Question

A bit (meaning binary digit) is 0 or \(1 .\) An ordered array of eight bits (such as 01101001) is a byte. How many different bytes are there? If you select a byte at random, what is the probability that you select \(11000010 ?\) What is the probability that you select a byte containing three 1 's and five 0 's?

Short answer

There are 256 different bytes. The probability of selecting 11000010 is \( \frac{1}{256} \). The probability of selecting a byte with three 1's and five 0's is \( \frac{7}{32} \).

Step-by-step solution

Determine the Total Number of Bytes

Each position in the byte can be either 0 or 1. With 8 positions, the total number of different bytes is given by the formula for combinations with repetition: \[ 2^8 = 256 \] So, there are 256 different bytes.

Calculate the Probability of Selecting a Specific Byte

The probability of selecting any specific byte (like 11000010) from the total number of bytes is given by the formula for probability:\[ P(\text{specific byte}) = \frac{1}{256} \]

Find the Number of Bytes with Three 1's and Five 0's

The number of ways to arrange three 1's and five 0's in eight positions is given by the binomial coefficient: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = 56 \]

Calculate the Probability of Selecting a Byte with Three 1's and Five 0's

The probability is given by the number of favorable outcomes divided by the total number of possible outcomes:\[ P(\text{3 1's and 5 0's}) = \frac{56}{256} = \frac{7}{32} \]

Key concepts

binary digits

Binary digits, commonly known as bits, are the most fundamental unit of information in computing and digital communications. A bit can have one of two values: 0 or 1. When these binary digits are arranged in sequences, they form larger data structures like bytes. A byte is a sequence of 8 bits. Think of a byte as an ordered array; for instance, 01101001 is an example of one such byte. There are precisely 256 unique configurations of a byte because each of the eight positions can either be a 0 or a 1. This is calculated using the formula for combinations, which is expressed as \[ 2^8 = 256 \]. The value of each bit within a byte affects the overall binary number, and combinations can range from 00000000 to 11111111. Understanding binary digits helps in grasping how data is stored and processed in computers.

binomial coefficient

The binomial coefficient is a crucial concept in combinatorics and probability. It is used to determine the number of ways to choose a specific number of elements from a larger set, regardless of the order in which they are chosen. The notation for the binomial coefficient is \(\binom{n}{k}\), and it’s read as 'n choose k'. It is calculated using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]. For example, if we want to find out how many different ways we can arrange three 1's and five 0's in a byte, we use the binomial coefficient \(\binom{8}{3}\). Plugging into the formula, we get: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = 56 \]. Therefore, there are 56 different combinations where a byte can have exactly three 1's and five 0's.

combinations with repetition

Combinations with repetition allow us to understand how to count arrangements where elements can be reused. In the context of binary digits within a byte, each position must be filled with either a 0 or a 1. Since each binary digit can be repeated across the byte's eight positions, we calculate combinations with repetition using the formula \( 2^n \). For a byte, this is \( 2^8 \), which equates to 256, confirming that there are 256 possible unique bytes. These combinations help us in understanding how data and memory are structured in computing systems.

probability calculation

Probability calculation is a pivotal concept in determining the likelihood of a certain event happening out of all possible events. To calculate the probability, we use the formula: \[ P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} \]. In the context of selecting a specific byte out of 256 possible bytes, the likelihood is \( \frac{1}{256} \). To find the probability of selecting a byte with three 1's and five 0's, we rely on the number of favorable configurations, which we previously calculated as 56. Using the probability formula: \[ P(\text{3 1's and 5 0's}) = \frac{56}{256} = \frac{7}{32} \]. This breaks down to approximately 0.21875 or 21.875%. Understanding probability allows us to predict and analyze outcomes in various fields of study.