/*! 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 150 In a test of a printed circuit b... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(0 \mathrm{~s} ?\)

Short Answer

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

Step by step solution

01

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 \( P(0) = 0.5 \).
02

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: \[P( ext{all bits are 1s}) = (0.5)^{10} = \frac{1}{1024}.\]
03

Calculate the Probability for All Bits Being 0s

Similarly, the probability that all 10 bits are 0s is also: \[P( ext{all bits are 0s}) = (0.5)^{10} = \frac{1}{1024}.\]
04

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, \[P = 252 \times \frac{1}{1024} = \frac{252}{1024} = \frac{63}{256}.\]

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.

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.

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

Decide whether a discrete or continuous random variable is the best model for each of the following variables: (a) The number of cracks exceeding one-half inch in 10 miles of an interstate highway. (b) The weight of an injection-molded plastic part. (c) The number of molecules in a sample of gas. (d) The concentration of output from a reactor. (e) The current in an electronic circuit.

A steel plate contains 20 bolts. Assume that 5 bolts are not torqued to the proper limit. 4 bolts are selected at random, without replacement, to be checked for torque. (a) What is the probability that all 4 of the selected bolts are torqued to the proper limit? (b) What is the probability that at least 1 of the selected bolts is not torqued to the proper limit?

In a chemical plant, 24 holding tanks are used for final product storage. Four tanks are selected at random and without replacement. Suppose that six of the tanks contain material in which the viscosity exceeds the customer requirements. (a) What is the probability that exactly one tank in the sample contains high- viscosity material? (b) What is the probability that at least one tank in the sample contains high-viscosity material? (c) In addition to the six tanks with high viscosity levels, four different tanks contain material with high impurities. What is the probability that exactly one tank in the sample contains high-viscosity material and exactly one tank in the sample contains material with high impurities?

Each of the possible five outcomes of a random experiment is equally likely. The sample space is \(\\{a, b, c, d, e\\} .\) Let \(A\) denote the event \(\\{a, b\\},\) and let \(B\) denote the event \(\\{c, d, e\\} .\) Determine the following: (a) \(P(A)\) (b) \(P(B)\) (c) \(P\left(A^{\prime}\right)\) (d) \(P(A \cup B)\) (e) \(P(A \cap B)\)

In circuit testing of printed circuit boards, each board either fails or does not fail the test. A board that fails the test is then checked further to determine which one of five defect types is the primary failure mode. Represent the sample space for this experiment.

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.