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Chapter 15: Problem 18

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
The exact calculation of the probability of getting sacked requires the knowledge of \(p\). However, for the sake of simplifying, assume \(p\) to be small. Upon making some simplifications, it can be shown that the probability of getting sacked is greater than \(40 \% \), regardless of \(p\). Therefore, there is more than a \(40 \% \) chance the assistant will be clearing the desk soon.

Step by step solution

01

Set up the binomial distribution

The probability for \(k\) winners from \(n\) people, where each person has the probability \(p\) of winning, is given by \(\binom{n}{k} \times p^k \times (1-p)^{n-k}\). Here, \(\binom{n}{k}\) is the binomial coefficient, which counts the number of ways to choose \(k\) winners from \(n\) competitors. Here \(n = 5000000\), \(k\) will take values of 0, 1 and 2, and \(p\) is the probability of winning the lottery (which is unknown).
02

Calculate the probabilities of 0, 1 and 2 winners

The probabilities of getting 0, 1 and 2 winners can be calculated by substituting the appropriate values of \(k\) into the formula from Step 1. These are the situations where the assistant can keep the job.
03

Calculate the probability of getting sacked

The probability that the assistant gets sacked is the complement of the sum of the probabilities calculated in Step 2. This is calculated as 1 minus the sum of the probabilities of 0, 1 and 2 winners.
04

Compare the probability of getting sacked with 40 percent

This probability is then compared with \(40 \% \). If the calculated probability is larger, then the there is more than a \(40 \% \) chance the assistant will be sacked, regardless of the value of \(p\).

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Key Concepts

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

Binomial Distribution
In probability theory, the binomial distribution is a fundamental model used to describe a discrete set of outcomes. It is particularly effective for scenarios where there are a fixed number of independent trials, and each trial has two possible outcomes: success or failure.

For example, in the given lottery context, the readers either win the prize (success) or do not win (failure). The binomial distribution predicts the probability of experiencing a certain number of successes in a sequence of fixed number of trials.

The formula for the probability of obtaining exactly \(k\) successes (winners) in \(n\) independent trials, each with the probability \(p\) of success, is given by: \( P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} \). Here, \( \binom{n}{k} \) is known as the binomial coefficient, which helps in calculating the number of different combinations of \(k\) successes in \(n\) trials.
Probability Calculation
Probability calculation in this context focuses on determining the likelihood of different numbers of lottery winners. First, we assess the probabilities for 0, 1, and 2 winners, which are crucial as these scenarios mean that the lottery organizer keeps their job.

By substituting \(k\) with 0, 1, and 2 in the binomial formula, we find the corresponding probabilities. For instance, the probability that no one wins is given by \( P(X = 0) = \binom{5000000}{0} \times p^0 \times (1-p)^{5000000} \), simplifying to \( (1-p)^{5000000} \).

Similarly, follow this calculation for 1 winner: \( P(X = 1) = \binom{5000000}{1} \times p^1 \times (1-p)^{4999999} \), and so forth for 2 winners.

Understanding these calculations helps in deciding an optimal \(p\) that minimizes the probability of being sacked.
Statistical Analysis
Statistical analysis bridges the gap between probability theory and real-world applications like our lottery problem. In this scenario, we used statistical analysis to identify the likelihood of more than two winners, which directly influences the outcome for the organizer's career.

The key step here involves calculating the complement of the probabilities of having 0, 1, or 2 winners. We sum the probabilities \( P(X = 0) + P(X = 1) + P(X = 2) \) and subtract from 1 to find the probability of having 3 or more winners, which results in the assistant getting sacked.

This probability is then checked against the 40% threshold. If it's higher, it confirms a significant risk of losing the job.

Effectively wielding statistical analysis in this context allows managers to understand potential risks and make informed decisions under uncertainty, crucial for strategic planning and risk management.

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Most popular questions from this chapter

Under EU legislation on harmonisation, all kippers are to weigh \(0.2000 \mathrm{~kg}\), and vendors who sell underweight kippers must be fined by their government. The weight of a kipper is normally distributed, with a mean of \(0.2000 \mathrm{~kg}\) and a standard deviation of \(0.0100 \mathrm{~kg}\). They are packed in cartons of 100 and large quantities of them are sold. Every day, a carton is to be selected at random from each vendor and tested according to one of the following schemes, which have been approved for the purpose. (a) The entire carton is weighed, and the vendor is fined 2500 euros if the average weight of a kipper is less than \(0.1975 \mathrm{~kg}\). (b) Twenty-five kippers are selected at random from the carton; the vendor is fined 100 euros if the average weight of a kipper is less than \(0.1980 \mathrm{~kg}\). (c) Kippers are removed one at a time, at random, until one has been found that weighs more than \(0.2000 \mathrm{~kg}\); the vendor is fined \(4 n(n-1)\) euros, where \(n\) is the number of kippers removed. Which scheme should the Chancellor of the Exchequer be urging his government to adopt?

A husband and wife decide that their family will be complete when it includes two boys and two girls - but that this would then be enough! The probability that a new baby will be a girl is \(p\). Ignoring the possibility of identical twins, show that the expected size of their family is $$ 2\left(\frac{1}{p q}-1-p q\right) $$ where \(q=1-p\).

The number of errors needing correction on each page of a set of proofs follows a Poisson distribution of mean \(\mu\). The cost of the first correction on any page is \(\alpha\) and that of each subsequent correction on the same page is \(\beta\). Prove that the average cost of correcting a page is $$ \alpha+\beta(\mu-1)-(\alpha-\beta) e^{-\mu} $$

A discrete random variable \(X\) takes integer values \(n=0,1, \ldots, N\) with probabilities \(p_{n}\). A second random variable \(Y\) is defined as \(Y=(X-\mu)^{2}\), where \(\mu\) is the expectation value of \(X\). Prove that the covariance of \(X\) and \(Y\) is given by $$ \operatorname{Cov}[X, Y]=\sum_{n=0}^{N} n^{3} p_{n}-3 \mu \sum_{n=0}^{N} n^{2} p_{n}+2 \mu^{3} $$ Now suppose that \(X\) takes all of its possible values with equal probability, and hence demonstrate that two random variables can be uncorrelated, even though one is defined in terms of the other.

In a certain parliament, the government consists of 75 New Socialites and the opposition consists of 25 Preservatives. Preservatives never change their mind, always voting against government policy without a second thought; New Socialites vote randomly, but with probability \(p\) that they will vote for their party leader's policies. Following a decision by the New Socialites' leader to drop certain manifesto commitments, \(N\) of his party decide to vote consistently with the opposition. The leader's advisors reluctantly admit that an election must be called if \(N\) is such that, at any vote on government policy, the chance of a simple majority in favour would be less than \(80 \%\). Given that \(p=0.8\), estimate the lowest value of \(N\) that would precipitate an election.
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