/*! 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 89 Individual A has a circle of fiv... [FREE SOLUTION] | 91Ó°ÊÓ

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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 B, C, D, E, and F? b. What is the probability that \(F\) 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?

Short Answer

Expert verified
a. \(\frac{1}{120}\), b. \(\frac{2}{15}\), c. \(\frac{1}{5}\).

Step by step solution

01

Probability of a Specific Order (Part a)

For all 5 people to hear the rumor in a specific order (B, C, D, E, and F), A first chooses B, then B chooses C out of the remaining 4, C chooses D out of the remaining 3, D chooses E from the remaining 2, and finally, E chooses F. The probability for each step is:\[ \text{Probability of B first} = \frac{1}{5}, \quad \text{Probability of C second} = \frac{1}{4}, \quad \text{Probability of D third} = \frac{1}{3}, \quad \text{Probability of E fourth} = \frac{1}{2}, \quad \text{Probability of F last} = 1. \]Thus, the total probability is:\[ \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{120}. \]
02

Probability F is Third (Part b)

Calculate the probability that F is the third person to hear the rumor. There are two scenarios: 1. A picks one of \( \{B, C, D, E\} \).2. If A picks B, B must pick F, then F picks among the remaining people. Each sequence inside this scenario has \(3!\) arrangements.Calculate the probability for each path (since multiple paths lead to F being third): A chooses any but F (\(\frac{1}{5}\)), the second person chooses F (\(\frac{1}{4}\)), and then the rest can be anyone (\(\frac{1}{6}\) for each 3 remaining ones): \[ \frac{4}{5} \times \frac{1}{4} \times \frac{1}{6} = \frac{1}{30}. \]Add the ways across the first person selected correctly, so total is: \(4 \, \text{ways} \times \frac{1}{30} = \frac{4}{30}=\frac{2}{15} \).
03

Probability F is Last (Part c)

Calculate the probability that F hears the rumor last. F should not be chosen by each of the first four people hearing the rumor.1. Probability for listeners 1-4: each can choose one from \(\{B, C, D, E\}\) \(= \frac{4}{4, 3, 2, 1}\).Probability F is chosen fourth in sequence by everyone \(\rightarrow \frac{1}{4}= \frac{1}{24} \text{ since the numerator 4 sequences divide completely once more i.e \(\frac{1}{5!} \)}. \)This multiplies the sequence possibilities into it since you set (e.g. stuck method) while not choosing F through first 4-person chain\(\frac{1}{120} same too .\)

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Key Concepts

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

Combinatorics
Combinatorics is a field of mathematics concerned with counting, arrangement, and combination of objects. In our context, combinatorics helps us determine the number of ways a rumor can circulate among a group of friends. It allows us to set up the problem by understanding how choices among a particular set can be made.
  • Combinatorics covers various concepts, such as permutations and combinations.
  • Permutations focus on the arrangement of objects in a specific order, which is precisely what we're looking at when determining the order the friends hear the rumor.
For example, in the problem, if we wanted to find the total number of ways to arrange the order in which five friends might hear a rumor, we would calculate this using factorial notation. Specifically, with five friends, the permutations would be represented by: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]This equation tells us that there are 120 different ways to arrange the sequence in which the rumor can be heard among friends B, C, D, E, and F.
Permutations
Permutations involve arranging items in a specific sequence or order. The key feature of permutations is that the order matters. In the given problem, every permutation signifies a unique sequence in which the rumor is passed.
  • The problem requires understanding that each choice by a person affects the subsequent choices and overall sequence.
  • It's important to realize that with each person hearing the rumor, there are progressively fewer people left to choose from.
For the step-by-step solution to part a, for example, we calculated the probability of the specific order B, C, D, E, and F by considering these selections:
  • The first choice has 5 options: \(\frac{1}{5}\)
  • The next choice out of four remaining: \(\frac{1}{4}\)
  • Then out of three: \(\frac{1}{3}\)
  • Out of two, and finally one: \(\frac{1}{2}\) and last as \(1\)
By multiplying these probabilities, we find that the probability of the exact sequence B, C, D, E, F occurring is \(\frac{1}{120}\). This reflects how permutations require specific ordered arrangements.
Conditional Probability
Conditional probability is the likelihood of an event happening given that another event has already occurred. In our exercise, we found conditional probabilities to be key when determining probabilities of different rumor sequences.
  • It first appears in part b, where we seek the probability that F is the third to hear the rumor given certain conditions are met.
  • Here, conditional probability helps understand how preceding events affect subsequent probabilities.
For part b, the event that F is the third listener is conditional on the choices of the first two people. We calculated it as: \[ \frac{4}{5} \times \frac{1}{4} \times \frac{3!}{6} = \frac{2}{15} \]This entails multiplying the probability of the first two selections and considering all possible permutations of the remaining people. Similarly, part c involves F being last, which depends on none of the first four people picking F. Through conditional probability, the intricate nature of how earlier choices condition following probabilities becomes clear, especially where the order and specific roles of friends in sequence matter.

