/*! 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 95 Individual A has a circle of fiv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Individual A has a circle of five close friends (B, C, D, E, and F). A has heard a certain rumor from outside the circle and has invited the five friends to a party to circulate the rumor. To begin, A selects one of the five at random and tells the rumor to the chosen individual. That individual then selects at random one of the four remaining individuals and repeats the rumor. Continuing, a new individual is selected from those not already having heard the rumor by the individual who has just heard it, until everyone has been told. a. What is the probability that the rumor is repeated in the order \(\mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}\), and \(\mathrm{F}\) ? b. What is the probability that \(\mathrm{F}\) is the third person at the party to be told the rumor? c. What is the probability that \(\mathrm{F}\) is the last person to hear the rumor? d. If at each stage the person who currently "has" the rumor does not know who has already heard it and selects the next recipient at random from all five possible individuals, what is the probability that \(\mathrm{F}\) has still not heard the rumor after it has been told ten times at the party?

Short Answer

Expert verified
(a) \(\frac{1}{120}\); (b) \(\frac{1}{10}\); (c) \(\frac{1}{60}\); (d) \(0.107\).

Step by step solution

01

Understanding the Problem Conditions

The rumor starts with individual A and is told to one out of five friends (B, C, D, E, F). Each person who hears the rumor then tells it to one of the remaining friends who have not heard it yet. Thus, each choice is independent and random among those who haven't heard the rumor.
02

Calculating Probability for Part (a)

We want the rumor to be told in the order of B, C, D, E, and then F. Initially, A selects one of the five friends; thus, the probability of picking B is \( \frac{1}{5} \). Next, B chooses C out of the four who haven't heard the rumor, with probability \( \frac{1}{4} \). C chooses D out of the three, with probability \( \frac{1}{3} \). D chooses E out of the two, with probability \( \frac{1}{2} \). Then E finally chooses F. Multiplying these probabilities gives the total probability.\[ \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times 1 = \frac{1}{120}.\]
03

Calculating Probability for Part (b)

To find the probability that F is third to hear the rumor, F must be chosen after two others. First person selected can be any of 5, second from 4, and F must be selected third from the 3 people left. Thus, the probability depends on this third choice:\[ \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}.\]
04

Calculating Probability for Part (c)

To determine the probability that F is last to hear the rumor, F must be chosen from the last 2 remaining people. The sequence before does not matter as long as F is told last:\[ \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{60}.\]
05

Computing Iterative Probability for Part (d)

Considering each person telling the rumor does not remember or know of previous recipients; if the rumor is told 10 times, we analyze if F has not heard it. F must not be chosen in the initial selections at any of these stages, the probability follows a geometric distribution where the number of times F isn't selected increases exponentially small as we recount the total distribution:\[ \left(\frac{4}{5}\right)^{10} \approx 0.107.\]

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.

