/*! 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 36 A box contains two defective Chr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 36

A box contains two defective Christmas tree lights that have been inadvertently mixed with cight nondefective lights. If the lights are selected one at a time without replacement and tested until both defective lights are found, what is the probability that both defective lights will be found after exactly three trials?

Short Answer

Expert verified
The probability that both defective lights will be found after exactly three trials is \(\frac{2}{45}\).

Step by step solution

01

Identify the events

There are two possible events that can help us find both defective lights in exactly three trials: 1. We find a defective light in the first trial and another defective light in the second trial. 2. We find a defective light in the first trial, a non-defective light in the second trial, and another defective light in the third trial. We need to find the probabilities of these events happening, and then sum them to get the total probability of finding both defective lights after exactly three trials.
02

Calculate the probability of each scenario

Let's calculate the probabilities of the two events mentioned above: 1. Probability of finding both defective lights in the first two trials: In this case, we first find the probability of finding a defective light in the first trial and then, given that we found a defective light in the first trial, the probability of finding another defective light in the second trial. P(Defective in trial 1) = \(\frac{2}{10}\) P(Defective in trial 2 | Defective in trial 1) = \(\frac{1}{9}\) So, the probability of finding both defective lights in the first two trials is: P(Defective in trial 1 and Defective in trial 2) = \(\frac{2}{10} \times \frac{1}{9} = \frac{1}{45}\) 2. Probability of finding a defective light in the first trial, a non-defective light in the second trial, and another defective light in the third trial: P(Defective in trial 1) = \(\frac{2}{10}\) P(Non-defective in trial 2 | Defective in trial 1) = \(\frac{8}{9}\) P(Defective in trial 3 | Non-defective in trial 2 and Defective in trial 1) = \(\frac{1}{8}\) So, the probability of this event is: P(Defective in trial 1, Non-defective in trial 2, and Defective in trial 3) = \(\frac{2}{10} \times \frac{8}{9} \times \frac{1}{8} = \frac{1}{45}\)
03

Sum the probabilities of the two events

Now, we will sum the probabilities of both events together to get the total probability of finding both defective lights after exactly three trials: P(Both are found in exactly three trials) = P(First two trials) + P(First and third trial) P(Both are found in exactly three trials) = \(\frac{1}{45}\) + \(\frac{1}{45}\) = \(\frac{2}{45}\) So, the probability that both defective lights will be found after exactly three trials is \(\frac{2}{45}\).

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 91影视!

Key Concepts

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

Defective Items
In probability problems, the term 'defective items' often refers to items that are not functioning as intended. In this exercise, the defective items are two malfunctioning Christmas tree lights mixed with eight working ones. Understanding that these items are distinct from non-defective items is crucial when calculating probabilities.
Defective items, in a given set, create unique outcomes compared to non-defective items. For example, in our scenario, finding both defective lights amidst the non-defective ones requires considering how often each type of item appears in various stages of selection.
  • Identifying defectives helps to define the events whose probabilities we are interested in.
  • Each draw from the set reduces the number of items available for selection, which affects subsequent probabilities.
Probability Calculation
Probability calculations involve determining the likelihood of a specific event occurring, such as finding both defective lights in a series of trials. Calculating these probabilities step by step ensures accuracy and contextual understanding.
In our problem, two types of successful trials were identified. To calculate each, we multiplied the probabilities of events in sequence:
  • Finding a defective light initially is \( \frac{2}{10} \)
  • Finding another defective next, given the first was defective, is \( \frac{1}{9} \)
These calculations are completed by multiplying each event鈥檚 probability, reflecting the sequence鈥檚 dependency.
This encapsulates the step-by-step nature of probability, where each event's occurrence affects the following ones, changing the available outcomes and calculations:
  • Sequential events mean looking at outcomes at each stage, adjusting each probability as the situation changes.
Without Replacement
The concept of 'without replacement' means that once an item is selected, it is not returned to the set. This significantly impacts probability calculations as it changes the pool of items after each trial:
  • The first item's selection alters the total number of items for the next trial.
  • For instance, selecting a light reduces the items from 10 to 9 items, with the same applying to future trials.
This method reflects real-world scenarios where selected items affect future possibilities. It鈥檚 essential to account for this adjustment in calculations, distinguishing problems without replacement from those with replacement.
In our exercise, without replacement changes probabilities with each draw, reflecting the dynamic nature of the scenario where items do not replenish. This approach models specific situations where resources or conditions change continuously, leading to unique probability scenarios.

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

Determine whether the given experiment has a sample space with equally likely outcomes. A loaded die is rolled, and the number appearing uppermost on the die is recorded.

In a survey to determine the opinions of Americans on health insurers, 400 baby boomers and 600 pre-boomers were asked this question: Do you believe that insurers are very responsible for high health costs? Of the baby boomers, 212 answered in the affirmative, whereas 198 of the pre-boomers answered in the affirmative. If a respondent chosen at random from those surveyed answered the question in the affirmative, what is the probability that he or she is a baby boomer? A pre-boomer?

Two hundred workers were asked: Would a better economy lead you to switch jobs? The results of the survey follow: $$ \begin{array}{lccccc} \hline \text { Answer } & \begin{array}{c} \text { Very } \\ \text { likely } \end{array} & \begin{array}{c} \text { Somewhat } \\ \text { likely } \end{array} & \begin{array}{c} \text { Somewhat } \\ \text { unlikely } \end{array} & \begin{array}{c} \text { Very } \\ \text { unlikely } \end{array} & \begin{array}{c} \text { Don't } \\ \text { know } \end{array} \\ \hline \text { Respondents } & 40 & 28 & 26 & 104 & 2 \\ \hline \end{array} $$ If a worker is chosen at random, what is the probability that he or she a. Is very unlikely to switch jobs? b. Is somewhat likely or very likely to switch jobs?

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A\) is a subset of \(B\) and \(P(B)=0\), then \(P(A)=0\).

If a 5-card poker hand is dealt from a well-shuffled deck of 52 cards, what is the probability of being dealt the given hand? Four of a kind
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation
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.