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Most popular questions from this chapter

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\\{\) the first board is green \(\\}\) and \(B=\\{\) the second board is green \(\\}\). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=\) \(P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think that an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

A college library has five copies of a certain text on reserve. Two copies ( 1 and 2 ) are first printings, and the other three \((3,4\), and 5\()\) are second printings. A student examines these books in random order, stopping only when a second printing has been selected. One possible outcome is 5 , and another is \(213 .\) a. List the outcomes in \(\mathscr{\text { . }}\) b. Let \(A\) denote the event that exactly one book must be examined. What outcomes are in \(A ?\) c. Let \(B\) be the event that book 5 is the one selected. What outcomes are in \(B ?\) d. Let \(C\) be the event that book 1 is not examined. What outcomes are in \(C\) ?

Allan and Beth currently have \(\$ 2\) and \(\$ 3\), respectively. A fair coin is tossed. If the result of the toss is \(\mathrm{H}\), Allan wins \(\$ 1\) from Beth, whereas if the coin toss results in \(\mathrm{T}\), then Beth wins \(\$ 1\) from Allan. This process is then repeated, with a coin toss followed by the exchange of \(\$ 1\), until one of the two players goes broke (one of the two gamblers is ruined). We wish to determine \(a_{2}=P\) (Allan is the winner \(\mid\) he starts with \(\$ 2\) ) To do so, let's also consider \(a_{i}=P\) (Allan wins | he starts with \(\$ i\) ) for \(i=0,1,3,4\), and 5 . a. What are the values of \(a_{0}\) and \(a_{5}\) ? b. Use the law of total probability to obtain an equation relating \(a_{2}\) to \(a_{1}\) and \(a_{3}\). [Hint: Condition on the result of the first coin toss, realizing that if it is a \(\mathrm{H}\), then from that point Allan starts with \$3.] c. Using the logic described in (b), develop a system of equations relating \(a_{i}(i=1,2,3,4)\) to \(a_{i-1}\) and \(a_{i+1}\). Then solve these equations. [Hint: Write each equation so that \(a_{i}-a_{i-1}\) is on the left hand side. Then use the result of the first equation to express each other \(a_{i}-a_{i-1}\) as a function of \(a_{1}\), and add together all four of these expressions \((i=2,3,4,5)\).] d. Generalize the result to the situation in which Allan's initial fortune is \(\$ a\) and Beth's is \(\$ b\). Note: The solution is a bit more complicated if \(p=P(\) Allan wins \(\$ 1) \neq .5 .\)

A particular iPod playlist contains 100 songs, of which 10 are by the Beatles. Suppose the shuffle feature is used to play the songs in random order (the randomness of the shuffling process is investigated in "Does Your iPod Really Play Favorites?" (The Amer. Statistician, 2009: 263 - 268)). What is the probability that the first Beatles song heard is the fifth song played?

Fifteen telephones have just been received at an authorized service center. Five of these telephones are cellular, five are cordless, and the other five are corded phones. Suppose that these components are randomly allocated the numbers 1 , \(2, \ldots, 15\) to establish the order in which they will be serviced. a. What is the probability that all the cordless phones are among the first ten to be serviced? b. What is the probability that after servicing ten of these phones, phones of only two of the three types remain to be serviced? c. What is the probability that two phones of each type are among the first six serviced?

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