Random Selection
Random selection is a fundamental principle in probability theory where each member of a set has an equal chance of being chosen. In the context of this exercise, random selection is applied each time a friend tells the rumor to another. This ensures that any individual who hasn't heard the rumor before has the same probability of hearing it next.
This concept is important because it determines the likelihood of each potential outcome. For example, when individual A initially selects one of the five friends, each friend (B, C, D, E, F) has a \( \frac{1}{5} \) chance of being told the rumor first. It's this principle of fairness and unpredictability that keeps each selection independent of past choices, as each selection does not affect the next.
  • Every choice is independent.
  • Each person who hasn't heard the rumor yet is equally likely to be chosen.
  • Probabilities remain constant with each new selection.
Independent Events
In probability theory, independent events are those whose outcomes do not affect each other. When one event occurs, it does not influence the likelihood of another event.
For this exercise, each time the rumor is passed on, the process of selection is independent of previous selections. This continues until all individuals have heard the rumor. The action of one friend telling the rumor doesn't influence the preference of the next choice.
Because each selection is independent:
  • The probability remains unaffected by prior selections.
  • Previous events do not alter the pool of choices except by reducing the number of unselected individuals.
  • Calculations rely on consistent probabilities given remaining options.
These principles simplify complex probability problems by allowing a breakdown into simple sequential steps.
Geometric Distribution
Geometric distribution describes the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials. It is applicable to scenarios where we repeat an experiment until a success occurs.
In the context of the rumor, as in part (d) of the problem, we consider how many times a friend can hear the rumor before F finally hears it. If we define "success" as F finally hearing the rumor, then the probability that F still hasn't heard the rumor after 10 announcements follows a geometric distribution.
The formula for this scenario is \( \left(\frac{4}{5}\right)^{10} \), reflecting the outcome where F is not chosen "successfully" after many repetitions. The geometric distribution helps calculate the declining probability as the number of trials increases, showing how unlikely it is for F to remain unnoticed over several turns.
Combinatorial Probability
Combinatorial probability combines basic counting principles with probability to study outcomes based on permutations and combinations.
In this problem, especially for parts (a) and (c), combinatorial probability is employed to calculate specific orderings and arrangements. To find the probability for sequence B, C, D, E, and then F, we apply combinatorial reasoning. The probability of receiving the rumor follows a series of decisions, each influenced by the reduced number of yet-to-hear friends.
Using simple permutations:
  • The choice of each subsequent friend depends on how many are left.
  • We multiply probabilities reflecting these decreasing choices.
  • Calculations consider only the sequence without repetition or influence by previous choices.
This type of probability calculation is crucial for understanding how different arrangements and sequences lead to different outcomes.

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

Again consider a Little League team that has 15 players on its roster. a. How many ways are there to select 9 players for the starting lineup? b. How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters? c. Suppose 5 of the 15 players are left-handed. How many ways are there to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players?

Consider the following information about travelers on vacation (based partly on a recent Travelocity poll): \(40 \%\) check work email, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them, \(23 \%\) both check work email and use a cell phone to stay connected, and \(51 \%\) neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, 88 out of every 100 who bring a laptop also check work email, and 70 out of every 100 who use a cell phone to stay connected also bring a laptop. a. What is the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected? b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected? c. If the randomly selected traveler checked work email and brought a laptop, what is the probability that he/she uses a cell phone to stay connected?

Consider randomly selecting a single individual and having that person test drive 3 different vehicles. Define events \(A_{1}\), \(\mathrm{A}_{2}\), and \(\mathrm{A}_{3}\) by \(A_{1}=\) likes vehicle \(\\# 1 \quad A_{2}=\) likes vehicle \(\\# 2\) \(A_{3}=\) likes vehicle \(\\# 3\) Suppose that \(P\left(A_{1}\right)=.55, P\left(A_{2}\right)=.65, P\left(A_{3}\right)=.70\), \(P\left(A_{1} \cup A_{2}\right)=.80, P\left(A_{2} \cap A_{3}\right)=.40\), and \(P\left(A_{1} \cup A_{2} \cup A_{3}\right)=.88\). a. What is the probability that the individual likes both vehicle #1 and vehicle #2? b. Determine and interpret \(P\left(A_{2} \mid A_{3}\right)\). c. Are \(\mathrm{A}_{2}\) and \(\mathrm{A}_{3}\) independent events? Answer in two different ways. d. If you learn that the individual did not like vehicle #1, what now is the probability that he/she liked at least one of the other two vehicles?

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\\{\) left ear tag is lost \(\\}\) and \(C_{2}=\\{\) right ear tag is lost \(\\}\). Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost, given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildlife Mgmt., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities: a. \(P\) (all of the next three vehicles inspected pass) b. \(P\) (at least one of the next three inspected fails) c. \(P\) (exactly one of the next three inspected passes) d. \(P\) (at most one of the next three vehicles inspected passes) e. Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)?

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.