/*! 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 15 a. How many strings of four hexa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

a. How many strings of four hexadecimal digits do not have any repeated digits? b. How many strings of four hexadecimal digits have at least one repeated digit? c. What is the probability that a randomly chosen string of four hexadecimal digits has at least one repeated digit?

Short Answer

Expert verified
a) 43680 strings have no repeated digits. b) 21856 strings have at least one repeated digit. c) The probability of a string having at least one repeated digit is \(\frac{1366}{4096}\).

Step by step solution

01

Identify total possible hexadecimal strings

Since there are 16 possible digits (0-9 and A-F), we have 16 options for the first digit, 16 options for the second digit, 16 options for the third digit, and 16 options for the fourth digit. The total number of possible strings is given by multiplying the number of options for each digit: Total possible strings = \(16^4 = 65536\)
02

Calculate the strings with no repeated digits

For a string with no repeated digits, there are 16 options (0-9 and A-F) for the first digit. Since we cannot repeat the first digit, we'll have 15 options for the second digit, 14 for the third, and 13 for the fourth. Using the counting principle: Strings with no repeated digits= \(16 \times 15 \times 14 \times 13 = 43680\) This is the answer for part (a).
03

Calculate the strings with at least one repeated digit

In order to find the strings with at least one repeated digit, we can subtract the number of strings with no repeated digits from the total possible strings: Strings with at least one repeated digit = Total possible strings - Strings with no repeated digits = 65536 - 43680 = 21856 This is the answer for part (b).
04

Calculate the probability of a string with at least one repeated digit

To find the probability of a string with at least one repeated digit, we can divide the number of strings with at least one repeated digit by the total number of possible strings: Probability = \(\frac{Strings\,with\,at\,least\,one\,repeated\,digit}{Total\,possible\,strings}\) = \(\frac{21856}{65536} = \frac{1366}{4096}\) This is the answer for part (c). So, a) There are 43680 strings with no repeated digits. b) There are 21856 strings with at least one repeated digit. c) The probability of having at least one repeated digit is \(\frac{1366}{4096}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Counting Principle
The counting principle is an essential concept in combinatorics, which is the branch of mathematics dealing with counting, combinations, and permutations. In essence, the counting principle states that if you have several sets of choices, the total number of ways to make one choice from each set is simply the product of the number of choices in each set.

For example, if you are trying to find out how many different outfits you can create with 4 shirts and 3 pants, the counting principle tells you to multiply the number of options for shirts (4) by the number of options for pants (3), thus giving you 12 different outfits. Similarly, in the exercise solution, using the counting principle showed that 16 choices for the first digit, combined with 15 for the second, 14 for the third, and 13 for the fourth, results in 43680 unique four-digit hexadecimal strings without any repeated digits.

The application of the counting principle is crucial for understanding basic probability and solving problems like the number of unique password combinations, arrangements of books on a shelf, or even the total number of different pizzas you can order with a given set of toppings.
Probability
Probability is the measure of how likely an event is to occur. It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In the context of our exercise, when considering a random event such as selecting a string of four hexadecimal digits, the probability of each possible event happening can be determined.

For instance, to calculate the probability of choosing a string with at least one repeated digit, first identify the total number of outcomes (which is the total number of possible strings) and then find how many of these outcomes are favorable (which is the number of strings with at least one repeated digit). The probability in part (c) of the exercise is found by dividing 21856 (strings with at least one repeated digit) by 65536 (total possible strings), which simplifies to \(\frac{1366}{4096}\).

Understanding probability is vital for various real-world applications ranging from predicting weather patterns to calculating insurance premiums and even to strategizing in games of chance such as poker or lotteries. It is also a foundational element in statistics, where it helps determine the likelihood of certain outcomes in a sample or population.
Permutations
Permutations are arrangements of objects in a specific order. When the order of these objects is significant, we turn to permutations to count the number of possible ways to arrange them. In general, if you have 'n' items and want to find out how many ways you can arrange 'r' of them, the formula for permutations is given by n! / (n-r)!, where 'n' is the total number of items and 'r' is the number of items to be arranged.

The concept of permutations is deeply intertwined with the previously discussed counting principle. For instance, with hexadecimal strings, arranging four digits without repetition is a permutation of those digits. In our exercise, there are 16 hexadecimal characters, and the permutation formula for arranging these digits without repetition for four places is: 16 x 15 x 14 x 13, which corresponds to the solution found for part (a) resulting in 43,680 unique strings.

Permutations are a fundamental aspect of combinatorial mathematics and are applied in a wide range of fields, including cryptography, game theory, and even in organizing the layout of computer networks. Understanding permutations enables students to solve complex arrangement and ordering problems, paving the way for more advanced studies in mathematics and computer science.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A coin is loaded so that the probability of heads is \(0.7\) and the probability of tails is \(0.3\). Suppose that the coin is tossed twice and that the results of the tosses are independent. a. What is the probability of obtaining exactly two heads? b. What is the probability of obtaining exactly one head? c. What is the probability of obtaining no heads? d. What is the probability of obtaining at least one head?

A camera shop stocks eight different types of batteries. a. How many ways can a total inventory of 30 batteries be distributed among the eight different types? b. Assuming that one of the types of batteries is A76, how many ways can a total inventory of 30 batteries be distributed among the eight different types if the inventory must include at least four A76 batteries? c. If an inventory of 30 batteries is selected at random from the cight different types, what is the probability that at least four A76 batteries will be included? d. If an inventory of 30 batteries is selected at random from the eight different types, what is the probability that exactly four A 76 batteries will be included?

An urn contains 25 red balls and 15 blue balls. Two are chosen at random, one after the other, without replacement. a. What is the probability that both balls are red? b. What is the probability that the second ball is red but the first ball is not? c. What is the probability that the second ball is red? d. What is the probability that at least one of the balls is red?

Six people attend the theater together and sit in a row with exactly six seats. a. How many ways can they be seated together in the row? b. Suppose one of the six is a doctor who must sit on the aisle in case she is paged. How many ways can the people be seated together in the row with the doctor in an aisle seat? c. Suppose the six people consist of three married couples and each couple wants to sit together with the husband on the left. How many ways can the six be seated together in the row?

Two difterent factories both produce a certain automobile part. The probability that a component from the first factory is defective is \(2 \%\), and the probability that a component from the second factory is defective is \(5 \%\). In a supply of 180 of the parts, 100 were obtained from the first factory and 80 from the second factory. a. What is the probability that a part chosen at random from the 180 is from the first factory? b. What is the probability that a part chosen at random from the 180 is from the second factory? c. What is the probability that a part chosen at random from the 180 is defective? d. If the chosen part is defective, what is the probability that it came from the first factory?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.