/*! 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 25 Professional basketball is now a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Professional basketball is now a reality for women basketball players in the United States. There are two conferences in the WNBA, each with six teams, as shown in the following table. \(^{3}\) $$ \begin{array}{ll} \hline \text { Western Conference } & \text { Eastern Conference } \\ \hline \text { Minnesota Lynx } & \text { Atlanta Dream } \\ \text { Phoenix Mercury } & \text { Indiana Fever } \\ \text { Dallas Wings } & \text { New York Liberty } \\ \text { Los Angeles Sparks } & \text { Washington Mystics } \\ \text { Seattle Storm } & \text { Connecticut Sun } \\ \text { San Antonio Stars } & \text { Chicago Sky } \end{array} $$ Two teams, one from each conference, are randomly selected to play an exhibition game. a. How many pairs of teams can be chosen? b. What is the probability that the two teams are Los Angeles and New York? c. What is the probability that the Western Conference team is not from California?

Short Answer

Expert verified
Answer: There are 36 possible pairs of teams for an exhibition game. The probability that the Los Angeles Sparks will play the New York Liberty is 1/36, and the probability that the Western Conference team is not from California is 2/3.

Step by step solution

01

Part a: Number of Pairs of Teams

To find the number of possible pairs of teams for an exhibition game, notice that each team in the Western Conference can be paired with each team in the Eastern Conference, making a total number of pairs equal to the product of the number of teams in each conference. There are 6 teams in each conference, so there are \(6\times6 = 36\) possible pairs of teams that can be chosen.
02

Part b: Probability of Los Angeles and New York Playing

To find the probability that the two teams playing are Los Angeles and New York, we will use the definition of probability: \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). There is only one favorable outcome, the pair of Los Angeles Sparks and New York Liberty out of 36 possible pairs of teams. So, the probability is \(\frac{1}{36}\).
03

Part c: Probability of the Western Conference Team Not Being from California

To find the probability that the Western Conference team is not from California, we need to find another way to express the desired probability. Note that the number of teams in the Western Conference not from California is 4 (Minnesota, Phoenix, Dallas, and San Antonio). Each of these 4 teams can be paired with any of the 6 teams in the Eastern Conference, yielding 24 favorable outcomes. The total number of outcomes is still 36, so the desired probability is \(\frac{24}{36} = \frac{2}{3}\).

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 91Ó°ÊÓ!

Key Concepts

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

Combinatorics
Combinatorics involves the study of counting and arrangement of events or objects. In sports, combinatorial techniques help determine the number of different possible matchups between teams. Let's consider the WNBA, where we have the Western and Eastern Conferences, each having six teams. When creating exhibition game pairings, each Western team can match with any Eastern team. This shows the essence of combinatorics in sports, as calculations are done in combinations and products. For instance, with six teams in each conference:
  • Number of Western combinations: 6
  • Number of Eastern combinations: 6
  • Total possible matchups: 6 × 6 = 36
The straightforward clinic of multiplication allows us to see the different pair combinations, helping leagues plan exhibitions or schedules effectively.
Conditional Probability
Conditional probability evaluates the probability of an event occurring, given that another event happens. It can be particularly useful in sports predictions. For instance, if we're interested in knowing the likelihood of the Los Angeles Sparks and New York Liberty playing together, this is about analyzing a specific case among all possible cases. To use conditional probability:
  • Identify favorable outcomes: There is only one - Los Angeles and New York.
  • Divide by total outcomes: Total of 36 matchups possible.
  • Probability: \( \frac{1}{36} \)
Thus, the probability of these two teams playing together in an exhibition match is \( \frac{1}{36} \), illustrating how conditional probability narrows down possibilities to specific scenarios.
WNBA
The Women's National Basketball Association (WNBA) has grown significantly, providing professional opportunities for women in the sport. Within the league, both Western and Eastern Conferences compete, each conference comprising six vibrant teams. These teams showcase some of the best basketball talents from across the globe, making the league a critical element in the promotion and development of women's basketball.
The WNBA utilizes complex scheduling and statistical methods, like those in questions of probability and combinatorics, to ensure the smooth delivery of matches. The ability to calculate pairings and match probabilities is crucial in arranging games and tournaments that are engaging and fair. Ultimately, this structured approach not only serves as a platform for player development but also fosters fan engagement, fueling the growth of basketball among diverse audiences.

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

Use a table of outcomes to display the simple events for the experiments in Exercises \(25-27 .\) A card is randomly drawn from a deck of 52 cards. You record the suit (spade, heart, diamond, or club) and whether the card is a face card \((\mathrm{J}, \mathrm{Q}, \mathrm{K},\) or \(\mathrm{A}) .\)

A college student frequents one of two coffee houses on campus, choosing Starbucks \(70 \%\) of the time and Peet's \(30 \%\) of the time. Regardless of where she goes, she buys a cafe mocha on \(60 \%\) of her visits. a. The next time she goes into a coffee house on campus, what is the probability that she goes to Starbucks and orders a cafe mocha? b. Are the two events in part a independent? Explain. c. If she goes into a coffee house and orders a cafe mocha, what is the probability that she is at Peet's? d. What is the probability that she goes to Starbucks or orders a cafe mocha or both?

During the inaugural season of Major League Soccer in the United States, the medical teams documented 256 injuries that caused a loss of playing time to the player. The results reported in The American Journal of Sports Medicine are shown in the table. If one individual is drawn at random from this group of 256 soccer players, find the following probabilities: a. \(P(A)\) b. \(P(G)\) c. \(P(A \cap G)\) d. \(P(G \mid A)\) e. \(P(G B)\) f. \(P(G C)\) g. \(P(C \mid P)\) h. \(P\left(B^{c}\right)\)

An investor has the option of investing in three of five recommended stocks. Unknown to her, only two will show a substantial profit within the next 5 years. If she selects the three stocks at random (giving every combination of three stocks an equal chance of selection), what is the probability that she selects the two profitable stocks? What is the probability that she selects only one of the two profitable stocks?

A certain manufactured item is visually inspected by two different inspectors. When a defective item comes through the line, the probability that it gets by the first inspector is \(.1 .\) Of those that get past the first inspector, the second inspector will "miss" 5 out of \(10 .\) What fraction of the defective items get by both inspectors?

See all solutions

Recommended explanations on Math Textbooks

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.