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An urn contains \(n\) balls, with ball \(i\) having weight \(w_{i}, i=1, \ldots, n .\) The balls are withdrawn from the urn one at a time according to the following scheme: When \(S\) is the set of balls that remains, ball \(i, i \in S\), is the next ball withdrawn with probability \(w_{i} / \sum_{j \in S} w_{j} .\) Find the expected number of balls that are withdrawn before ball \(i, i=1, \ldots, n\)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The expected number of balls withdrawn before each ball i can be calculated using the following formula: \[E_i = \sum_{j=1, j \neq i}^{n} \frac{w_j}{\sum_{k \neq i} w_k},\] where \(i = 1, 2, \ldots,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According to the problem, when a ball is withdrawn, the probabilities of the remaining balls being withdrawn next are proportional to their weights.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Define the expected number of balls withdrawn

The expected number of balls withdrawn before ball i can be defined as the sum of the probabilities of withdrawing each ball j before ball i (j ≠ i): \(E_i = \sum_{j \neq i} P(j|i)\).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Calculate the expected number of balls withdrawn before each ball i

Now we need to calculate the expected number of balls withdrawn before each ball i using the formula we derived in the previous step.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Find the expected number of balls withdrawn before each ball i

By calculating the expected number of balls withdrawn before each ball i using the above formula, you will get the expected number of balls withdrawn before each of the balls in the urn.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.

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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.

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Most popular questions from this chapter

An urn contains three white, six red, and five black balls. Six of these balls are randomly selected from the urn. Let \(X\) and \(Y\) denote respectively the number of white and black balls selected. Compute the conditional probability mass function of \(X\) given that \(Y=3\). Also compute \(E[X \mid Y=1]\).

Suppose \(p(x, y, z)\), the joint probability mass function of the random variables \(X\), \(Y\), and \(Z\), is given by $$ \begin{array}{ll} p(1,1,1)=\frac{1}{8}, & p(2,1,1)=\frac{1}{4} \\ p(1,1,2)=\frac{1}{8}, & p(2,1,2)=\frac{3}{16} \\ p(1,2,1)=\frac{1}{16}, & p(2,2,1)=0 \\ p(1,2,2)=0, & p(2,2,2)=\frac{1}{4} \end{array} $$ \text { What is } E[X \mid Y=2] ? \text { What is } E[X \mid Y=2, Z=1] ?

Suppose there are \(n\) types of coupons, and that the type of each new coupon obtained is independent of past selections and is equally likely to be any of the \(n\) types. Suppose one continues collecting until a complete set of at least one of each type is obtained. (a) Find the probability that there is exactly one type \(i\) coupon in the final collection. Hint: Condition on \(T\), the number of types that are collected before the first type \(i\) appears. (b) Find the expected number of types that appear exactly once in the final collection.

You have two opponents with whom you alternate play. Whenever you play \(A\), you win with probability \(p_{A}\); whenever you play \(B\), you win with probability \(p_{B}\), where \(p_{B}>p_{A}\). If your objective is to minimize the expected number of games you need to play to win two in a row, should you start with \(A\) or with \(B\) ? Hint: Let \(E\left[N_{i}\right]\) denote the mean number of games needed if you initially play \(i\). Derive an expression for \(E\left[N_{A}\right]\) that involves \(E\left[N_{B}\right] ;\) write down the equivalent expression for \(E\left[N_{B}\right]\) and then subtract.

Consider \(n\) multinomial trials, where each trial independently results in outcome \(i\) with probability \(p_{i}, \sum_{i=1}^{k} p_{i}=1 .\) With \(X_{i}\) equal to the number of trials that result in outcome \(i\), find \(E\left[X_{1} \mid X_{2}>0\right]\)

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