/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 83 An urn contains \(n\) balls, wit... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An urn contains \(n\) balls, with ball \(i\) having weight \(w_{i}, i=1, \ldots, n .\) The balls are withdrawn from the urn one at a time according to the following scheme: When \(S\) is the set of balls that remains, ball \(i, i \in S\), is the next ball withdrawn with probability \(w_{i} / \sum_{j \in S} w_{j} .\) Find the expected number of balls that are withdrawn before ball \(i, i=1, \ldots, n\)

Short Answer

Expert verified
The expected number of balls withdrawn before each ball i can be calculated using the following formula: \[E_i = \sum_{j=1, j \neq i}^{n} \frac{w_j}{\sum_{k \neq i} w_k},\] where \(i = 1, 2, \ldots, n\) and \(w_j\) represents the weight of each ball. The result will highlight the relationship between the weights of the balls and their expected waiting time.

Step by step solution

01

Define the notation

Let's denote the probability of withdrawing a ball i as \(P(i)\). Similarly, we denote the probability of withdrawing a specific ball j before ball i as \(P(j|i)\).
02

Calculate \(P(j|i)\) given problem specifications

According to the problem, when a ball is withdrawn, the probabilities of the remaining balls being withdrawn next are proportional to their weights. So, the probability of withdrawing a ball j before ball i can be calculated as follows: \(P(j|i) = \frac{w_j}{\sum_{k \neq i} w_k}\).
03

Define the expected number of balls withdrawn

The expected number of balls withdrawn before ball i can be defined as the sum of the probabilities of withdrawing each ball j before ball i (j ≠ i): \(E_i = \sum_{j \neq i} P(j|i)\).
04

Calculate the expected number of balls withdrawn before each ball i

Now we need to calculate the expected number of balls withdrawn before each ball i using the formula we derived in the previous step. For each \(i=1,2,\ldots,n\), we will have: \[ E_i = \sum_{j=1, j \neq i}^{n} \frac{w_j}{\sum_{k \neq i} w_k}. \]
05

Find the expected number of balls withdrawn before each ball i

By calculating the expected number of balls withdrawn before each ball i using the above formula, you will get the expected number of balls withdrawn before each of the balls in the urn. This result will show a clear pattern of the relationship between the weights of the balls and their expected waiting time.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Models
When encountered with circumstances where outcomes are uncertain, we use probability models to represent and analyze the situation. In the case of the urn with weighted balls, a probability model consists of all possible outcomes (which ball is drawn) and the probabilities associated with these outcomes. The probabilities are determined by the specific conditions of our problem: the weights of the balls and the sum of weights of all remaining balls. This model provides a structured way to predict the likelihood of each ball being drawn next.

For students or anyone looking to deepen their understanding of these models, it's essential to grasp that probability models can range from very simple to exceedingly complex. They are used across various fields, from game theory to economics to weather forecasting. The key is to set up the model to reflect the real-world situation accurately, which in this exercise, means recognizing that the probability of drawing a ball is weighted by the ball’s individual weight relative to the total weight of all remaining balls.
Weighted Probabilities
The concept of weighted probabilities comes into play when not all outcomes in a probability model have the same likelihood of occurring. Just as some lottery tickets have better odds based on purchase quantity, or certain events have a higher likelihood of happening due to various factors, the balls in our urn exercise have different probabilities of being drawn based on their weights.

This difference is accounted for by applying the weighted probability formula, which adjusts the chance of each event according to its weight. Typically, the weight represents the significance or influence of each outcome. In our urn example, a ball’s weight directly impacts its probability of being selected. It's a fundamental concept that extends beyond textbook exercises, affecting real-world financial markets, sports, and risk assessments. Therefore, understanding how to correctly establish and calculate weighted probabilities is a valuable skill.
Expected Value Calculation
The expected value calculation is a powerful mathematical tool used to predict the average outcome of a random event over the long run. It's calculated as the sum of all possible outcomes, each multiplied by its corresponding probability. In other words, it’s a type of weighted average. For the urn exercise, we're interested in the expected number of withdrawals before a particular ball is drawn which is a practical application of the expected value.

To clarify this concept, let's consider a simpler example: imagine you flip a fair coin. The expected value of earning \(1 for heads and nothing for tails is \)0.50. Although you’ll never actually receive \(0.50 in any single coin flip, over many flips, the average earnings per flip approach \)0.50. Applying this reasoning to our exercise, we sum the weighted probabilities of each event (ball withdrawal) to calculate the average or 'expected' number before a certain ball is withdrawn, providing insights into the most likely scenarios in uncertain conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the match problem, say that \((i, j), i

A prisoner is trapped in a cell containing three doors. The first door leads to a tunnel that returns him to his cell after two days of travel. The second leads to a tunnel that returns him to his cell after three days of travel. The third door leads immediately to freedom. (a) Assuming that the prisoner will always select doors 1,2, and 3 with probabilities \(0.5,0.3,0.2\), what is the expected number of days until he reaches freedom? (b) Assuming that the prisoner is always equally likely to choose among those doors that he has not used, what is the expected number of days until he reaches freedom? (In this version, for instance, if the prisoner initially tries door 1 , then when he returns to the cell, he will now select only from doors 2 and 3.) (c) For parts (a) and (b) find the variance of the number of days until the prisoner reaches freedom.

\(A\) and \(B\) roll a pair of dice in turn, with \(A\) rolling first. A's objective is to obtain a sum of 6 , and \(B\) 's is to obtain a sum of 7 . The game ends when either player reaches his or her objective, and that player is declared the winner. (a) Find the probability that \(A\) is the winner. (b) Find the expected number of rolls of the dice. (c) Find the variance of the number of rolls of the dice.

The number of coins that Josh spots when walking to work is a Poisson random variable with mean 6 . Each coin is equally likely to be a penny, a nickel, a dime, or a quarter. Josh ignores the pennies but picks up the other coins. (a) Find the expected amount of money that Josh picks up on his way to work. (b) Find the variance of the amount of money that Josh picks up on his way to work. (c) Find the probability that Josh picks up exactly 25 cents on his way to work.

A set of \(n\) dice is thrown. All those that land on six are put aside, and the others are again thrown. This is repeated until all the dice have landed on six. Let \(N\) denote the number of throws needed. (For instance, suppose that \(n=3\) and that on the initial throw exactly two of the dice land on six. Then the other die will be thrown, and if it lands on six, then \(N=2 .\) ) Let \(m_{n}=E[N]\). (a) Derive a recursive formula for \(m_{n}\) and use it to calculate \(m_{i}, i=2,3,4\) and to show that \(m_{5} \approx 13.024\). (b) Let \(X_{i}\) denote the number of dice rolled on the \(i\) th throw. Find \(E\left[\sum_{i=1}^{N} X_{i}\right]\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.