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Chapter 3: Problem 107

A computer system uses passwords that are exactly six characters and each character is one of the 26 letters \((\mathrm{a}-\mathrm{z})\) or 10 integers \((0-9)\). Suppose that 10,000 users of the system have unique passwords. A hacker randomly selects (with replacement) one billion passwords from the potential set, and a match to a user's password is called a hit. (a) What is the distribution of the number of hits? (b) What is the probability of no hits? (c) What are the mean and variance of the number of hits?

Short Answer

Expert verified
The hits follow a binomial distribution. The probability of no hits is approximately zero, with a mean and variance of about 4,595.18.

Step by step solution

01

Determine total possible passwords

Each character in the password can either be a letter (26 options) or a digit (10 options), totaling 36 options per character. The password is 6 characters long, so the total number of possible passwords is:\[36^6\]
02

Formula for total number of possible passwords

Calculate the total number of possible passwords using the formula from Step 1.\[36^6 = 2,176,782,336\]Thus, there are 2,176,782,336 possible passwords.
03

Define random variable for hits

Define the random variable \(X\) as the number of hits (matches of randomly selected passwords with user passwords). The hacker selects one billion passwords. The distribution of \(X\) follows a binomial distribution with parameters \(n = 1,000,000,000\) (number of trials) and \(p = \frac{10,000}{2,176,782,336}\) (probability of a hit).
04

Calculate probability of no hits

The probability of zero hits in a binomial distribution is given by:\[P(X = 0) = (1 - p)^n\]Substituting \(p\) and \(n\):\[P(X = 0) = \left(1 - \frac{10,000}{2,176,782,336}\right)^{1,000,000,000}\]
05

Use Poisson approximation for no hits

Since \(n\) is large and \(p\) is small, apply the Poisson approximation:\(-np\) also known as \(\lambda\). Here, \(\lambda = np = \left(1,000,000,000\right)\left(\frac{10,000}{2,176,782,336}\right) = 4,595.18\). The probability of zero hits is:\[P(X = 0) \approx e^{-\lambda} = e^{-4,595.18}\]This is extremely small, approaching zero.
06

Calculate mean and variance of hits

For a binomial distribution, the mean \(\mu\) is \(np\), and the variance \(\sigma^2\) is \(np(1-p)\). Calculating these:\[\text{Mean } \mu = \lambda = 4,595.18\]\[\text{Variance } \sigma^2 = np(1-p) \approx \lambda = 4,595.18 \] Since \(p\) is very small, \(1-p\) approximately equals 1.

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Key Concepts

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

Binomial Distribution
The Binomial Distribution is a fundamental probability distribution in statistics. It models the number of successes in a fixed number of independent and identically distributed Bernoulli trials. Each trial results in either success or failure.
In our exercise involving hackers and passwords, each password guess by the hacker is a trial. The success (or 'hit') is when a guessed password matches one of the 10,000 user passwords.
The key parameters of the binomial distribution are:
  • n: the number of trials. In this scenario, it is 1,000,000,000 because the hacker makes a billion guesses.
  • p: the probability of success in each trial. This is calculated as the ratio of unique passwords to total possible passwords, which is \( \frac{10,000}{2,176,782,336} \).
The distribution of hits follows a binomial distribution because each password trial is independent, with a fixed number of trials, and each trial has the same probability of resulting in a hit.
Poisson Approximation
When dealing with a binomial distribution where the number of trials is very large and the probability of success is quite small, it can be computationally cumbersome. In such scenarios, the Poisson Distribution serves as a useful approximation.
The Poisson distribution simplifies the problem by focusing on the rate of occurrence \((\lambda)\). For our exercise, the parameter \( \lambda = np \) embodies the expected number of hits. This approximation is practical because it requires less computational power and provides results with reasonable accuracy. Here:
  • Given \( n = 1,000,000,000 \) and \( p = \frac{10,000}{2,176,782,336} \), we calculate \( \lambda = 4,595.18 \).
  • Using the Poisson approximation, the probability of no hits is \( P(X = 0) \approx e^{-\lambda} \). In this case, the result approaches zero.
Overall, the Poisson approximation allows us to estimate probabilities in large-sample scenarios effectively.
Mean and Variance of Binomial Distribution
Understanding the mean and variance helps in predicting the average outcome and the expected spread or variability of the results.
For a binomial distribution:
  • Mean (\( \mu \)): The expected number of successes, calculated as \( np \). It's the average outcome you'd expect over many trials.
  • Variance (\( \sigma^2 \)): Measures the spread or variability of the distribution, calculated as \( np(1-p) \).
In our scenario, both the mean and variance are approximately equal due to the small value of \( p \), meaning \( 1-p \) is nearly 1. Thus, both values calculate to \( \lambda = 4,595.18 \). This symmetry arises because most trials result in a failure, given the low probability of guessing a password correctly.
Password Combinatorics
Password Combinatorics explores the number of possible combinations for passwords in the context of probability and security.
In this exercise, the key is understanding the size of the password space. Each character in the password is a combination of 26 letters and 10 digits, giving us 36 options per character. For a six-character password:
  • The formula to calculate the total number of possible combinations is \( 36^6 \).
  • This results in 2,176,782,336 distinctive passwords.
Larger combinatorial spaces increase the difficulty for hackers. The sheer number of options makes brute-force attacks鈥攔andomly guessing passwords鈥攎ore challenging and time-consuming. Understanding combinatorics helps design stronger systems by calculating the complexity more effectively.

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