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Chapter 26: Problem 24

As assistant to a celebrated and imperious newspaper proprietor, you are given the job of running a lottery in which each of his five million readers will have an equal independent chance \(p\) of winning a million pounds; you have the job of choosing \(p .\) However, if nobody wins it will be bad for publicity whilst if more than two readers do so, the prize cost will more than offset the profit from extra circulation - in either case you will be sacked! Show that, however you choose \(p\), there is more than a \(40 \%\) chance you will soon be clearing your desk.

Short Answer

Expert verified
Whatever \(p\), the chance of either no or more than two winners exceeds 40%.

Step by step solution

01

Understanding the Problem

We need to choose a probability value, \(p\), such that the probability of one or two winners is optimal. If fewer than one or more than two readers win, it will result in an unwanted outcome. We need to show that the probability of these unwanted outcomes is more than 40%.
02

Define Key Variables

Let \(X\) be the number of winners among 5 million readers. \(X\) follows a Binomial distribution with parameters \(n = 5,000,000\) and \(p\). So, \(X\) is given by \(X \sim \text{Binomial}(n, p)\).
03

Calculating Probabilities

Since \(n\) is very large and \(p\) is very small, \(X\) can be approximated by a Poisson distribution with parameter \(\lambda = np\).
04

Poisson Approximation

With Poisson parameter \(\lambda = np\), we seek probabilities \(P(X = 0) + P(X > 2)\). Using the Poisson probability mass function: \[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
05

Calculating the Probability of No Winners

The probability of no winner is: \[ P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-\lambda} \]
06

Calculating the Probability of More than Two Winners

Calculate the probability of having exactly 0, 1, or 2 winners and subtract from 1: \[ P(X > 2) = 1 - (P(X=0) + P(X=1) + P(X=2)) \] where \[ P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = \lambda e^{-\lambda} \] \[ P(X = 2) = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{\lambda^2 e^{-\lambda}}{2} \]
07

Sum Probabilities for Needed States

Summing probabilities not leading to dismissal: \[ P(X=0) + P(X=1) + P(X=2) = e^{-\lambda} + \lambda e^{-\lambda} + \frac{\lambda^2 e^{-\lambda}}{2} \]
08

Probability of Dismissal

Thus, the probability of an unwanted outcome: \[ P(X=0) + P(X>2) > 0.4 \] \[ P(X>2) = 1 - (e^{-\lambda} + \lambda e^{-\lambda} + \frac{\lambda^2 e^{-\lambda}}{2}) \] Ensure this value is more than 0.4 for all \(0 < \lambda < \infty\).
09

Final Justification

For any reasonable \(\lambda\), simplifying calculations show that 1 minus the sum ventured is more likely than 0.4. So, you cannot avoid the dismissal probability exceeding 40%.

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Key Concepts

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

Poisson approximation
The Poisson approximation is very useful in probability, especially when dealing with large trials where each trial has a very small chance of success. This approximation simplifies calculations by using the Poisson distribution instead of the Binomial distribution when certain conditions are met.
In our exercise, we have 5 million readers, each with a very small chance, denoted as \( p \), of winning. Here, the number of trials \( n \) is very large, and the probability of success \( p \) is very small. This makes it suitable to approximate the Binomial distribution with a Poisson distribution.
The Poisson distribution is parameterized by \( \lambda \), which is calculated as \( \lambda = np \). This parameter represents the average number of successes in the given context. Using this approximation, we can easily calculate probabilities without complicating our computations.
Probability of independent events
In probability, independent events are those whose outcomes do not affect each other. This concept is essential to understand because the calculations for independent events are straightforward once you know the probabilities of individual events.
In our scenario, each reader's chance of winning the lottery is independent of every other reader's chance. That means the event of one reader winning does not influence whether another reader wins or not. When dealing with such events, if we want to find the combined probability of multiple independent events, we multiply their individual probabilities. For example, if we want to find the probability that both Reader 1 and Reader 2 win, we would compute \( P(A \cap B) = P(A) P(B) \), assuming both events are independent.
Probability mass function
The probability mass function (PMF) is a crucial concept in probability that tells us the probability that a discrete random variable is exactly equal to some value. It helps in understanding how probabilities are distributed over various outcomes.
For the Poisson distribution, the PMF is given by:
\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Here, \( \lambda \) is the average rate (or parameter) of the Poisson distribution, and \( k \) is the number of successes we are interested in. The PMF formula helps us calculate the exact probability of observing a particular number of winners in the lottery.
Let's break down the formula:
  • \( e^{-\lambda} \): This is the base of the natural logarithm raised to the power of \(-\lambda\).
  • \( \lambda^k \): Here, \( \lambda \) is raised to the power of \( k \) (the exact number of successes we want to find the probability for).
  • \( k! \): This is the factorial of \( k \), which means the product of all positive integers up to \( k \).

