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Chapter 3: Problem 25

There are \(2 n\) balls in an urn; the balls are numbered \(1,2, \ldots, 2 n\). They are withdrawn at random without replacement. What is the probability that (a) For no integer \(j\), the \(2 j\) th ball drawn bears the number \(2 j\) ? (b) For no integer \(j\), the ball bearing the number \(j+1\) is removed next after the ball bearing the number \(j\) ?

Short Answer

Expert verified
(a) \(\frac{1}{e}\); (b) \(\frac{1}{e^2}\)

Step by step solution

01

Understanding the Problem

We are dealing with a classic probability problem that involves drawing balls from an urn without replacement. We want to calculate specific probabilities related to the numbering and ordering of the balls drawn.
02

Set Up the Scenario

Consider an urn with balls numbered from 1 to \(2n\). We will analyze two probabilities: (a) that a ball with number \(2j\) is not drawn as the \(2j\)-th ball, and (b) that for no integer \(j\), the ball bearing number \(j+1\) is drawn immediately after \(j\).
03

Calculate Probability for Part (a)

We are tasked with finding the probability that for no integer \(j\), the \(2j\)-th ball drawn is numbered \(2j\). For a sequence of draws, permutations where no ball lands on its 'correct spot' involve the concept of derangements. The number of derangements \(!k\) of \(k\) items is defined as \(!k = k! \sum_{i=0}^{k} \frac{(-1)^i}{i!}.\) For our scenario, we apply this derangement concept over the \(2n\) sequence. The probability is given by the formula \(\frac{!2n}{(2n)!},\) which is approximately \(\frac{1}{e},\) where \(e\) is the base of the natural logarithm.
04

Calculate Probability for Part (b)

In this part, we need to determine the probability that for no integer \(j\), the ball numbered \(j+1\) is drawn immediately after the ball numbered \(j\). We use a similar logic as in part (a). Define this condition as a non-successive pair for all \(j\). The problem becomes determining the number of permutations where no two consecutive numbers \((j, j+1)\) appear consecutively. This again relates to a derangement-like problem in consecutive segments, where the probability is influenced by not allowing any pairs of the form \((j, j+1)\). The final probability will approach \(\frac{1}{e^2}\), based on similar principles with a higher level of successive pair blocking.

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Key Concepts

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

Derangements
A derangement is a permutation of a set in which none of the elements appear in their original positions. Imagine having a set of numbers, and you need to rearrange it such that no number remains in its initial spot. This concept is crucial when dealing with problems where specific positions must not be occupied by particular items, like in our problem with the urn and balls.
  • The formula to find the number of derangements of a set of size \(k\) is \[!k = k! \sum_{i=0}^{k} \frac{(-1)^i}{i!}\]This formula counts the number of ways to completely shuffle items, ensuring no item stays in its original spot.
  • In our context, derangements help us calculate the probability of no ball being drawn as the 2\(j\)-th ball that matches the number 2\(j\).
Derangements frequently appear in problems involving selection or organization, highlighting their versatile nature in probability and combinatorics. They offer an elegant solution to complex scenarios, making the process more approachable.
Permutation
Permutations are arrangements of objects in a specific sequence or order. In probability, permutations allow us to explore different possibilities of arranging elements, such as numbering or ordering based on draw or selection criteria.
  • When discussing permutations, we are typically interested in how we can order a set of items uniquely. A set of \(n\) distinct items can be arranged in \(n!\) (n factorial) ways. This is the essence of permutations.
  • In our problem, we consider permutations where the sequence of draws results in arrangements that do not satisfy certain conditions, thereby incorporating the idea of derangements.
Permutation theory underlies many probability calculations by giving us a framework for counting possibilities and aligning them with our constraints for specific outcomes. This foundational concept enables the calculation of probabilities for more tailored, restriction-oriented scenarios, such as in our ball drawing scenario.
Probability without Replacement
Probability without replacement refers to scenarios where once an item is drawn, it is not returned to the pool for further selection. This is a common principle in problems where drawn items influence the remaining options, especially impacting probabilities.
  • In cases of drawing from an urn, this concept is crucial as the total number of items reduces after each draw, changing the odds for subsequent selections.
  • For our exercise, since once a ball is drawn it is not replaced, we need to consider this cumulative reduction when calculating probabilities, such as not drawing prediction number sequences consecutively.
Understanding probability without replacement helps in properly modeling real-world problems and ensuring accurate probability calculations by accounting for decreasing sample spaces. This aspect introduces a layer of complexity, as probabilities change dynamically with each draw, as evidenced by our urn problem.

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Most popular questions from this chapter

Let \(A, B, C, D\) be the vertices of a tetrahedron. A beetle is initially at \(A\); it chooses any of the edges leaving \(A\) and walks along it to the next vertex. It continues in this way; at any vertex, it is equally likely to choose to go to any other vertex next. What is the probability that it is at \(A\) when it has traversed \(n\) edges?

Let \(a_{n}(n=2,3, \ldots)\) denote the number of distinct ways the expression \(x_{1} x_{2} \ldots x_{n}\) can be bracketed so that only two quantities are multiplied together at any one time. [For example, when \(n=2\) there is only one way, \(\left(x_{1} x_{2}\right)\), and when \(n=3\) there are two ways, \(\left(x_{1}\left(x_{2} x_{3}\right)\right)\) and \(\left(\left(x_{1} x_{2}\right) x_{3}\right)\).] Prove that \(a_{n+1}=a_{n}+a_{2} a_{n-1}+a_{3} a_{n-2}+\cdots+a_{n-2} a_{3}+a_{n-1} a_{2}+a_{n}\). Defining \(A(x)=x+a_{2} x^{2}+a_{3} x^{3}+\cdots\) prove that \((A(x))^{2}=A(x)-x\). Deduce that \(A(x)=\frac{1}{2}\left(1-(1-4 x)^{\frac{1}{2}}\right)\), and show that $$ a_{n}=\frac{1.3 \ldots(2 n-3)}{n !} 2^{n-1} $$

The \(n\) passengers for an \(n\)-seat plane have been told their seat numbers. The first to board chooses a seat at random. The rest, boarding successively, sit correctly unless their allocated seat is occupied, in which case they sit at random. Let \(p_{n}\) be the probability that the last to board finds her seat free. Find \(p_{n}\), and show that \(p_{n} \rightarrow \frac{1}{2}\), as \(n \rightarrow \infty\).

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 $$ \mathbf{P}\left(A_{k}\right)=\frac{4^{-k}}{2 k-1}\left(\begin{array}{c} 2 k-1 \\ k \end{array}\right) $$

Find the number of distinguishable ways of colouring the faces of a solid regular tetrahedron with: (a) At most three colours (red, blue, and green); (b) Exactly four colours (red, blue, green, and yellow); (c) At most four colours (red, blue, green, and yellow).
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