91Ó°ÊÓ

An orienteer runs on the rectangular grid through the grid points \((m, n), m, n=0,1,2, \ldots\) of a Cartesian plane. On reaching \((m, n)\), the orienteer must next proceed either to \((m+1, n)\) or \((m, n+1)\). (a) Show the number of different paths from \((0,0)\) to \((n, n)\) equals the number from \((1,0)\) to \((n+1, n)\) and that this equals \(\left(\begin{array}{c}2 n \\ n\end{array}\right)\), where \(\left(\begin{array}{l}k \\ r\end{array}\right)=\frac{k !}{r !(k-r) !}\). (b) Show that the number of different paths from \((1,0)\) to \((n+1, n)\) passing through at least one of the grid points \((r, r)\) with \(1 \leq r \leq n\) is equal to the total number of different paths from \((0,1)\) to \((n+1, n)\) and that this equals \(\left(\begin{array}{c}2 n \\ n-1\end{array}\right)\). (c) Suppose that at each grid point the orienteer is equally likely to choose to go to either of the two possible next grid points. Let \(A_{k}\) be the event that the first of the grid points \((r, r), r \geq 1\), to be visited is \((k, k)\). Show that $$ \mathbf{P}\left(A_{k}\right)=\frac{4^{-k}}{2 k-1}\left(\begin{array}{c} 2 k-1 \\ k \end{array}\right) $$

Step by step solution

01

Understanding Pathways on the Grid

The orienteer starts from (0,0) and can only move right or up at each step. To get from (0,0) to (n,n), the orienteer will need to take exactly n right steps and n up steps, totaling 2n steps.

02

Binomial Coefficient Explanation

To choose n right steps out of 2n total steps, we calculate the number of combinations as \(\binom{2n}{n}\). This represents all distinct ways of arranging n right and n up steps in sequence.

03

Perspective Shift and Equivalent Path Recalculation

From (1,0) to (n+1,n), the orienteer still needs n right and n up moves. Consider the additional initial right step as part of the same sequence, confirming the paths count as still \(\binom{2n}{n}\).

04

Total Paths Verification for Other Start Points

For paths from (0,1) to (n+1,n), the orienteer needs n right steps and n-1 up steps. The number of distinct paths is thus \(\binom{2n}{n-1}\), mirroring the passage through at least one grid point (r,r).

05

Event Probability Calculation

For event \(A_k\), first reaching (k, k), calculate the probability. Prior pathways feature k-1 rights and k-1 ups in various orders to (k-1,k-1), followed by two choices to a single (k,k). Use probability "\(\mathbf{P}(A_k) = \frac{4^{-k}}{2k-1}\binom{2k-1}{k}\)."

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Key Concepts

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

Binomial Coefficient

When you're dealing with paths on a grid, the binomial coefficient plays a crucial role in counting possible routes. The binomial coefficient is denoted as \( \left( \begin{array}{c} k \ r \end{array} \right) \), and it represents the number of ways to choose \( r \) elements from a set of \( k \) elements without considering the order. To calculate it, use the formula: \( \left( \begin{array}{c} k \ r \end{array} \right) = \frac{k!}{r!(k-r)!} \), where \( ! \) denotes factorial, the product of all positive integers up to that number.

In our exercise, we look at pathways from \((0,0)\) to \((n,n)\). For every path, the orienteer needs to take \( n \) right (R) steps and \( n \) up (U) steps. Altogether, it amounts to choosing \( n \) right steps from \( 2n \) total steps, making the calculation equivalent to the binomial coefficient \( \left( \begin{array}{c} 2n \ n \end{array} \right) \).

This calculation helps us understand that there are many ways to arrange these steps, which is fundamental when considering all possible distinct paths on the grid.

Grid Points

Grid points are specific intersections where lines of a grid cross in the Cartesian plane. Think about it like coordinates \((m,n)\), where \(m\) and \(n\) denote the horizontal and vertical positions, respectively. In our exercise, an orienteer travels through these points, and the challenge involves finding pathways through them.

The orienteer has a choice at each grid point, to move in one of two possible directions - either to the right or upwards. This movement between grid points creates what we call *combinatorial paths*. As the orienteer continues, these choices systematically lead to the next grid point, continuing the journey through the grid.

Understanding the journey through grid points is essential. It breaks down a complex path into smaller, easily understandable steps. Each choice at a point influences the rest of the journey, offering a simple yet powerful way to view probability and counting in mathematical problems.

Cartesian Plane

The Cartesian plane is a coordinate system that specifies each point uniquely using pairs of numerical coordinates, in the form \((x, y)\). Think of it as a two-dimensional graph that helps visualize mathematical concepts like paths or geometrical shapes.

In relation to our exercise, imagine the orienteer moving on this plane, traveling from one coordinate \((m, n)\) to another. This setting allows for visualizing the paths in a structured way, making it easier to grasp the points of movement and direction.

Visualizing the paths on a Cartesian plane highlights the calculation's geometry. Here, points \((0,0)\) and \((n,n)\) are significant, as they represent start and end locations. Moving across the Cartesian plane involves understanding this grid layout, helping solve and explain numerous mathematical problems, including those associated with probability calculations and binomial coefficients.

Probability Calculation

Probability calculation involves determining the likelihood of various outcomes. In our scenario, it's about determining the probability of the orienteer reaching certain grid points first on a path. When the orienteer randomly chooses between two directions at each step, this probability reflects the system's inherent randomness.

The event \(A_k\) occurs when the orienteer first reaches the grid point \((k, k)\). This can be calculated by considering paths that lead precisely to \((k,k)\). The journey to \((k, k)\) involves arranging \( k-1 \) right moves and \( k-1 \) up moves in any order, before making one of two choices that finally lead to this specific point.

The probability of reaching \((k,k)\) "first" uses the formula: \( \mathbf{P}(A_k) = \frac{4^{-k}}{2k-1}\left( \begin{array}{c} 2k-1 \ k \end{array} \right) \). This reflects both the binomial arrangement of steps and the randomness inherent in each step choice. By understanding these calculations, one gains insights into combinatorial probabilities, which are an integral part of many mathematics problems.