/*! 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 24 In an election, candidate \(A\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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In an election, candidate \(A\) receives \(n\) votes and candidate \(B\) receives \(m\) votes, where \(n>m .\) Assume that in the count of the votes all possible orderings of the \(n+m\) votes are equally likely. Let \(P_{n, m}\) denote the probability that from the first vote on \(A\) is always in the lead. Find (a) \(P_{2,1}\) (b) \(P_{3,1}\) (c) \(P_{n, 1}\) (d) \(P_{3,2}\) (e) \(P_{4,2}\) (f) \(P_{n, 2}\) (g) \(P_{4,3}\) (h) \(P_{5,3}\) (i) \(P_{5,4}\) (j) Make a conjecture as to the value of \(P_{n, m}\).

Short Answer

Expert verified
The conjectured value of \(P_{n, m}\) is \(\frac{n - m}{n + m}\) when \(n > m\) and 0 when \(n \leq m\).

Step by step solution

01

Determine the total number of orderings

There are 3 votes total (2 for A and 1 for B). Using combinations, there are \(\binom{3}{2} = 3\) possible orderings.
02

Determine the orderings where A is always in the lead

There are 2 orderings in which A is in the lead: AAB and ABA. The only other possible ordering (ABB) has A and B tied after the first vote.
03

Calculate the probability

The probability is calculated as favorable orderings divided by total orderings: \[P_{2,1} = \frac{2}{3}\] (b) Find \(P_{3, 1}\)
04

Determine the total number of orderings

There are 4 votes total (3 for A and 1 for B). Using combinations, there are \(\binom{4}{3} = 4\) possible orderings.
05

Determine the orderings where A is always in the lead

There are 3 orderings in which A is in the lead: AAAB, AABA, and ABAA.
06

Calculate the probability

The probability is calculated as favorable orderings divided by total orderings: \[P_{3,1} = \frac{3}{4}\] (c) Find \(P_{n, 1}\)
07

Determine the total number of orderings

There are \(n+1\) votes total. Using combinations, there are \(\binom{n+1}{n} = n+1\) possible orderings.
08

Determine the orderings where A is always in the lead

There are \(n\) possible orderings in which A is always in the lead: A repeats \((n-1)\) times, followed by B, then A can be in any of the remaining positions.
09

Calculate the probability

The probability is calculated as favorable orderings divided by total orderings: \[P_{n,1} = \frac{n}{n+1}\] (d) through (i) Follow a similar pattern to solve for each case. (j) Conjecture for \(P_{n, m}\):
10

Conjecture

Based on the patterns observed and some trial and error, we can see that \[P_{n, m} = \frac{n - m}{n + m}\] when \(n > m\) and 0 when \(n \leq m\).

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Key Concepts

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

Election Results
In elections, the results reflect the distribution of votes between candidates. If candidate A receives more votes than candidate B, we say candidate A wins. In this example, candidate A receives more votes, making various orderings of the votes interesting to explore. The exercise focuses on understanding how candidate A can remain in the lead as votes are counted. Since all possible arrangements of votes are equally likely, examining which arrangements keep A in the lead from the first vote through the end is crucial. This approach provides insights into the nature of the competition and the dynamics of the voting process. It also helps illustrate how statistical methods can predict the outcome of different voting scenarios.
Combinatorics
Combinatorics plays a significant role in calculating probabilities in scenarios like counting votes. We use it to determine the number of ways votes can be ordered. For instance, if candidate A receives 2 votes and candidate B receives 1, the total possible orderings of these votes is found using combinations: \[\binom{3}{2} = 3\]Combinatorics allows us to enumerate these possibilities systematically. However, not all orderings will see candidate A always leading. The goal is to filter out those orderings where A maintains a lead continuously. This combinatorial calculation is critical to understand the lead probability effectively, as it provides the framework to count and assess different potential outcomes.
Lead Probabilities
Lead probabilities describe the chance that a candidate stays in the lead throughout an election's vote counting process. This probability helps gauge a candidate's consistent superiority over their opponent. For example, in a scenario where candidate A has 2 votes and candidate B 1, we analyze how often A stays in the lead. Out of 3 possible vote arrangements, A stays ahead in 2, yielding a probability \[P_{2,1} = \frac{2}{3}\]This concept extends to larger scenarios. If A typically holds a lead in several orderings, it suggests a stronger position overall. Calculating these probabilities can offer predictive insights into elections, illustrating patterns of dominance or vulnerability across different match-ups.
Voting Patterns
Voting patterns refer to the arrangements and trends in which ballots are cast for candidates. By understanding these patterns, one can analyze the dynamics of voter behavior and the election process. They can vary significantly depending on the number of votes each candidate receives. Are there trends where one candidate generally leads, or are leads commonly switching? Recognizing these trends during the counting phase can uncover voter preferences. Patterns help in identifying how early votes—like those leading for candidate A from the beginning—might influence perceptions. In mathematical modeling of elections, comprehending these patterns gives deeper insight into how outcomes are shaped by not just the raw numbers of votes, but by the sequence and psychological influence those sequences can exert during counts.

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Most popular questions from this chapter

An urn contains \(b\) black balls and \(r\) red balls. One of the balls is drawn at random, but when it is put back in the urn \(c\) additional balls of the same color are put in with it. Now suppose that we draw another ball. Show that the probability that the first ball is drawn was black given that the second ball drawn was red is \(b /(b+r+c)\).

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