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Chapter 7: Problem 8

A total of \(n\) balls, numbered 1 through \(n\), are put into \(n\) urns, also numbered 1 through \(n\) in such a way that ball \(i\) is equally likely to go into any of the ums \(1,2, \ldots, i\). Find (a) the expected number of urns that are empty; (b) the probability that none of the urns is empty.

Short Answer

Expert verified
In summary, the expected number of empty urns is given by \(E_{total} = n - H_n\), and the probability that none of the urns is empty is given by \(P(\text{no urns empty}) = \frac{S(n, n)}{n!}\).

Step by step solution

01

(a) Find the expected number of empty urns

To find the expected number of empty urns, we will first find the probability of each urn being empty and then sum up those probabilities using the linearity of expectation. Let \(E_i\) denote the expected number of empty urns for \(1 \leq i \leq n\). Also, let \(P_i\) denote the probability that urn \(i\) is empty after all \(n\) balls are put into the urns. We know that ball \(i\) is equally likely to go into any of the urns \(1,2,\ldots,i\). So, the probability of it going into urn \(i\) itself is \(\frac{1}{i}\). This implies that the probability of urn \(i\) being empty is \(1 - \frac{1}{i} = \frac{i-1}{i}\). Thus, \(P_i = \frac{i-1}{i}\). Using the linearity of expectation, we can write the total expected number of empty urns as: \[ E_{total} = \sum_{i=1}^{n} E_i = \sum_{i=1}^{n} P_i \] Now, substitute the values for \(P_i\) and calculate the sum to determine the expected value: \[ E_{total} = \sum_{i=1}^{n} \frac{i-1}{i} \] We can rewrite this sum as follows: \[ E_{total} = \sum_{i=1}^{n} (1 - \frac{1}{i}) = \sum_{i=1}^{n} 1 - \sum_{i=1}^{n} \frac{1}{i} \] The second sum on the right-hand side has a well-known sum, namely the harmonic number \(H_n = \sum_{i=1}^{n} \frac{1}{i}\). Thus, \[ E_{total} = n - H_n \] So, the expected number of empty urns is \(n - H_n\).
02

(b) Find the probability that none of the urns is empty

To determine the probability that none of the urns is empty, we will count the total possible ways of arranging the balls and urns and the ways when none of the urns is empty. Each ball \(i\) can be placed into any of the urns \(1,2,\ldots,i\). Thus, there are \(1\times2\times \ldots \times n = n!\) total possible ways to arrange the balls in the urns. Now we will focus on the ways when no urn is empty. For this, we will use a combinatorial technique called the Stirling Number of the Second Kind, denoted by \(S(n,n)\). It defines the number of ways to partition a set of \(n\) elements into \(n\) non-empty subsets. The probability that none of the urns is empty is given by: \[ P(\text{no urns empty}) = \frac{\text{number of arrangements with no empty urns}}{\text{total possible arrangements}} \] We can write this as: \[P(\text{no urns empty}) = \frac{S(n, n)}{n!}\] So, the probability that none of the urns is empty is given by \(\frac{S(n, n)}{n!}\).

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Key Concepts

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

Expected Value
The expected value in probability theory is a fundamental concept used to predict the average outcome of a random event over time. When discussing balls and urns, the expected value allows us to calculate what we anticipate as the average number of empty urns when balls are randomly placed. It is a theoretical mean which helps in making predictions about potential outcomes, even if each individual experiment might yield different results.

An essential trait of the expected value is that it accounts for all possible events and their probabilities. By multiplying each possible outcome by its probability and summing these products, we acquire the expected value for the given random event. This approach provides a reliable expectation that helps in decision-making and predictions, especially useful when repetitively executing similar tasks, like in numerous experiments or runs.
Linearity of Expectation
Linearity of Expectation is a principle that greatly simplifies calculations involving expected values. It states that the expected value of a sum of random variables is the sum of their respective expected values, regardless of any dependencies between them. This property can be mathematically represented as follows:

- If you have two random variables, say X and Y, then the expected value of their sum, E(X + Y), is equal to E(X) + E(Y). This principle holds true even if X and Y are not independent. In our urn example, each urn's chance of being empty acts as an independent component, allowing us to sum the expected values of emptiness.

Through this approach, we bypass complex joint probability calculations and instead focus on simple individual expected values. This makes calculations more efficient and manageable, especially with large datasets or numerous events.
Harmonic Number
Harmonic numbers appear frequently in problems involving probability and number theory. They are defined as the sum of the reciprocals of the first n natural numbers:

- The nth Harmonic Number is given by the formula:
\[ H_n = \sum_{i=1}^{n} \frac{1}{i} \]

In the context of our exercise, the harmonic number helps determine the expected number of empty urns. By summing the probabilities of each urn being occupied, we derive the complement, which gives us the expected number of empty urns as \( n - H_n \).

