/*! 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 6 An electronic product contains 4... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electronic product contains 40 integrated circuits. The probability that any integrated circuit is defective is \(0.01,\) and the integrated circuits are independent. The product operates only if there are no defective integrated circuits. What is the probability that the product operates?

Short Answer

Expert verified
The probability that the product operates is approximately 0.6703.

Step by step solution

01

Understanding the Problem

We have a product containing 40 integrated circuits, each with a 0.01 probability of being defective, and they are independent of each other. We need to find the probability that none of these integrated circuits are defective for the product to operate.
02

Identifying Success Criteria

For the product to operate successfully, each integrated circuit must be non-defective. If the probability of a defect is 0.01, the probability of a single circuit being non-defective is given by the complement, which is 1 - 0.01 = 0.99.
03

Calculating the Probability of No Defects

Since the circuits are independent, the probability that all 40 circuits are non-defective is the product of each circuit being non-defective. Therefore, the probability is calculated as:\[(0.99)^{40}\]
04

Computing the Result

We compute \((0.99)^{40}\) to find the probability that the product operates without any defective circuits. By calculation:\[(0.99)^{40} \approx 0.6703\]

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.

Independent Events
In probability theory, independent events are those whose outcomes do not affect each other. This means the occurrence of one event does not change the likelihood of another event happening.

When we consider integrated circuits within a product, describing them as independent implies that whether one circuit is defective does not influence any other circuit.
  • Independence is a key assumption in many probability calculations, simplifying the process of calculating overall probabilities.
  • If two or more events are independent, the probability of all events occurring together is the product of their individual probabilities.
For example, if you have event A and event B, and both are independent, the probability of both events occurring is given by:

\( P(A \text{ and } B) = P(A) \times P(B) \)

In this exercise, each integrated circuit failing is independent of the others. Hence, the probability calculation involves multiplying the probability of each being non-defective.
Defective Probability
Defective probability refers to the likelihood of a component failing or being defective. In our exercise, each integrated circuit has a defective probability of 0.01.

This probability is crucial when determining the reliability of a system, especially in electronic products. A small probability of defect does not guarantee absence of defects but does suggest it's less likely.
  • Understanding defective probability is important for making informed decisions in quality control and risk assessment strategies.
  • Often, manufacturers aim for negligible defective probabilities to ensure robust and reliable performance.
When dealing with multiple components, such as the 40 circuits in our problem, it's essential to compute the combined effect of these individual probabilities. This helps in assessing the overall defectiveness risk of the product.
Complement Rule
The complement rule is a fundamental concept in probability, helping to find the probability of the complement of an event. The complement of an event is the event not occurring.

Mathematically, it can be expressed as:
  • The probability of an event occurring, plus the probability of it not occurring, equals 1.

    That is: \( P( ext{Event}) + P( ext{Not Event}) = 1 \)
In our scenario, we applied the complement rule to determine the probability of an integrated circuit being non-defective.

With a defective probability of 0.01, its complement is:

\( P( ext{Non-defective}) = 1 - P( ext{Defective}) = 1 - 0.01 = 0.99 \)

Using the complement rule makes calculations more straightforward by focusing on the likelihood of a component functioning correctly, which is central to establishing the product's operability.

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

For each of the following exercises, determine the range (possible values) of the random variable. A batch of 500 machined parts contains 10 that do not conform to customer requirements. Parts are selected successively, without replacement, until a nonconforming part is obtained. The random variable is the number of parts selected.

A player of a video game is confronted with a series of opponents and has an \(80 \%\) probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to contest opponents. a. What is the probability mass function of the number of opponents contested in a game? b. What is the probability that a player defeats at least two opponents in a game? c. What is the expected number of opponents contested in a game? d. What is the probability that a player contests four or more opponents in a game? e. What is the expected number of game plays until a player contests four or more opponents?

Samples of 20 parts from a metal punching process are selected every hour. Typically, \(1 \%\) of the parts require rework. Let \(X\) denote the number of parts in the sample of 20 that require rework. A process problem is suspected if \(X\) exceeds its mean by more than 3 standard deviations. a. If the percentage of parts that require rework remains at \(1 \%,\) what is the probability that \(X\) exceeds its mean by more than 3 standard deviations? b. If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds \(1 ?\) c. If the rework percentage increases to \(4 \%,\) what is the probability that \(X\) exceeds 1 in at least one of the next five hours of samples?

Assume that a random variable is normally distributed with a mean of 24 and a standard deviation of 2 . Consider an interval of length one unit that starts at the value \(a\) so that the interval is \([a, a+1] .\) For what value of \(a\) is the probability of the interval greatest? Does the standard deviation affect that choice of interval?

A utility company might offer electrical rates based on time-of-day consumption to decrease the peak demand in a day. Enough customers need to accept the plan for it to be successful. Suppose that among 50 major customers, 15 would accept the plan. The utility selects 10 major customers randomly (without replacement) to contact and promote the plan. a. What is the probability that exactly two of the selected major customers accept the plan? b. What is the probability that at least one of the selected major customers accepts the plan? c. Instead of 15 customers, what is the minimum number of major customers that would need to accept the plan to meet the following objective? The probability that at least one selected major customer accepts the plan is greater than or equal to 0.95

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.