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Chapter 26: Problem 13

A set of \(2 N+1\) rods consists of one of each integer length \(1,2, \ldots, 2 N, 2 N+1\) Three, of lengths \(a, b\) and \(c\), are selected, of which \(a\) is the longest. By considering the possible values of \(b\) and \(c\), determine the number of ways in which a nondegenerate triangle (i.e. one of non-zero area) can be formed (i) if \(a\) is even, and (ii) if \(a\) is odd. Combine these results appropriately to determine the total number of non-degenerate triangles that can be formed with the \(2 N+1\) rods, and hence show that the probability that such a triangle can be formed from a random selection (without replacement) of three rods is $$ \frac{(N-1)(4 N+1)}{2\left(4 N^{2}-1\right)} $$

Short Answer

Expert verified
The probability is \( \frac{(N-1)(4N+1)}{2(4N^2-1)} \).

Step by step solution

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01

- Understanding the Triangle Inequality

To form a non-degenerate triangle with sides of lengths a, b, and c, where a is the longest side, the triangle inequality must be satisfied: \(a < b + c\).
02

- Evaluating the Maximum for Side a

Given the rod lengths are \(1, 2, \ldots, 2N, 2N+1\), let’s assume \(a\) is the longest side. If \(a = 2k\) (even), the possible scenarios for \(b\) and \(c\) are \(b < a\) and \(c < a\).
03

- Counting the Possibilities for Even a

For \(a = 2k\), we have \(b\) and \(c\) such that \(b + c > 2k\). This implies different cases based on the values \(b\) and \(c\) can take. We'll need to count valid combinations.
04

- Evaluating the Maximum for Side a if Odd

If \(a = 2k+1\) (odd), similar constraints of \(a\) still hold: \(a < b + c\) with \(b < a\) and \(c < a\).
05

- Counting the Possibilities for Odd a

Use similar methods to tally combinations for \(b\) and \(c\) such that \(b + c > 2k+1\) while ensuring both \(b, c \leq 2k\).
06

- Summing Up Even and Odd Cases

Sum combinations for both even and odd \(a\) to find the total number of ways to select three lengths \(a, b, c\) to form non-degenerate triangles.
07

- Calculating Total Possible Selections

The total possible ways to select any 3 rods from \(2N+1\) is \(\binom{2N+1}{3}\).
08

- Finding the Probability

Calculate the probability that a randomly selected set of 3 rods form a non-degenerate triangle by dividing the number of valid triangles by the total possible selections.
09

- Final Expression for Probability

Using above steps, derive and simplify the probability formula to show that it equals \(\frac{(N-1)(4N+1)}{2(4N^2-1)}\).

Key Concepts

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

Triangle Inequality
To form a non-degenerate triangle (one with a non-zero area), we need to satisfy the triangle inequality condition: for any chosen sides of lengths \(a\), \(b\), and \(c\) with \(a \geq b \geq c\), the following must hold true:
\[ a < b + c \]
This means the length of the longest side must be less than the sum of the other two sides. This rule is important because, without it, the 'triangle' would just be a straight line or not possible at all. When working on problems involving rods or segments, always remember this fundamental rule to determine if a valid triangle can form.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, combination, and permutation. Here, we use combinatorics to determine the number of ways to select sets of rods that can form a triangle. Given a set of \(2N + 1\) rods with lengths ranging from 1 to \(2N + 1\), we need to count the valid combinations of three rods that satisfy the triangle inequality. To do so:
  • Identify all possible combinations of three rods using \(\binom{2N+1}{3}\).
  • Filter combinations where the chosen rods meet the triangle inequality.
By considering cases where the longest rod \(a\) is either even or odd, we can simplify the counting process. This allows us to methodically evaluate different scenarios without missing any valid combinations.
Probability Calculation
After determining the number of valid ways to select three rods forming a triangle, we compute the probability. Here is the step-by-step process:
  • Calculate the total number of ways to select any three rods from \(2N + 1\) rods using the combinations formula: \(\binom{2N+1}{3}\).
  • Determine the number of valid combinations that form a non-degenerate triangle.
  • Divide the number of valid combinations by the total combinations to get the probability.
Thus, if \(V\) is the number of valid triangles, and \(T\) is the total combinations, the probability \(P\) is given by:
\[ P = \frac{V}{T} \] For our specific case, the probability formula is simplified to:
\[ \frac{(N-1)(4N+1)}{2(4N^2-1)} \] This step-by-step approach ensures clarity and methodically breaks down the problem.
Integer Length Rods
In this problem, the rods have integer lengths from 1 to \(2N + 1\). The key steps to solve the problem involve understanding how:
  • Each rod’s length impacts the formation of a triangle.
  • The parity (odd or even nature) of the longest rod affects the combinations.
If the longest rod, \(a\), is even:
  • Consider lengths for \(b\) and \(c\) such that \(b + c > a\).
  • Evaluate maximum possible pairings without exceeding \(a\).
Similarly, if the longest rod, \(a\), is odd:
  • Parameters for \(b\) and \(c\) change slightly but still must satisfy \(b + c > a\).
  • Count pairings systematically.
Working through the problem systematically by considering even and odd cases separately helps in managing the complexity and ensuring all valid combinations are counted.

