/*! 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 11 A shipment of 7 television sets ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 11

A shipment of 7 television sets contains 2 defective sets. A hotel makes a random purchase of; 3 of the sets. If \(x\) is the number of defective sets purchased by the hotel, find the probability distribution of \(X .\) Express the results graphically as a probability histogram.

Short Answer

Expert verified
The probability distribution is: \(P(x=0)=\frac{C(5,3)*C(2,0)}{C(7,3)}\), \(P(x=1)=\frac{C(5,2)*C(2,1)}{C(7,3)}\), \(P(x=2)=\frac{C(5,1)*C(2,2)}{C(7,3)}\), and \(P(x=3)=0\). These can be represented graphically as a probability histogram.

Step by step solution

01

Identify the different outcomes

The number of defective TVs purchased, \(x\), can take on values of 0, 1, 2 or 3, since these are the possible numbers of defective sets the hotel might end up with.
02

Calculate Probability for \(x=0\)

This is the scenario where none of the purchased TVs is defective. It means the hotel has picked all three sets from the five non-defective TVs. We calculate this probability as follows: \(P(x=0)=\frac{C(5,3)*C(2,0)}{C(7,3)}\), where \(C(n,k)\) is the combination function, that calculates how many ways \(k\) items can be chosen out of \(n\).
03

Calculate Probability for \(x=1\)

This scenario is when one of the purchased TVs is defective. We calculate this probability as follows: \(P(x=1)=\frac{C(5,2)*C(2,1)}{C(7,3)}\). This calculates how likely it is that the hotel picks exactly one defective TV and two non-defective sets out of the total shipment.
04

Calculate Probability for \(x=2\)

This is the scenario where two of purchased TVs are defective. We calculate this probability as follows: \(P(x=2)=\frac{C(5,1)*C(2,2)}{C(7,3)}\). This measures the chance the hotel picks out exactly two defective TVs and one non-defective TV from the shipment.
05

Calculate Probability for \(x=3\)

This is the scenario where all purchased TVs are defective. However, since there are only 2 defective sets within the original shipment, it's impossible for all 3 purchased TVs to be defective. Thus, \(P(x=3) = 0\).
06

Summarize the probability distribution

The probability distribution for \(x\) can now be summarized. Lastly, to present this distribution graphically, a probability histogram would be plotted with the number of defective sets (0,1,2,3) on the x-axis and the corresponding probabilities on the y-axis.

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.

Combinations in Probability
Understanding how to use combinations in probability is crucial when dealing with exercises like determining the likelihood of purchasing defective television sets. Combinations, represented as \(C(n, k)\), help us calculate the number of ways we can select \(k\) items from a larger set of \(n\) without considering the order. In our example, this math concept is applied to depict the various scenarios in which a hotel could buy defective sets from a shipment.

For instance, to determine the probability of selecting no defective TVs (\(P(x=0)\)), we look at the combinations of choosing 3 non-defective sets from 5 available (\(C(5,3)\)), and then multiply by the combinations of choosing 0 defective sets from the 2 defective, \(C(2,0)\). The total number of purchasing combinations is the combinations of choosing any 3 TVs from the entire shipment of 7, given by \(C(7,3)\). This approach is similarly used to calculate the probability for selecting exactly 1 or 2 defective TVs by varying the values of \(k\) in the combination formula.
Probability Histogram
A probability histogram is a graphical representation that displays the probability distribution of a discrete random variable. When plotting a probability histogram, the x-axis represents the different possible outcomes, and the y-axis shows the probability of each outcome.

In our textbook problem concerning the TV sets, the x-axis would list the possible outcomes: 0, 1, and 2 defective sets. (Remember, 3 is not possible because there are only 2 defective sets in the shipment). The y-axis would show the respective probabilities for each of these outcomes, calculated using combinations as previously discussed. A bar is drawn for each possible outcome with its height reflecting the probability of that number of defective sets being purchased. This type of histogram provides a quick visual understanding of the likelihoods and is an effective tool for summarizing a probability distribution in a clear and concise manner.
Discrete Random Variables
In probability, a discrete random variable is a variable that can take on a finite or countably infinite number of values. Each of these values has an associated probability that is determined based on the scenario being examined.

In our exercise, the discrete random variable in question is \(x\), which represents the number of defective TV sets that the hotel might purchase. Since the number of defective sets the hotel can purchase is countable and does not continue indefinitely, \(x\) fits the definition of a discrete random variable. Knowing how to work with such variables allows us to build the entire probability distribution, which in turn provides a comprehensive picture of all potential outcomes and their likelihoods.

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

Find the probability distribution for the number of jazz CDs when 4 CDs are selected at random from a collection consisting of 5 jazz CDs, 2 classical CDs, and 3 rock CDs. Express your results by means of a formula.

A coin is flipped until 3 heads in succession occur. List only those elements of the sample space that require 6 or less tosses. Is this a discrete sample space? Explain.

The joint density function of the random variables \(X\) and \(Y\) is $$ f(x, y)=\left\\{\begin{array}{ll} 6 x, & 0< x < 1,0 < y < 1-x, \\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Show that \(X\) and \(Y\) are not independent. (b) Find \(P(X>0.3 \quad Y=0.5)\).

Magnetron tulses are produced from an automated assembly line. A sampling plan is used periodically to assess quality on the lengths of the tubes. This measurement is subject to uncertainty. It is thought that the probability that a random tube meets length specification is \(0.99 .\) A sampling plan is used in which the lengths of 5 random tubes are measured. (a) Show that the probability function of \(Y\), the number out of 5 that meet length specification, is given by the following discrete probability function $$ \begin{aligned} f(y) &=\frac{5 !}{y !(5-y) !}(0.99)^{y}(0.01)^{5-y.} \\ \text { for } y=0.1,2,3,4,5. \end{aligned} $$ (b) Suppose random selections are made off the line and 3 are outside specifications. Use \(f(y)\) above either to support or refute the conjecture that the probability is 0.99 that a single tube meets specifications.

A candy company distributes boxes of chocolates with a mixture of creams, toffees, and cordials. Suppose that the weight of each box is 1 kilogram, but the individual weights of the creams, toffees, and cordials vary from box to box. For a randomly selected box, let \(X\) and \(Y\) represent the weights of the creams and the toffees, respectively, and suppose that the joint density function of these variables is $$ f(x, y)=\left\\{\begin{array}{ll} 24 x y, & 0 \leq x \leq 1, \quad 0 \leq y \leq 1, \\ & x+y \leq 1 ,\\\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Find the probability that in a given box the cordials account for more than \(1 / 2\) of the weight. (b) Find the marginal density for the weight of the creams. (c) Find the probability that the weight of the toffees in a box is less than \(1 / 8\) of a kilogram if it is known that creams constitute \(3 / 4\) of the weight.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Calculus

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.