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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

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 \(\frac{(N-1)(4N+1)}{2(4N^2-1)}\).

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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

Under EU legislation on harmonisation, all kippers are to weigh \(0.2000 \mathrm{~kg}\) and vendors who sell underweight kippers must be fined by their government. The weight of a kipper is normally distributed with a mean of \(0.2000 \mathrm{~kg}\) and a standard deviation of \(0.0100 \mathrm{~kg}\). They are packed in cartons of 100 and large quantities of them are sold. Every day a carton is to be selected at random from each vendor and tested according to one of the following schemes, which have been approved for the purpose. (a) The entire carton is weighed and the vendor is fined 2500 euros if the average weight of a kipper is less than \(0.1975 \mathrm{~kg}\). (b) Twenty-five kippers are selected at random from the carton; the vendor is fined 100 euros if the average weight of a kipper is less than \(0.1980 \mathrm{~kg}\). (c) Kippers are removed one at a time, at random, until one has been found that weighs more than \(0.2000 \mathrm{~kg}\); the vendor is fined \(n(n-1)\) euros, where \(n\) is the number of kippers removed.

(a) Gamblers \(A\) and \(B\) each roll a fair six-faced die, and \(B\) wins if his score is strictly greater than \(A\) 's. Show that the odds are 7 to 5 in \(A\) 's favour. (b) Calculate the probabilities of scoring a total \(T\) from two rolls of a fair die for \(T=2,3, \ldots, 12 .\) Gamblers \(C\) and \(D\) each roll a fair die twice and score respective totals \(T_{C}\) and \(T_{D}, D\) winning if \(T_{D}>T_{C} .\) Realising that the odds are not equal, \(D\) insists that \(C\) should increase her stake for each game. \(C\) agrees to stake \(£ 1.10\) per game, as compared to \(D\) 's \(£ 1.00\) stake. Who will show a profit?

\(X_{1}, X_{2}, \ldots, X_{n}\) are independent identically distributed random variables drawn from a uniform distribution on \([0,1] .\) The random variables \(A\) and \(B\) are defined by $$ A=\min \left(X_{1}, X_{2}, \ldots, X_{n}\right), \quad B=\max \left(X_{1}, X_{2}, \ldots, X_{n}\right) $$ For any fixed \(k\) such that \(0 \leq k \leq \frac{1}{2}\), find the probability \(p_{n}\) that both $$ A \leq k \quad \text { and } \quad B \geq 1-k $$ Check your general formula by considering directly the cases (a) \(k=0,\left(\right.\) b) \(k=\frac{1}{2}\), (c) \(n=1\) and \((\) d) \(n=2\)

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 \(V[X]=\alpha(1-\alpha)^{-2}\).

As assistant to a celebrated and imperious newspaper proprietor, you are given the job of running a lottery in which each of his five million readers will have an equal independent chance \(p\) of winning a million pounds; you have the job of choosing \(p .\) However, if nobody wins it will be bad for publicity whilst if more than two readers do so, the prize cost will more than offset the profit from extra circulation - in either case you will be sacked! Show that, however you choose \(p\), there is more than a \(40 \%\) chance you will soon be clearing your desk.

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