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Most popular questions from this chapter

For a non-negative integer random variable \(X\), in addition to the probability generating function \(\Phi_{X}(t)\) defined in equation (26.71) it is possible to define the probability generating function $$ \Psi_{X}(t)=\sum_{n=0}^{\infty} g_{n} t^{n} $$ where \(g_{n}\) is the probability that \(X>n\). (a) Prove that \(\Phi_{X}\) and \(\Psi_{X}\) are related by $$ \Psi_{X}(t)=\frac{1-\Phi_{X}(t)}{1-t} $$ (b) Show that \(E[X]\) is given by \(\Psi_{X}(1)\) and that the variance of \(X\) can be expressed as \(2 \Psi_{X}^{\prime}(1)+\Psi_{X}(1)-\left[\Psi_{X}(1)\right]^{2}\) (c) For a particular random variable \(X\), the probability that \(X>n\) is equal to \(\alpha^{n+1}\) with \(0<\alpha<1\). Use the results in \((\mathrm{b})\) to show that \(V[X]=\alpha(1-\alpha)^{-2}\).

A particle is confined to the one-dimensional space \(0 \leq x \leq a\) and classically it can be in any small interval \(d x\) with equal probability. However, quantum mechanics gives the result that the probability distribution is proportional to \(\sin ^{2}(n \pi x / a)\), where \(n\) is an integer. Find the variance in the particle's position in both the classical and quantum mechanical pictures and show that, although they differ, the latter tends to the former in the limit of large \(n\), in agreement with the correspondence principle of physics.

In the game of Blackball, at each turn Muggins draws a ball at random from a bag containing five white balls, three red balls and two black balls; after being recorded, the ball is replaced in the bag. A white ball earns him \(\$ 1\) whilst a red ball gets him \(\$ 2\); in either case he also has the option of leaving with his current winnings or of taking a further turn on the same basis. If he draws a black ball the game ends and he loses all he may have gained previously. Find an expression for Muggins' expected return if he adopts the strategy to drawing up to \(n\) balls if he has not been eliminated by then. Show that, as the entry fee to play is \$3, Muggins should be dissuaded from playing Blackball, but if that cannot be done what value of \(n\) would you advise him to adopt?

A tennis tournament is arranged on a straight knockout basis for \(2^{n}\) players and for each round, except the final, opponents for those still in the competition are drawn at random. The quality of the field is so even that in any match it is equally likely that either player will win. Two of the players have surnames that begin with ' \(Q^{\prime}\). Find the probabilities that they play each other (a) in the final, (b) at some stage in the tournament.

The variables \(X_{i}, i=1,2, \ldots, n\), are distributed as a multivariate Gaussian, with means \(\mu_{i}\) and a covariance matrix \(\mathrm{V} .\) If the \(X_{i}\) are required to satisfy the linear constraint \(\sum_{i-1}^{n} c_{i} X_{i}=0\), where the \(c_{i}\) are constants (and not all equal to zero), show that the variable $$ \chi_{n}^{2}=(\mathrm{x}-\mu)^{\mathrm{T}} \mathrm{V}^{-1}(\mathrm{x}-\mu) $$ follows a chi-squared distribution of order \(n-1 .\)
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