answer

Expert verified
(a) \( \frac{26^8}{52^8} \); (b) \( 1 - \frac{26^8}{52^8} \); (c) \( \frac{5^8}{10^8} \)

Step by step solution

01

Determine Total Possible Passwords

To find the total number of possible passwords, we consider each character position in the password. Each character can be one of the 52 letters (26 lowercase + 26 uppercase) or 10 digits. Therefore, there are 62 possible characters per position in the password. Since the password is exactly eight characters long, the total number of possible passwords is given by the formula:\[ |\Omega| = 62^8 \]
02

Calculate (a) Password Only with Letters

If a password contains only letters, it can either be lowercase or uppercase but not numbers. Thus, each position in the password has 52 possible choices (26 lowercase + 26 uppercase). Therefore, the total number of such passwords is:\[ 52^8 \]
03

Compute (a) Probability of All Lowercase Letters

For a password of only lowercase letters, each position has 26 choices. Therefore, there are:\[ 26^8 \] total passwords with only lowercase letters. The probability is the ratio of all lowercase passwords to all letter-only passwords:\[ P(A) = \frac{26^8}{52^8} \]
04

Compute (b) Probability of At Least One Uppercase Letter

To find the probability of having at least one uppercase letter, we first calculate the probability of having no uppercase letters, i.e., all lowercase. Using the complement rule:\[ P(B) = 1 - \frac{26^8}{52^8} \]
05

Calculate (c) Password with All Numbers

Given that the password consists only of numbers, each position has 10 possible digits. Therefore, there is a total of:\[ 10^8 \] passwords consisting of all numbers.
06

Compute (c) Probability of All Even Numbers

A password containing only even numbers (0, 2, 4, 6, 8) gives 5 choices per position. Thus, the total number of such passwords is:\[ 5^8 \] The probability of having only even numbers given a number-only password is:\[ P(C) = \frac{5^8}{10^8} \]

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Key Concepts

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

Combinatorics
Combinatorics is all about counting and arranging different things. It's like figuring out how many different ways you can sort things or choose different items. In password security, combinatorics helps us understand how many different passwords we can create.
For example, if we're creating a password that can be 8 characters long, and each character can be a letter or a digit, we use combinatorics to figure out the total combinations. Each character has 62 possibilities (26 lowercase letters + 26 uppercase letters + 10 digits), so the total number of password combinations is computed as \(62^8\). This equation tells us there are many possible passwords to choose from.
  • Each character position is independent and has the same set of possible characters.
  • Combinatorics is useful in ensuring password security by having a vast number of potential passwords.
Understanding how to count these combinations is essential in designing secure systems.
Conditional Probability
Conditional probability is about finding the likelihood of an event happening, given that another event has already occurred. In other words, it's asking, "What are the chances of this happening if we know something else is true?"
In the context of passwords, consider the scenario where we want to find the probability that a password contains only lowercase letters, given that it contains only letters (either lowercase or uppercase). It's a conditional case because we are already given that the password has only letters.
The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( A \) and \( B \) are two events. Here, \( A \) is the event "password contains all lowercase letters" and \( B \) is the event "password only contains letters."
  • This formula allows us to calculate how likely it is for a password to be all lowercase under the given condition.
  • It's widely used in real-world problems where conditions or constraints apply.
Using conditional probability helps us make more informed predictions based on existing information.
Permutations
Permutations involve arranging items in a specific order, which is key in understanding password strengths. It focuses on sequence and the order of items, making it distinct from just selecting items.
The idea of permutations applies when creating passwords where the order of characters matters. For instance, let's say you can choose lowercase letters or digits for creating your password. Every unique arrangement of these characters gives us a different password, and that's where permutations come in handy.
Although the password exercise earlier doesn't directly calculate permutations (since the order in a password position sequence is fixed), the concept is about realizing that the order significantly impacts the security or possibility of certain outcomes.
  • In permutations, changing the order of characters results in a different password.
  • Understanding permutations helps you appreciate why every character position in a password is crucial.
Permutations help reinforce why careful arrangement contributes to the strength and uniqueness of passwords.

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Most popular questions from this chapter

In a chemical plant, 24 holding tanks are used for final product storage. Four tanks are selected at random and without replacement. Suppose that six of the tanks contain material in which the viscosity exceeds the customer requirements. (a) What is the probability that exactly one tank in the sample contains high- viscosity material? (b) What is the probability that at least one tank in the sample contains high-viscosity material? (c) In addition to the six tanks with high viscosity levels, four different tanks contain material with high impurities. What is the probability that exactly one tank in the sample contains high-viscosity material and exactly one tank in the sample contains material with high impurities?

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