/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 The National Basketball Associat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 15

The National Basketball Association (NBA) draft lottery involves the 11 teams that had the worst won-lost records during the year. A total of 66 balls are placed in an urn. Each of these balls is inscribed with the name of a team: Eleven have the name of the team with the worst record, 10 have the name of the team with the second-worst record, 9 have the name of the team with the thirdworst record, and so on (with 1 ball having the name of the team with the 11 th-worst record). A ball is then chosen at random, and the team whose name is on the ball is given the first pick in the draft of players about to enter the league. Another ball is then chosen, and if it "belongs" to a team different from the one that received the first draft pick, then the team to which it belongs receives the second draft pick. (If the ball belongs to the team receiving the first pick, then it is discarded and another one is chosen; this continues until the ball of another team is chosen.) Finally, another ball is chosen, and the team named on the ball (provided that it is different from the previous two teams) receives the third draft pick. The remaining draft picks 4 through 11 are then awarded to the 8 teams that did not "win the lottery," in inverse order of their won-lost records. For instance, if the team with the worst record did not receive any of the 3 lottery picks, then that team would receive the fourth draft pick. Let \(X\) denote the draft pick of the team with the worst record. Find the probability mass function of \(X\)

Short Answer

Expert verified
The probability mass function of \(X\), the draft pick of the team with the worst record, is: \(P(X=1) = \frac{11}{66}\) \(P(X=2) = \frac{55}{66} \times \frac{11}{65}\) \(P(X=3) = \frac{55}{66} \times \frac{54}{65} \times \frac{11}{64}\) \(P(X=4) = 1 - (\frac{11}{66} + \frac{55}{66} \times \frac{11}{65} + \frac{55}{66} \times \frac{54}{65} \times \frac{11}{64})\)

Step by step solution

01

Case 1: Worst team gets the first pick

The worst team has 11 balls in the urn, out of a total of 66 balls. The probability of the worst team getting the first pick is \(\frac{11}{66}\).
02

Case 2: Worst team gets the second pick

To find the probability of the worst team getting the second pick, we first need to find the probability of the worst team not getting the first pick. This probability is \(\frac{55}{66}\), as there are 55 balls that don't belong to the worst team. Then, out of the remaining 65 balls, 11 belong to the worst team. So, the probability of the worst team getting the second pick is \(\frac{55}{66} \times \frac{11}{65}\).
03

Case 3: Worst team gets the third pick

In this case, the worst team does not get the first or second pick. The probability of not getting the first pick is still \(\frac{55}{66}\). After that, out of the 65 balls left, 54 don't belong to the worst team. So, the probability of not getting the second pick is \(\frac{54}{65}\). Then, out of the remaining 64 balls, 11 belong to the worst team. So, the probability of the worst team getting the third pick is \(\frac{55}{66} \times \frac{54}{65} \times \frac{11}{64}\).
04

Case 4: Worst team doesn't get any lottery picks

In this case, the worst team doesn't get any of the first three picks. We can find the probability of this situation by subtracting the probabilities of Cases 1, 2, and 3 from 1 (since the total probability should add up to 1). The probability of the worst team not getting any lottery picks (thus getting the fourth pick) is \(1 - (\frac{11}{66} + \frac{55}{66} \times \frac{11}{65} + \frac{55}{66} \times \frac{54}{65} \times \frac{11}{64})\). Now, we can calculate the probability mass function of X: \(P(X=1) = \frac{11}{66}\) \(P(X=2) = \frac{55}{66} \times \frac{11}{65}\) \(P(X=3) = \frac{55}{66} \times \frac{54}{65} \times \frac{11}{64}\) \(P(X=4) = 1 - (\frac{11}{66} + \frac{55}{66} \times \frac{11}{65} + \frac{55}{66} \times \frac{54}{65} \times \frac{11}{64})\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

NBA Draft Lottery
Understanding the NBA draft lottery is essential for comprehending how draft picks are awarded in professional basketball. In essence, the lottery is a system used by the National Basketball Association to determine the order of selections for the first 14 picks in the annual draft. Teams that do not make the playoffs in the previous year are entered into the lottery, as their chances of winning are weighted based on their regular-season records鈥攖he worse the record, the higher the chance of securing a top pick.

