91Ó°ÊÓ

This exercise extends the hypergeometric distribution to multiple variables. Consider a population with \(N\) items of \(k\) different types. Assume there are \(N_{1}\) items of type \(1, N_{2}\) items of type \(2, \ldots, N_{k}\) items of type \(k\) so that \(N_{1}+N_{2}+\ldots+\ldots, N_{\mathrm{k}}=N\). Suppose that a random sample of size \(\mathrm{n}\) is selected, without replacement, from the population. Let \(X_{1}, X_{2}, \ldots, X_{\mathrm{k}}\) denote the number of items of each type in the sample so that \(X_{1}+X_{2},+\ldots\) \(+\ldots+X_{\mathrm{k}}=n .\) Show that for feasible values of \(n, \mathrm{x}_{1}\) \(\mathrm{x}_{2}, \ldots, x_{\mathrm{k}}, N_{1}, N_{2}, \ldots, N_{\mathrm{k}},\) the probability is $$ P\left(X_{1}=x_{1}, X_{2}=x_{2}, \ldots, X_{\mathrm{k}}=x_{k}\right)=\frac{\left(\begin{array}{l} N_{1} \\ x_{1} \end{array}\right)\left(\begin{array}{l} N_{2} \\ x_{2} \end{array}\right) \cdots\left(\begin{array}{l} N_{k} \\ x_{n} \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)} $$

Step by step solution

01

Understand the Multivariate Hypergeometric Distribution

The probability of selecting specific numbers of each type of item from a finite population without replacement can be described using the multivariate hypergeometric distribution. It extends the hypergeometric distribution to multiple variables (types). This is useful for understanding the distribution of several outcomes when sampling without replacement.

02

Define the Parameters

We have a population with a total of \(N\) items, each belonging to one of \(k\) types. Let \(N_i\) denote the number of items of type \(i\) where \(N_1 + N_2 + \cdots + N_k = N\). A sample of size \(n\) is drawn, and \(X_1, X_2, \ldots, X_k\) are the number of each type in the sample, such that \(X_1 + X_2 + \cdots + X_k = n\).

03

Multivariate Hypergeometric Probability Formula

The probability of drawing \(x_1\), \(x_2\), ..., \(x_k\) items of types 1, 2, ..., k, respectively, is given by the formula: \[P(X_1 = x_1, X_2 = x_2, \ldots, X_k = x_k) = \frac{\binom{N_1}{x_1} \binom{N_2}{x_2} \cdots \binom{N_k}{x_k}}{\binom{N}{n}}\]. This formula calculates the probability of a specific composition of a sample.

04

Derive the Formula's Components

For each type \(i\), \(\binom{N_i}{x_i}\) is the number of ways to choose \(x_i\) items from \(N_i\). The denominator, \(\binom{N}{n}\), is the total number of ways to choose a sample of size \(n\) from the entire population. The product of combinations for each type accounts for all the ways to select \(x_1\), \(x_2\), ..., \(x_k\) items simultaneously.

05

Understand Feasibility of Values

The values of \(x_1, x_2, \ldots, x_k\) must satisfy \(x_1 + x_2 + \cdots + x_k = n\) while being feasible. Each \(x_i\) should not exceed \(N_i\), ensuring the selection is valid within the population's constraints. This ensures that the sum of chosen items matches the sample size \(n\).

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Key Concepts

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

Probability Theory

In probability theory, we study how likely events are to happen. It's like predicting the weather, but instead of rain, we predict outcomes from different possible events. The multivariate hypergeometric distribution is part of this field. It helps understand the probability of drawing specific numbers of different types of items from a population. For example, if you have a bag of marbles in different colors, probability theory helps you calculate the chances of picking a particular mix of colors when you draw a few marbles without looking. This kind of problem is solved using probability formulas, which take into account the total number of items and their types.

Combinatorics

Combinatorics deals with counting and arranging things systematically. It's like finding out how many different ways you can arrange books on a shelf. In the context of the multivariate hypergeometric distribution, it focuses on the different ways items can be chosen. For instance, if you have 5 red marbles and you want to pick 3, combinatorics gives you the formula to calculate how many different combinations are possible. This is vital for the formula used in our original problem: \(P(X_1 = x_1, X_2 = x_2, \ldots, X_k = x_k)\). Here, \(\binom{N_i}{x_i}\) represents the number of ways to pick \(x_i\) items from \(N_i\) available items of a specific type. Understanding combinatorics helps break down complex choices into simple, manageable calculations.

Sampling Without Replacement

Sampling without replacement means once an item is selected, it's not put back into the population. This drastically changes the probabilities involved. Imagine a box of chocolates; once you eat one, it's gone, and the rest are fewer in number. This is different from sampling with replacement, where the population stays the same size after each pick. Without replacement makes calculations dynamic because the chance of picking another item depends on the previous selections. For our multivariate hypergeometric distribution exercise, sampling without replacement ensures that each draw affects the next. It's a key concept because it leads to decreased total options with every choice made, represented by the denominator \(\binom{N}{n}\) of the probability formula, which shrinks as the sample comparison intensifies.