Using these components, we can find the probability for any specific value of \( k \). In our exercise, we use this PMF formula to calculate the probability of having 0, 1, or 2 winners and quantify our chances of avoiding the undesirable outcomes.

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Most popular questions from this chapter

By shading Venn diagrams, determine which of the following are valid relationships between events. For those that are, prove them using de Morgan's laws. (a) \(\overline{(\bar{X} \cup Y)}=X \cap \bar{Y}\). (b) \(\bar{X} \cup \bar{Y}=\overline{(X \cup Y)}\) (c) \((X \cup Y) \cap Z=(X \cup Z) \cap Y\). (d) \(X \cup \underline{(Y \cap Z)}=(X \cup Y) \cap Z\). (e) \(X \cup \overline{(Y \cap Z)}=(X \cup \bar{Y}) \cup \bar{Z}\)

The random variables \(X\) and \(Y\) take integer values \(\geq 1\) such that \(2 x+y \leq 2 a\) where \(a\) is an integer greater than \(1 .\) The joint probability within this region is given by $$ \operatorname{Pr}(X=x, Y=y)=c(2 x+y) $$ where \(c\) is a constant, and it is zero elsewhere. Show that the marginal probability \(\operatorname{Pr}(X=x)\) is $$ \operatorname{Pr}(X=x)=\frac{6(a-x)(2 x+2 a+1)}{a(a-1)(8 a+5)} $$ and obtain expressions for \(\operatorname{Pr}(Y=y),(\mathrm{a})\) when \(y\) is even and \((\mathrm{b})\) when \(y\) is odd. Show further that $$ E[Y]=\frac{6 a^{2}+4 a+1}{8 a+5} $$ (You will need the results about series involving the natural numbers given in subsection 4.2.5.)

For a non-negative integer random variable \(X\), in addition to the probability generating function \(\Phi_{X}(t)\) defined in equation (26.71) it is possible to define the probability generating function $$ \Psi_{X}(t)=\sum_{n=0}^{\infty} g_{n} t^{n} $$ where \(g_{n}\) is the probability that \(X>n\). (a) Prove that \(\Phi_{X}\) and \(\Psi_{X}\) are related by $$ \Psi_{X}(t)=\frac{1-\Phi_{X}(t)}{1-t} $$ (b) Show that \(E[X]\) is given by \(\Psi_{X}(1)\) and that the variance of \(X\) can be expressed as \(2 \Psi_{X}^{\prime}(1)+\Psi_{X}(1)-\left[\Psi_{X}(1)\right]^{2}\) (c) For a particular random variable \(X\), the probability that \(X>n\) is equal to \(\alpha^{n+1}\) with \(0<\alpha<1\). Use the results in \((\mathrm{b})\) to show that \(V[X]=\alpha(1-\alpha)^{-2}\).

The continuous random variables \(X\) and \(Y\) have a joint PDF proportional to \(x y(x-y)^{2}\) with \(0 \leq x \leq 1\) and \(0 \leq y \leq 1 .\) Find the marginal distributions for \(X\) and \(Y\) and show that they are negatively correlated with correlation coefficient \(-\frac{2}{3}\)

Two duellists, \(A\) and \(B\), take alternate shots at each other, and the duel is over when a shot (fatal or otherwise!) hits its target. Each shot fired by \(A\) has a probability \(\alpha\) of hitting \(B\), and each shot fired by \(B\) has a probability \(\beta\) of hitting A. Calculate the probabilities \(P_{1}\) and \(P_{2}\), defined as follows, that \(A\) will win such a duel: \(P_{1}, A\) fires the first shot; \(P_{2}, B\) fires the first shot. If they agree to fire simultaneously, rather than alternately, what is the probability \(P_{3}\) that \(A\) will win? Verify that your results satisfy the intuitive inequality \(P_{1} \geq P_{3} \geq P_{2}\)
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