This system is designed to ensure competitiveness by giving struggling teams the opportunity to select the most promising young players. In the NBA draft lottery, probabilities come into play as discrete outcomes (in this case, draft picks) are assigned different chances of occurring, much like in our exercise, where the probability mass function determines the likelihood of each draft pick for the team with the worst record.
Discrete Probability Distribution
A discrete probability distribution describes the probability of occurrence of each value of a discrete random variable. It can be represented with a table, a graph, or a mathematical formula, and it is a cornerstone of studying random processes in statistics.

In simple terms, think of a discrete probability distribution as a summary of the probability of various outcomes in a list or a graph. Each possible outcome has a probability associated with it, which must be between 0 (the event definitely will not occur) and 1 (the event definitely will occur). Importantly, when you add up the probabilities of all possible outcomes, they must equal 1, signaling all possible outcomes have been accounted for. An NBA draft lottery is an excellent example, where the sum of the probabilities for getting each pick totals to 1.
Random Experiment
A random experiment is a process that leads to the occurrence of one and only one of several possible observations. The term 'random' implies that the outcome is unpredictable and varies every time we perform the experiment. In probability theory, random experiments are key concepts since they embody situations where outcomes are uncertain.

The NBA draft lottery is a classic example: selecting a team's ball from the urn is a random experiment because there is no way to predict which ball will be selected. Each drawing of a ball is an independent trial, and the outcome cannot be foreseen, fulfilling the criteria of a random experiment. This unpredictability is quantified using the probability mass function, which assigns a probability to each potential outcome of this random process, such as the likelihood of the worst team receiving any given draft pick.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a roulette wheel consisting of 38 numbers 1 through \(36,0,\) and double \(0 .\) If Smith always bets that the outcome will be one of the numbers 1 through \(12,\) what is the probability that (a) Smith will lose his first 5 bets; (b) his first win will occur on his fourth bet?

A game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers 1 through \(80 .\) A player can select from 1 to 15 numbers; a win occurs if some fraction of the player's chosen subset matches any of the 20 numbers drawn by the house. The payoff is a function of the number of elements in the player's selection and the number of matches. For instance, if the player selects only 1 number, then he or she wins if this number is among the set of \(20,\) and the payoff is \(\$ 2.2\) won for every dollar bet. (As the player's probability of winning in this case is \(\frac{1}{4},\) it is clear that the "fair" payoff should be \(\$ 3\) won for every \(\$ 1\) bet.) When the player selects 2 numbers, a payoff (of odds) of \(\$ 12\) won for every \(\$ 1\) bet is made when both numbers are among the 20 (a) What would be the fair payoff in this case? Let \(P_{n, k}\) denote the probability that exactly \(k\) of the \(n\) numbers chosen by the player are among the 20 selected by the house. (b) Compute \(P_{n, k}\) (c) The most typical wager at Keno consists of selecting 10 numbers. For such a bet the casino pays off as shown in the following table. Compute the expected payoff: $$\begin{array}{cc} \hline \multicolumn{2}{c} {\text { Keno Payoffs in 10 Number Bets }} \\ \hline \text { Number of matches } & \text { Dollars won for each \$1 bet } \\\ \hline 0-4 & -1 \\ 5 & 1 \\ 6 & 17 \\ 7 & 179 \\ 8 & 1,299 \\ 9 & 2,599 \\ 10 & 24,999 \\ \hline \end{array}$$

The suicide rate in a certain state is 1 suicide per 100,000 inhabitants per month. (a) Find the probability that, in a city of 400,000 inhabitants within this state, there will be 8 or more suicides in a given month. (b) What is the probability that there will be at least 2 months during the year that will have 8 or more suicides? (c) Counting the present month as month number \(1,\) what is the probability that the first month to have 8 or more suicides will be month number \(i, i \geq 1 ?\) What assumptions are you making?

Suppose that 10 balls are put into 5 boxes, with each ball independently being put in box \(i\) with probability \(p_{i}, \sum_{i=1}^{5} p_{i}=1\) (a) Find the expected number of boxes that do not have any balls. (b) Find the expected number of boxes that have exactly 1 ball.

It is known that diskettes produced by a certain company will be defective with probability \(.01,\) independently of each other. The company sells the diskettes in packages of size 10 and offers a money-back guarantee that at most 1 of the 10 diskettes in the package will be defective. The guarantee is that the customer can return the entire package of diskettes if he or she finds more than one defective diskette in it. If someone buys 3 packages, what is the probability that he or she will return exactly 1 of them?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.