91Ó°ÊÓ

A computer system uses passwords constructed from the 26 letters \((a-z)\) or 10 integers \((0-9)\). Suppose that 10,000 users of the system have unique passwords. A hacker randomly selects (with replacement) passwords from the potential set. (a) Suppose that 9900 users have unique six-character passwords and the hacker randomly selects six-character passwords. What are the mean and standard deviation of the number of attempts before the hacker selects a user password? (b) Suppose that 100 users have unique three-character passwords and the hacker randomly selects three-character passwords. What are the mean and standard deviation of the number of attempts before the hacker selects a user password? (c) Comment on the security differences between six- and three-character passwords.

Step by step solution

01

Understanding the Problem

We need to find the mean and standard deviation of the number of attempts required before the hacker correctly guesses a user's password from a given set. This is a geometric distribution problem where success is defined as correctly guessing a password.

02

Calculating the Password Pool for Six-Character Passwords

Each password is made up of 26 letters or 10 numbers, giving a total of 36 possible characters. For a six-character password, the total number of possible combinations is \( 36^6 \).

03

Probability of Guessing a Six-Character Password

The probability \( p \) that a hacker guesses a six-character password correctly on one attempt is \( \frac{9900}{36^6} \).

04

Mean and Standard Deviation for Six-Character Passwords

For a geometric distribution, the mean \( \mu \) is \( \frac{1}{p} \) and the standard deviation \( \sigma \) is \( \frac{\sqrt{1-p}}{p} \). Calculate these using the probability from Step 3.

05

Calculating the Password Pool for Three-Character Passwords

For a three-character password, the total number of possible combinations is \( 36^3 \).

06

Probability of Guessing a Three-Character Password

The probability \( p \) that a hacker guesses a three-character password correctly on one attempt is \( \frac{100}{36^3} \).

07

Mean and Standard Deviation for Three-Character Passwords

Again, use the geometric distribution formulas where mean \( \mu = \frac{1}{p} \) and standard deviation \( \sigma = \frac{\sqrt{1-p}}{p} \) to compute these values using the probability from Step 6.

08

Security Comparison

The mean and standard deviations calculated illustrate that more attempts are needed on average to guess a six-character password compared to a three-character password, indicating higher security.

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Key Concepts

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

Password Security

In today's digital age, password security is vital to protect personal data and sensitive information. When creating passwords, it is crucial that they are complex enough to deter unauthorized access.

Passwords can be composed of various characters, including numbers and letters. The blend of these characters and their arrangement significantly impacts the strength of the password.

• Six-character passwords, for example, have more combinations than three-character ones, thus making them inherently more secure.
• Longer passwords take more time to crack due to the larger number of potential combinations.

To ensure robust password security, always opt for longer and more complex passwords, incorporating a mix of numbers, letters, and symbols.

Probability of Success

In scenarios where a hacker tries to guess a password, the probability of success calculates how likely it is to guess correctly on a single try.

This probability is a critical factor in determining the difficulty of hacking into a system.

• For instance, if there are 9900 unique six-character passwords, the probability that a hacker guesses the right one in a single attempt is given by the formula: \( p = \frac{9900}{36^6} \).
• If a hacker targets a pool of 100 three-character passwords, the probability becomes \( p = \frac{100}{36^3} \).

The lower the probability, the more secure the password is, as it signifies there are more possible password combinations to guess from, requiring more attempts.

Mean of Geometric Distribution

The mean of a geometric distribution in the context of guessing passwords tells us the average number of attempts required for a successful guess.

For a password guessing scenario, this mean can be calculated using the formula: \[ \mu = \frac{1}{p} \]where \( p \) is the probability of success.

• A higher mean implies that more attempts are needed on average, reflecting a more secure password.
• For six-character passwords with a calculated probability \( p \), the mean is significantly higher than that for three-character passwords.

Understanding this helps in assessing the effectiveness of a password's security.

Standard Deviation of Geometric Distribution

Standard deviation in a geometric distribution gives insight into the variation and consistency of the number of trials needed to succeed.

It is calculated with the formula: \[ \sigma = \frac{\sqrt{1-p}}{p}, \]where \( p \) represents the probability of success on a single try.

• A larger standard deviation indicates more inconsistency in the number of attempts needed.
• For more secure six-character passwords, the standard deviation will be larger than for three-character passwords, showing greater spread because of the low probability \( p \).

Considering both mean and standard deviation helps in understanding both the average case and the variability in attempts, providing a clear picture of password security effectiveness.