Problem 150 - Applied Statistics and Probability for Engineers

Question

In a test of a printed circuit board using a random test pattern, an array of 10 bits is equally likely to be 0 or 1 . Assume the bits are independent. (a) What is the probability that all bits are \(1 \mathrm{~s}\) ? (b) What is the probability that all bits are 0 s? (c) What is the probability that exactly 5 bits are \(1 \mathrm{~s}\) and 5 bits are \(0 \mathrm{~s} ?\)

Step-by-Step Solution

Answer

(a) \( \frac{1}{1024} \), (b) \( \frac{1}{1024} \), (c) \( \frac{63}{256} \).

Step 1: Understand the Probability of a Single Event

For this problem, each bit is independent and can be either 0 or 1 with equal probability. Therefore, the probability of any single bit being 1 is \( P(1) = 0.5 \) and the probability of any bit being 0 is \( P(0) = 0.5 \).

Step 2: Calculate the Probability for All Bits Being 1s

Since each bit is independent, the probability that all 10 bits are 1s is the product of the individual probabilities: \[P( ext{all bits are 1s}) = (0.5)^{10} = \frac{1}{1024}.\]

Step 3: Calculate the Probability for All Bits Being 0s

Similarly, the probability that all 10 bits are 0s is also: \[P( ext{all bits are 0s}) = (0.5)^{10} = \frac{1}{1024}.\]

Step 4: Calculate the Probability for Exactly 5 Bits Being 1s and 5 Bits Being 0s

This is a binomial probability problem where \( n = 10 \) trials and you are selecting \( k = 5 \) out of those to be 1s. Use the binomial formula: \[P( ext{exactly 5 ones and 5 zeros}) = \binom{10}{5} \times (0.5)^{10},\]where \( \binom{10}{5} = 252 \) is the number of ways to choose 5 ones out of 10 bits. Thus, \[P = 252 \times \frac{1}{1024} = \frac{252}{1024} = \frac{63}{256}.\]

Key Concepts

Binomial Probability

When dealing with events that have two possible outcomes, such as tossing a coin or observing bits as 0 or 1, binomial probability is a useful tool. In this scenario, the printed circuit board being tested can yield two outcomes for each bit: 0 or 1.

When we talk about binomial probability, we're often interested in determining the probability of a certain number of successes (let's say 1s in this case) in a given number of trials (10 bits). We utilize the binomial probability formula, given by:

• \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where
• \( n \) is the total number of trials,
• \( k \) is the number of successful trials,
• \( p \) is the probability of success in a single trial, and
• \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successful trials out of \( n \).

In our exercise, 10 trials mean 10 bits, trying to find out probability for different outcomes using equal success probability of 0.5 for each bit.

By applying the binomial formula, for instance, finding the probability that exactly 5 bits are 1s involves computing \( \binom{10}{5} (0.5)^{10} \). This results in 252 different combinations of five 1s and five 0s, leading to a probability of \( \frac{63}{256} \). This showcases how binomial probability helps in solving problems involving multiple trials.

Independent Events

In probability theory, independent events are those whose outcomes do not influence each other. When we say that the bits in the exercise are independent, we mean the result of one bit does not affect the result of another.

For our circuit board example, each bit acts independently and can either be a 0 or a 1 with an equal probability of 0.5. Therefore, calculating the probability for all bits being 1s (or all being 0s) involves understanding and multiplying the probabilities of each bit being 1 (or 0) individually.

• The formula for calculating the probability of all bits being 1s when they are independent is the product of their individual probabilities: \((0.5)^{10}\).
• The same applies for all bits being 0s.

With independent events, knowing the outcome of one event doesn’t give any information about another event. Thus, probability calculations using products of individual probabilities are valid, which simplifies the process for larger datasets as seen in this exercise.

Random Variables

In the world of probability theory, random variables are used to represent the possible outcomes of a probabilistic event. In the context of our exercise, each bit can be modeled as a random variable, taking on a value of either 0 or 1.

Random variables help us formalize the process of determining probabilities for different outcomes. They can be discrete, with distinct outcomes, such as the bit values here. Each bit is a random variable with a probability distribution where each of the two outcomes (0 or 1) has an equal chance of occurring.

When dealing with multiple random variables, such as 10 bits, we can model complex events, like getting exactly 5 ones and 5 zeros, using these variables. The probability distribution of these events is outlined by principles of independent events and binomial distributions.

By considering these bits as random variables, it makes it easier to break down the probability calculations into manageable parts, leveraging their properties, and ultimately solve the main problem of the exercise with clear insights.