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Chapter 3: Problem 85

Twenty pairs of individuals playing in a bridge toumament have been seeded \(1, \ldots, 20\). In the first part of the toumament, the 20 are randomly divided into 10 east-west pairs and 10 northsouth pairs. a. What is the probability that \(x\) of the top 10 pairs end up playing east- west? b. What is the probability that all of the top five pairs end up playing the same direction? c. If there are \(2 n\) pairs, what is the pmf of \(X=\) the number among the top \(n\) pairs who end up playing east-west? What are \(E(X)\) and \(V(X)\) ?

Short Answer

Expert verified
a) \(P(X = x) = \frac{\binom{10}{x} \binom{10}{10-x}}{\binom{20}{10}}\). b) \(\frac{\binom{10}{5}}{\binom{20}{10}}\). c) PMF: \(P(X = x) = \frac{\binom{n}{x} \binom{n}{n-x}}{\binom{2n}{n}}\), \(E(X) = \frac{n}{2}\), \(V(X) = \frac{n}{4} \cdot \frac{n}{2n - 1}\).

Step by step solution

01

Understanding the Problem

We have 20 pairs in total, and our task is to determine probabilities based on the arrangement of the top pairs in the initial setup. We will break it down into three parts: focusing first on when some pairs are east-west, then all are in the same direction, and finally extending the logic for a general case with an arbitrary even number of pairs.
02

Probability of x Pairs Playing East-West (Part a)

To find the probability that \(x\) of the top 10 pairs end up playing east-west, notice that choosing \(x\) top pairs to be east-west is a combination problem. The total number of ways to choose \(x\) pairs from the top 10 is \(\binom{10}{x}\). Next, consider the remaining 10 pairs to be chosen from the remaining 10 pairs, which is \(\binom{10}{10-x}\). The total ways to choose 10 east-west pairs from 20 pairs is \(\binom{20}{10}\). Therefore, the probability \(P(X = x)\) is given by:\[ P(X = x) = \frac{\binom{10}{x} \binom{10}{10-x}}{\binom{20}{10}}. \]
03

Probability of All Top Five Pairs Playing Same Direction (Part b)

For the top five pairs to be all in the same direction, they must either be all east-west or all north-south. There are \(2\) choices for the arrangement: either all east-west or all north-south.In these scenarios, either 5 top or 5 bottom pairs take one direction, and the remaining 15 pairs are arranged in the other. For each case, the total possible ways to choose these directions is \(\binom{10}{5}\) for either set. Thus, the probability is:\[ P(\text{All in Same Direction}) = \frac{\binom{10}{5}}{\binom{20}{10}}. \]
04

General Case for 2n Pairs (Part c)

We need the probability distribution of \(X\), which represents the number of east-west pairs among the top \(n\). As in part (a), this aligns with the hypergeometric distribution:\[ P(X = x) = \frac{\binom{n}{x} \binom{n}{n-x}}{\binom{2n}{n}}. \]Also calculate the expectation and variance for the hypergeometric distribution:\[ E(X) = \frac{n}{2}, \]\[ V(X) = \frac{n}{2} \left(1 - \frac{n}{2n}\right) \left(\frac{2n-n}{2n-1}\right) = \frac{n}{4} \cdot \frac{n}{2n - 1}. \]
05

Solution Conclusions

These results collectively illustrate the nature of the pairing probability, use of combinations and application within probability in tournaments or games structured in a symmetric, random manner.
06

Final Calculation Check

Ensure consistent calculation evaluation for each part across all given scenarios confirming hypergeometric or combinatorial identities are adequately applied, checking against sample sizes and directions.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, combinations, and permutations of sets. In the context of this exercise, combinatorics helps to determine how many ways we can select groups from a larger set. When tasked with finding the number of top pairs playing east-west, it fundamentally becomes a problem of choosing combinations.
For example, if we need to select a specific number of pairs from the top 10 to play east-west, we can use the binomial coefficient formula:
  • \( \binom{n}{k} \) represents "n choose k", the number of ways to choose a subset of \( k \) elements from a larger set of \( n \) elements without regard to the order of selection.
  • In the case of choosing top pairs for a specific direction, the first step is calculating \( \binom{10}{x} \).
Dividing combinations must also consider the other group, calculated similarly or as the remaining pairs after choosing the first. These combinatorial calculations are essential for understanding and determining probabilities in such scenarios.
Hypergeometric Distribution
The hypergeometric distribution arises in scenarios where we are interested in the probabilities of obtaining a certain number of successes in draws, without replacement, from a finite set. It's particularly applicable when the size of the sample is large, compared to the population, like in this tournament example. When considering pairs playing east-west, if you want to choose \( x \) pairs from the top \( n \) pairs, the hypergeometric probability is given by:
  • \( P(X = x) = \frac{\binom{n}{x} \binom{n}{n-x}}{\binom{2n}{n}} \)
  • This structure reflects drawing \( n \) pairs, dividing them between the two groups, and favorably counting the ones of interest.
The denominator, \( \binom{2n}{n} \), serves as the total number of ways to draw \( n \) items from \( 2n \), providing the normalization factor of the distribution. This method accurately models tournament settings, where outcomes depend on specific groupings from an overall pool.
Expectation and Variance in Statistics
Expectation (or expected value) and variance are crucial concepts in statistics, providing measures for the central tendency and spread of a distribution. For hypergeometric distributions, the expectation is the mean number of successful outcomes, calculated as:
  • \( E(X) = \frac{n}{2} \), which means on average half of the selected \( n \) playing in one direction are expected from the top \( n \) pairs.
Variance measures how much the values spread out from the mean, which for the hypergeometric distribution is:
  • \( V(X) = \frac{n}{4} \cdot \frac{n}{2n - 1} \).
This formula considers the dependency and finite adjustment from drawing without replacement. Variance here reflects likely variability in outcomes across such trials. Together, these metrics offer a detailed view of the expected and typical deviations in results when engaging with combinatorial probability distributions.

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Most popular questions from this chapter

Some parts of Califomia are particularly earthquake-prone. Suppose that in one such area, \(30 \%\) of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let \(X\) denote the number among the four who have earthquake insurance. a. Find the probability distribution of \(X\). [Hint: Let \(S\) denote a homeowner who has insurance and \(F\) one who does not. One possible outcome is SFSS, with probability \((.3)(.7)(.3)(.3)\) and associated \(X\) value 3 . There are 15 other outcomes.] b. Draw the corresponding probability histogram. c. What is the most likely value for \(X\) ? d. What is the probability that at least two of the four selected have earthquake insurance?

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