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Chapter 4: Problem 27

For a hypergeometric random variable, determine $$ P\\{X=k+1\\} / P\\{X=k\\} $$

Short Answer

Expert verified
\(\frac{P(X=k+1)}{P(X=k)} = \frac{\binom{M}{k+1}\binom{N-M}{n-(k+1)}}{\binom{M}{k}\binom{N-M}{n-k}}\)

Step by step solution

01

Understand the Hypergeometric Probability Mass Function

The formula for the probability mass function of a hypergeometric random variable is given by: \[ P(X=k) = \frac{\binom{M}{k}\binom{N-M}{n-k}}{\binom{N}{n}} \] where - \(N\) is the total number of items, - \(M\) is the number of items with the desired characteristic, - \(n\) is the number of items we choose randomly, and - \(k\) is the number of items with the desired characteristic in our random sample. Now we need to apply this formula to both cases: when \(X=k+1\) and when \(X=k\).
02

Calculate the Probability for X=k and X=k+1 Cases

First, let's calculate the probability for each case using the hypergeometric PMF formula: \(P(X=k) = \frac{\binom{M}{k}\binom{N-M}{n-k}}{\binom{N}{n}}\) \(P(X=k+1) = \frac{\binom{M}{k+1}\binom{N-M}{n-(k+1)}}{\binom{N}{n}}\)
03

Find the Ratio of the Probabilities

Now, we want to calculate the ratio of \(P(X=k+1)\) to \(P(X=k)\), which can be represented as: \(\frac{P(X=k+1)}{P(X=k)} = \frac{\frac{\binom{M}{k+1}\binom{N-M}{n-(k+1)}}{\binom{N}{n}}}{\frac{\binom{M}{k}\binom{N-M}{n-k}}{\binom{N}{n}}}\) By simplifying the fraction, we get: \(\frac{P(X=k+1)}{P(X=k)} = \frac{\binom{M}{k+1}\binom{N-M}{n-(k+1)}}{\binom{M}{k}\binom{N-M}{n-k}}\)
04

Final Result

After calculating the ratio of the probabilities, we have: \(\frac{P(X=k+1)}{P(X=k)} = \frac{\binom{M}{k+1}\binom{N-M}{n-(k+1)}}{\binom{M}{k}\binom{N-M}{n-k}}\) This is the final result for the given problem.

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Key Concepts

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

Probability Mass Function
The Probability Mass Function (PMF) is a fundamental concept in probability theory for discrete random variables. A discrete random variable is one that can only take on a finite or countably infinite set of values. The PMF defines the probability that a random variable equals each of its possible values.

In mathematical terms, for a discrete random variable \( X \), the PMF \( P(X=x) \) gives the probability that \( X \) will take the value \( x \). It's critical in understanding distributions as it lays out the landscape of all possible outcomes and their likelihoods, allowing us to calculate probabilities for different scenarios.

For the hypergeometric distribution, the PMF differs from other distributions like binomial or Poisson since it deals with sampling without replacement from a finite population. This PMF is particularly useful when evaluating scenarios where the probability of success changes after each trial because the population size diminishes.
Combinatorics
Combinatorics, the branch of mathematics dealing with combinations, permutations, and counting, plays a crucial role in calculating probabilities for the hypergeometric distribution. At the core of the hypergeometric PMF are binomial coefficients, which are combinatorial expressions that count the number of ways to choose a subset of items from a larger set, frequently denoted as \( \binom{n}{k} \). These coefficients are key to calculating different probability scenarios in the hypergeometric distribution.

Understanding the basics of combinatorics, especially how to calculate binomial coefficients, is essential for working with most probability distributions that involve a finite sample space. This knowledge allows one to see beyond mere formulas and grasp the reasoning behind different probabilities, such as why the chances of drawing a specific number of successful outcomes changes as the sample size or the number of successes in the population changes.
Probability Theory
Probability theory is the mathematical framework that deals with uncertainty and quantifies the likelihood of events. It underpins all of probability and statistics and is fundamental to various disciplines, including science, engineering, finance, and even philosophy.

In the context of a hypergeometric random variable, probability theory helps us understand how to model situations where we are randomly sampling from a finite population without replacement. It allows us to define the hypergeometric distribution, determine its PMF, and calculate the chances of having exactly \( k \) successes in our sample. This area of mathematics is rich with concepts like independence, random variables, distributions, expectation, variance, and more, all of which are essential tools for analyzing random phenomena.
Statistical Distribution
A statistical distribution represents how the values of a random variable are distributed. It tells us which values are more likely and which are less so. There are many different types of statistical distributions, each tailored to model different types of data and situations.

The hypergeometric distribution is one such statistical distribution used when the samples are drawn without replacement from a finite population. It's different from the binomial distribution, which assumes sampling with replacement, thereby keeping the probabilities constant across trials. The importance of statistical distributions lies in their ability to provide a model that closely matches real-world phenomena, helping us to make predictions and informed decisions based on the characteristics of the data we observe.

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Most popular questions from this chapter

Balls numbered 1 through \(N\) are in an urn. Suppose that \(n, n \leq N\), of them are randomly selected without replacement. Let \(Y\) denote the largest number selected. (a) Find the probability mass function of \(Y\). (b) Derive an expression for \(E[Y]\) and then use Fermat's combinatorial identity (see Theoretical Exercise 11 of Chapter 1) to simplify.

From a set of \(n\) randomly chosen people let \(E_{i j}\) denote the event that persons \(i\) and \(j\) have the same birthday. Assume that each person is equally likely to have any of the 365 days of the year as his or her birthday. Find (a) \(P\left(E_{3,4} \mid E_{1,2}\right)\); (b) \(P\left(E_{1,3} \mid E_{1,2}\right) ;\) (c) \(P\left(E_{2,3} \mid E_{1,2} \cap E_{1,3}\right)\). What can you conclude from the above about the independence of the \(\left(\begin{array}{l}n \\ 2\end{array}\right)\) events \(E_{i j} ?\)

An interviewer is given a list of potential people she can interview. If the interviewer needs to interview 5 people and if each person (independently) agrees to be interviewed with probability \(\frac{2}{3}\), what is the probability that her list of potential people will enable her to obtain her necessary number of interviews if the list consists of (a) 5 people and (b) 8 people? For part (b) what is the probability that the interviewer will speak to exactly (c) 6 people and (d) 7 people on the list?

An urn contatns \(2 n\) balls, of which 2 are numbered 1,2 are numbered \(2, \ldots\) and 2 are numbered \(n\). Balls are successively withdrawn 2 at a time without replacement. Let \(T\) denote the first selection in which the balls withdrawn have the same number (and let it equal infinity if none of the pairs withdrawn has the same number). For \(0<\alpha<1\) we want to show that $$ \lim _{n} P\\{T>\alpha n\\}=e^{-\alpha / 2} $$ To verify the above, let \(M_{k}\) denote the number of pairs withdrawn in the first \(k\) selections, \(k=1, \ldots, n\) (a) Argue that when \(n\) is large, \(M_{k}\) can be regarded as the number of successes in \(k\) (approximately) independent trials. (b) When \(n\) is large, approximate \(P\left\\{M_{k}=0\right\\}\). (c) Write the event \(\\{T>\alpha n\\}\) in terms of the value of one of the variables \(M_{k}\). (d) Verify the limiting probability above.

If \(X\) has distribution function \(F\), what is the distribution function of the random variable \(\alpha X+\beta\), where \(\alpha\) and \(\beta\) are constants, \(\alpha \neq 0\) ?
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