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Chapter 2: Problem 36

An academic department with five faculty members narrowed its choice for department head to either candidate \(A\) or candidate \(B\). Each member then voted on a slip of paper for one of the candidates. Suppose there are actually three votes for \(A\) and two for \(B\). If the slips are selected for tallying in random order, what is the probability that \(A\) remains ahead of \(B\) throughout the vote count (e.g., this event occurs if the selected ordering is \(A A B A B\), but not for \(A B B A A)\) ?

Short Answer

Expert verified
The probability is \( \frac{1}{5} \).

Step by step solution

01

Understanding the Problem

We are tasked with finding the probability that candidate \( A \) stays ahead of candidate \( B \) throughout the vote count when slips are drawn randomly. Given there are 3 votes for \( A \) and 2 votes for \( B \), the challenge is to create all possible sequences where \( A \) always leads \( B \).
02

Total Possible Sequences

The total number of possible sequences of announcing votes is determined by how many different ways we can reorder three A's and two B's. This is given by combinations: \( \binom{5}{3} = 10 \).
03

Calculating Favorable Sequences

To ensure \( A \) always stays ahead of \( B \), the sequence must never allow B to tie or surpass A at any point during the counting process. In our example, sequences like \( AABAB \) or \( AAABB \) are valid. Using the Catalan number, there's only 1 valid sequence: \( AABAB \).
04

Probability Calculation

Now, divide the number of favorable sequences by the total possible sequences to get the probability. With 1 favorable sequence and 10 total sequences, the probability that \( A \) stays ahead is \( \frac{1}{5} \).

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of objects. It provides tools for determining how items can be arranged or grouped under specific constraints. In the context of our voting problem, combinatorics helps us understand how to arrange the votes by using combinations.
  • A combination specifically refers to selecting items from a group, where the order does not matter.
  • In the exercise, to figure out how voters cast their ballots, we compute the total number of ways to arrange the votes for candidates \(A\) and \(B\).
  • The formula for combinations is \(\binom{n}{k}\), which represents the number of ways to choose \(k\) items from \(n\) items without regard to the order.
For the given problem, we calculate \(\binom{5}{3}\), which tells us there are 10 ways to arrange the votes with three for \(A\) and two for \(B\), supporting our analysis.
Catalan numbers
Catalan numbers are a sequence of natural numbers that have found their way into various combinatorial math problems, such as polygon division and path counting. They are especially useful in scenarios like the voting problem where certain conditions must be maintained in the arrangement.
  • In our problem, Catalan numbers help identify the sequences where \(A\) stays ahead of \(B\).
  • The \(n\)-th Catalan number, \(C_n\), can be given by the formula \(C_n = \frac{1}{n+1} \binom{2n}{n}\).
  • For the voting scenario, where there are 3 votes for \(A\) and 2 for \(B\), we use the Catalan distribution to confirm that only certain sequences maintain \(A\)'s lead.
Here, the application of Catalan numbers reduces the complexity of listing each sequence by ruling out those that allow \(B\) to tie or surpass \(A\).
Voting problem
The voting problem ensures that one candidate always stays ahead in the tallying process. This problem is commonly studied using concepts from combinatorics and probability to illustrate "staying ahead" sequences.
  • In this specific problem, candidate \(A\) needs to maintain a lead through all counts of votes.
  • Given 3 votes for \(A\) and 2 for \(B\), the sequences like \(AABAB\) where \(A\) stays ahead need to be identified.
  • Each alternative order of counting votes is systematically evaluated to determine which maintains the lead for \(A\).
The probability of \(A\) staying ahead is the number of favorable sequences divided by the total possible sequences. The use of Catalan numbers shows that there is precisely one such sequence, confirming the probability as \(\frac{1}{5}\). This demonstrates the power of combining mathematical concepts to solve practical problems in probability.

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Most popular questions from this chapter

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered \(1,2, \ldots, 6\), then one outcome consists of computers 1 and 2 , another consists of computers 1 and 3 , and so on). a. What is the probability that both selected setups are for laptop computers? b. What is the probability that both selected setups are desktop machines? c. What is the probability that at least one selected setup is for a desktop computer? d. What is the probability that at least one computer of each type is chosen for setup?

Disregarding the possibility of a February 29 birthday, suppose a randomly selected individual is equally likely to have been born on any one of the other 365 days. a. If ten people are randomly selected, what is the probability that all have different birthdays? That at least two have the same birthday? b. With \(k\) replacing ten in part (a), what is the smallest \(k\) for which there is at least a \(50-50\) chance that two or more people will have the same birthday? c. If ten people are randomly selected, what is the probability that either at least two have the same birthday or at least two have the same last three digits of their Social Security numbers? [Note: The article "Methods for Studying Coincidences" (F. Mosteller and P. Diaconis, J. Amer: Stat. Assoc., 1989: 853鈥�861) discusses problems of this type.]

Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose that \(16 \%\) of all ticks in a certain location carry Lyme disease, \(10 \%\) carry HGE, and \(10 \%\) of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease?

One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. a. What is the probability that a red ball is selected from the first box and a red ball is selected from the second box? b. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

A company uses three different assembly lines- \(A_{1}, A_{2}\), and \(A_{3}\)-to manufacture a particular component. Of those manufactured by line \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components need rework and \(10 \%\) of \(A_{3}\) 's need rework. Suppose that \(50 \%\) of all components are produced by line \(A_{1}, 30 \%\) are produced by line \(A_{2}\), and \(20 \%\) come from line \(A_{3}\). If a randomly selected component needs rework, what is the probability that it came from line \(A_{1}\) ? From line \(A_{2}\) ? From line \(A_{3}\) ?
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