/*! 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 26 My wife has organized a monthly ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

My wife has organized a monthly neighborhood party. Five people are involved in the group: Yolanda (my wife), Lorrie, Laura, Kim, and Anne Marie. They decide to randomly select the first and second home that will host the party. What is the probability that my wife hosts the first party and Lorrie hosts the second? Note: Once a home has hosted, it cannot host again until all other homes have hosted.

Short Answer

Expert verified
The probability is \( \frac{1}{20} \).

Step by step solution

01

- Total number of possible outcomes

Calculate the total number of ways to choose the first and second home from 5 homes. For the first home, there are 5 choices. After choosing the first home, there are 4 choices left for the second home. So, the total number of possible outcomes is: \[ 5 \times 4 = 20 \]
02

- Number of favorable outcomes

Determine the number of ways for Yolanda to host the first party and Lorrie to host the second. There is only 1 possible outcome where Yolanda hosts the first party and Lorrie hosts the second.
03

- Calculate the probability

Divide the number of favorable outcomes by the total number of possible outcomes. The probability that Yolanda hosts the first party and Lorrie hosts the second is: \[ \frac{1}{20} \]

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.

Event Probability
Understanding probability helps us determine how likely an event is to occur. In this exercise, the event in question is whether Yolanda will host the first party and Lorrie the second.

To calculate the probability of an event, we use the formula: \( P(event) = \frac{number\ of\ favorable\ outcomes}{total\ number\ of\ possible\ outcomes} \)

Here, the 'favorable outcome' is Yolanda hosting the first and Lorrie hosting the second. The 'total possible outcomes' include all combinations of who can host the first and second parties.

Using this approach, we determine that the probability of Yolanda and Lorrie hosting sequentially is \( \frac{1}{20} \). This formula is essential for understanding various scenarios where certain outcomes are preferred over others.
Combinatorics
Combinatorics is the field of mathematics that deals with counting combinations and arrangements of objects. In this problem, it's used to figure out how many different ways two homes can be selected from five for hosting the party.

Consider that when choosing the first home, there are 5 different choices. After one home is chosen, only 4 homes remain available for the second choice. Using this logic, the total number of combinations is calculated as: \[ 5 \times 4 = 20 \]

This means there are 20 different possible hosting sequences. Combinatorics often involves factorial calculations and other techniques to determine possible arrangements, but here a simple multiplication suffices. This field is crucial in many aspects of probability and statistics.
Random Selection
Random selection ensures that each subject or item has an equal chance of being chosen. In our problem, selecting the hosts randomly means there is no favoritism or predetermined sequence.

When Yolanda's home is randomly chosen among the five, her chance of being selected first is \( \frac{1}{5} \). After she's chosen, Lorrie's chance of being selected from the remaining four homes is \( \frac{1}{4} \).

These probabilities combine to reflect the event's likelihood: \[ P(Yolanda \ first \ and \ Lorrie \ second) = \frac{1}{5} \times \frac{1}{4} = \frac{1}{20} \]

Random selection is a fundamental concept in probability theory, ensuring fairness and unbiased outcomes in various processes.

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

The probability that a randomly selected 40-year-old female will live to be 41 years old is 0.99855 according to the National Vital Statistics Report, Vol. 56, No. \(9 .\) (a) What is the probability that two randomly selected 40 -yearold females will live to be 41 years old? (b) What is the probability that five randomly selected 40 -yearold females will live to be 41 years old? (c) What is the probability that at least one of five randomly selected 40 -year-old females will not live to be 41 years old? Would it be unusual if at least one of five randomly selected 40-year-old females did not live to be 41 years old?

According to the National Center for Health Statistics, there is a \(20.3 \%\) probability that a randomly selected resident of the United States aged 18 years or older is a smoker. In addition, there is a \(44.5 \%\) probability that a randomly selected resident of the United States aged 18 years or older is female, given that he or she smokes. What is the probability that a randomly selected resident of the United States aged 18 years or older is female and smokes? Would it be unusual to randomly select a resident of the United States aged 18 years or older who is female and smokes?

Suppose that a shipment of 120 electronic components contains 4 defective components. To determine whether the shipment should be accepted, a quality- control engineer randomly selects 4 of the components and tests them. If 1 or more of the components is defective, the shipment is rejected. What is the probability that the shipment is rejected?

According to a Gallup Poll, about \(17 \%\) of adult Americans bet on professional sports. Census data indicate that \(48.4 \%\) of the adult population in the United States is male. (a) Assuming that betting is independent of gender, compute the probability that an American adult selected at random is male and bets on professional sports. (b) Using the result in part (a), compute the probability that an American adult selected at random is male or bets on professional sports. (c) The Gallup poll data indicated that \(10.6 \%\) of adults in the United States are males and bet on professional sports. What does this indicate about the assumption in part (a)? (d) How will the information in part (c) affect the probability you computed in part (b)?

Suppose that you just received a shipment of six televisions and two are defective. If two televisions are randomly selected, compute the probability that both televisions work. What is the probability that at least one does not work?

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.