Problem 90 - Applied Statistics and Probability for Engineers

Question

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) \(P(A \cup B)\)

Step-by-Step Solution

Answer

(a) \(0.1389\), (b) \(0.1389\), (c) \(0.0193\), (d) \(0.2546\).

Step 1: Determine Total Possible Passwords

Each character in the password can be one of 36 possible options (26 letters and 10 numbers). Since there are six characters, the total number of possible passwords is \(36^6\).

Step 2: Calculate Probability P(A)

Event \(A\) is that the password begins with a vowel. There are 5 vowels and each of the remaining 5 positions can be any of the 36 characters. Therefore, there are \(5 \times 36^5\) passwords starting with a vowel. The probability is \(P(A) = \frac{5 imes 36^5}{36^6} = \frac{5}{36} \approx 0.1389\).

Step 3: Calculate Probability P(B)

Event \(B\) is that the password ends with an even number. There are 5 possible even numbers, and each of the other 5 characters can be any of the 36 options, so \(B\) has \(36^5 \times 5\) favorable outcomes. The probability is \(P(B) = \frac{36^5 \times 5}{36^6} = \frac{5}{36} \approx 0.1389\).

Step 4: Calculate Probability P(A ∩ B)

Event \(A \cap B\) is the password starts with a vowel and ends with an even number. The first position has 5 vowels, the last has 5 even numbers, and the other 4 can be any of 36 characters. So, it has \(5 \times 36^4 \times 5\) favorable outcomes. Hence, \(P(A \cap B) = \frac{5 \times 36^4 \times 5}{36^6} = \frac{25}{36^2} = \frac{25}{1296} \approx 0.0193\).

Step 5: Calculate Probability P(A ∪ B)

Use the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). With \(P(A) = \frac{5}{36}\), \(P(B) = \frac{5}{36}\), and \(P(A \cap B) = \frac{25}{1296}\), compute \(P(A \cup B) = 2 \times \frac{5}{36} - \frac{25}{1296} = \frac{55}{216} \approx 0.2546\).

Key Concepts

Combinatorial Analysis

Combinatorial analysis helps us count the number of ways things can happen. In our password exercise, it's used to compute possible combinations for password creation.
Imagine you need a six-character password. Each character can be one of the 36 options (26 letters and 10 numbers).
This forms a "combinatorial" problem because it involves combinations of characters.

• Total Combinations: Since each of the 6 characters can independently be one of 36 options, the total number of possible passwords is computed by multiplying these options. This is mathematically expressed as \(36^6\).
• Vowels Combination: For passwords starting with a vowel, you focus first on the 5 vowel options, with the remaining positions taking any of the 36 characters, hence, \(5 \times 36^5\).
• Even Numbers Combination: Similarly, passwords ending with an even number involve 5 number choices for the last character, leading to \(36^5 \times 5\).

Combinatorial analysis can simplify the complexity involved in counting large varieties, making complex probability calculations possible.

Event Probability

Event probability is about finding how likely an event is to happen. This is a foundational concept in probability theory.
For the password exercise, probability calculates how likely passwords have certain features, like starting with a vowel or ending with an even number.
Event probability uses the formula:

• \(P(A)\): The probability that event \(A\) happens. For instance, to find passwords that start with a vowel, it's \(\frac{5 \times 36^5}{36^6}\). This reduces to \(\frac{5}{36}\), about 0.1389.
• \(P(B)\): The probability that event \(B\) occurs. For our exercise, it’s the probability of ending with an even number, calculated the same way as \(P(A)\): \(\frac{5}{36}\).
• \(P(A \cap B)\): This represents the probability of both events happening—starting with a vowel and ending with an even number. It's smaller because it requires meeting both criteria.
• \(P(A \cup B)\): Entails the probability of either event happening. Use the formula \(P(A) + P(B) - P(A \cap B)\). This formula adjusts for double-counting instances where both events happen.

Understanding event probability provides insights into predicting specific outcomes.

Password Security

Password security is crucial for keeping digital data safe. It blends complex combinations to enhance security, making it harder for unauthorized access. In our exercise, a password consisting of 6 random characters from 36 possible options provides a large set of combinations, potentially tough to crack.

• Complexity: Increasing character length or adding more character options (upper and lowercase letters, symbols) can exponentially increase possible combinations, thus elevating security.
• Randomization: The calculations we perform ensure passwords are random. Using combinatorial analysis ensures a well-distributed probability, making it harder to guess.
• Vulnerabilities: Simple patterns, like only using numbers or common words, reduce complexity. Education about the importance of mixed character sets and avoiding predictable sequences is vital.

Understanding how password security is intertwined with probability theory helps design more secure systems capable of better protecting information.

Even Numbers

Even numbers are integers divisible by 2 without a remainder. These include examples like 0, 2, 4, 6, and 8.
In the context of the password exercise, they play a role when considering which passwords end with an even digit.

• Role in Probability: Since a number has ten possible digits in passwords, only half of them are even. Thus, for a password ending with an even number, there are 5 feasible choices.
• Calculation Perspective: In mathematical calculations, knowing how often specific conditions occur, like even numbers, helps in designing potential secure passwords.
• Foundation in Number Theory: Even numbers form a basic concept in mathematics, arising frequently in divisibility, algebra, and probabilistic calculations.

Recognizing the place of even numbers in these calculations enriches understanding how specific criteria influence outcomes in probability deductions.