This number not only provides insights in probability theory but also connects to logarithmic approximations and appears in various fields such as analysis and number theory. Understanding harmonic numbers enriches any study of summative probability distributions.
Stirling Number of the Second Kind
Stirling Numbers of the Second Kind are significant in combinatorics, denoting the number of ways to partition a set of n items into k non-empty subsets. They are particularly useful in arrangements and distribution problems. In our exercise, they help calculate the probability that none of the urns remains empty.

When we use Stirling Numbers of the Second Kind, denoted as \( S(n, n) \), we're identifying how many ways n distinct balls can be distributed into n urns such that no urn is left void. This focuses on ensuring each subset is filled, following the constraints set by the problem.

This concept is useful far beyond combinatorics and probability, finding applications in various fields like computer science and symbolic analysis, where set partitioning is a common requirement.

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Most popular questions from this chapter

Consider an um containing a large number of coins and suppose that each of the coins has some probability \(p\) of turning up heads when it is flipped. However, this value of \(p\) varies from coin to coin. Suppose that the composition of the um is such that if a coin is selected at random from the urn, then its \(p\)-value can be regarded as being the value of a random variable that is uniformly distributed over \([0,1]\). If a coin is selected at random from the urn and flipped twice, compute the probability that (a) the first flip is a head; (b) both flips are heads.

The best quadratic predictor of \(Y\) with respect to \(X\) is \(a+b X+c X^{2}\), where \(a, b\), and \(c\) are chosen to minimize \(E\left[\left(Y-\left(a+b X+c X^{2}\right)\right)^{2}\right]\). Determine \(a, b\), and \(c\).

Consider a population consisting of individuals able to produce offspring of the same kind. Suppose that each individual will, by the end of its lifetime, have produced \(j\) new offspring with probability \(P_{j}, j \geq 0\), independently of the number produced by any other individual. The number of individuals initially present, denoted by \(X_{0}\), is called the size of the zeroth generation. All offspring of the zeroth generation constitute the first generation, and their number is denoted by \(X_{1} .\) In general, let \(X_{n}\) denote the size of the \(n\)th generation. Let \(\mu=\sum_{j=0}^{x} j P_{j}\) and \(\sigma^{2}=\sum_{j=0}^{x}(j-\mu)^{2} P_{j}\) denote, respectively, the mean and the variance of the number of offspring produced by a single individual. Suppose that \(X_{0}=1\) - that is, initially there is a single individual in the population. (a) Show that $$ E\left[X_{n}\right]=\mu E\left[X_{n-1}\right] $$ (b) Use part (a) to conclude that $$ E\left[X_{n}\right]=\mu^{n} $$ (c) Show that $$ \operatorname{Var}\left(X_{n}\right)=\sigma^{2} \mu^{n-1}+\mu^{2} \operatorname{Var}\left(X_{n-1}\right) $$ (d) Use part (c) to conclude that $$ \operatorname{Var}\left(X_{n}\right)= \begin{cases}\sigma^{2} \mu^{n-1}\left(\frac{\mu^{n}-1}{\mu-1}\right) & \text { if } \mu \neq 1 \\ n \sigma^{2} & \text { if } \mu=1\end{cases} $$ The case described above is known as a branching process, and an important question for a population that evolves along such lines is the probability that the population will eventually die out. Let \(\pi\) denote this probability when the population starts with a single individual. That is, $$ \pi=P\left\\{\text { population eventually dies out } \mid X_{0}=1\right. \text { ) } $$ (e) Argue that \(\pi\) satisfies $$ \pi=\sum_{j=0}^{\alpha} P_{j} \pi^{j} $$ HINT: Condition on the number of offspring of the initial member of the population.

Type \(i\) light bulbs function for a random amount of time having mean \(\mu_{i}\) and standard deviation \(\sigma_{i}, i=1,2\). A light bulb randomly chosen from a bin of bulbs is a type 1 bulb with probability \(p\), and a type 2 bulb with probability \(1-p\). Let \(X\) denote the lifetime of this bulb. Find (a) \(E[X]\) (b) \(\operatorname{Var}(X)\).

Consider a list of \(m\) names, where the same name may appear more than once on the list. Let \(n(i)\) denote the number of times that the name in position \(i\) appears on the list, \(i=1, \ldots, m\), and let \(d\) denote the number of distinct names on the list. (a) Express \(d\) in terms of the variables \(m, n_{i}, i=1, \ldots, m\). Let \(U\) be a uniform \((0,1)\) random variable, and let \(X=[m U]+1\). (b) What is the probability mass function of \(X ?\) (c) Argue that \(E[m / n(X)]=d\).
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