Question

In: Statistics and Probability

In a university games tournament, 64 students are about to participate in a chess knockout competition....

In a university games tournament, 64 students are about to participate in a chess knockout competition.

The first round consists of 32 games, with two students per game.

The 32 winners of the first round get to play in the second round, which consists of 16 games, and so on, until an overall winner is declared in the sixth round. (In the case of a draw on any game, a coin is tossed to determine the winner.)

(a) In how many different ways can the 64 participating students be paired up on the first round? (Do not consider the order in which students can be paired up.)

(b) Suppose that the 64 participating players are of equal ability, and pairing up is purely random on each round. Find the probability that Eva and Brett (two of the 64 students) will get to play each other at some stage during the knockout competition

(very appreciate write in details, thank you very much)

Expert Solution

(a)

Number of ways 64 participating students be paired up = ⁶⁴C₂

= 64! / (64-2)! * 2!

= 64! / (62! * 2!)

= (64 * 63) / (2 * 1)

= 2016

(b)

Probability of Eva and Brett (two of the 64 students) will get to play each other in first round = 1 / ⁶⁴C₂

= 1/ 2016

Probability of Eva and Brett will not get to play each other in first round = 1 - 1/2016

Probability of winning any round for a player = 1/2

Probability that both Eva and Brett win 1st round = (1/2) * (1/2) = (1/2)²

Now, in 1st round 32 students will participate.

Probability of Eva and Brett will get to play each other in second round = 1 / ³²C₂

= 1/ 496

Probability of Eva and Brett will not get to play each other in second round = 1 - 1/496

In Similar way we can show that,

Probability that both Eva and Brett win kth round = (1/2²)^k

Probability of Eva and Brett will get to play each other in 1st , 2nd, 3rd, 4th , 5th and 6th round are 1 / ⁶⁴C₂, 1 / ³²C₂,1 / ¹⁶C₂,1 / ⁸C₂,1 / ⁴C₂and 1 / ²C₂respectively.

Probability that Eva and Brett will get to play each other at some stage

= Probability of Eva and Brett will get to play each other in first round +

Probability of Eva and Brett will not get to play each other in first round * Probability of Eva and Brett will get to play each other in second round * Probability that both win 1st round +

Probability of Eva and Brett will not get to play each other in first round * Probability of Eva and Brett will not get to play each other in second round * Probability of Eva and Brett will get to play each other in third round * Probability that both win till 2nd round +

Probability of Eva and Brett will not get to play each other in first round * Probability of Eva and Brett will not get to play each other in second round * Probability of Eva and Brett will not get to play each other in third round * Probability of Eva and Brett will get to play each other in 4th round * Probability that both win till 3rd round +

Probability of Eva and Brett will not get to play each other in first round * Probability of Eva and Brett will not get to play each other in second round * Probability of Eva and Brett will not get to play each other in third round * Probability of Eva and Brett will not get to play each other in 4th round * Probability of Eva and Brett will get to play each other in 5th round * Probability that both win till 4th round +

Probability of Eva and Brett will not get to play each other in first round * Probability of Eva and Brett will not get to play each other in second round * Probability of Eva and Brett will not get to play each other in third round * Probability of Eva and Brett will not get to play each other in 4th round * Probability of Eva and Brett will not get to play each other in 5th round * Probability of Eva and Brett will get to play each other in 6th round * Probability that both win till 4th round

= 1 / ⁶⁴C₂+ (1 - 1 / ⁶⁴C₂ ) * 1 / ³²C₂ * (1/2²)¹ + (1 - 1 / ⁶⁴C₂ ) * (1 - 1 / ³²C₂) * 1 / ¹⁶C₂ * (1/2²)² +

(1 - 1 / ⁶⁴C₂ ) * (1 - 1 / ³²C₂) * (1 - 1 / ¹⁶C₂ ) * 1 / ⁸C₂ * (1/2²)³ + (1 - 1 / ⁶⁴C₂ ) * (1 - 1 / ³²C₂) * (1 - 1 / ¹⁶C₂ ) * (1 - 1 / ⁸C₂ ) * 1 / ⁴C₂ * (1/2²)⁴

+ (1 - 1 / ⁶⁴C₂ ) * (1 - 1 / ³²C₂) * (1 - 1 / ¹⁶C₂ ) * (1 - 1 / ⁸C₂ ) * (1 - 1 / ⁴C₂ ) * 1 / ²C₂ * (1/2²)⁵

= 1/2016 + (1 - 1/2016) * 1/496 * 1/4 + (1 - 1/2016) * (1 - 1/496) * 1/120 * 1/16 + (1 - 1/2016) * (1 - 1/496) * (1 - 1/120) * 1/28 * 1/64 + (1 - 1/2016) * (1 - 1/496) * (1 - 1/120) * (1 - 1/28) * 1/6 * 1/256 + (1 - 1/2016) * (1 - 1/496) * (1 - 1/120) * (1 - 1/28) * (1 -1/6) * 1 * 1/1024

= 0.003468574513

orchestra answered 2 years ago

You are about to play a series of 9 chess games online against an opponent called...

You are about to play a series of 9 chess games online against an opponent called EliteChampion. The first to win 5 games wins the series. (Ignore the possibility of a draw.) You know that EliteChampion is most likely your friend, Jenna. There’s a 20% chance EliteChampion is your Mom, and that’s the only other possibility. You beat your Mom 70% of the time, but you only beat Jenna 40% of the time. Given that you won the first game,...

Question 3 At a local university, a sample of 64 evening students was selected in order...

Question 3 At a local university, a sample of 64 evening students was selected in order to determine whether the average age of the evening students is significantly different from 21. The average age of the students in the sample was 23 years. The population standard deviation is known to be 4.5 years. Determine whether or not the average age of the evening students is significantly different from 21. Use a 0.05 level of significance

A random sample of 64 students at a university showed an average age of 20 years...

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The 90% confidence interval for the true average age of all students in the university is 19.50 to 20.50 19.36 to 20.38 19.18 to 20.49 19.02 to 20.59

The faculty of a university mathematics department is concerned about the performance of students in the...

The faculty of a university mathematics department is concerned about the performance of students in the introductory calculus offered by the department and required of all science and engineering majors. Historically, class averages on the test have been about 75, a passing grade but indicative that students may not be learning the material as well as they need to in order to go on to the next course. The chair would like to raise the average to at least 80....

10. A university is concerned about the proportion of students that graduate. To address the issue,...

10. A university is concerned about the proportion of students that graduate. To address the issue, it does the following. First, it gets a random sample of incoming students. Some of the students in this random sample will be required to have extra meetings with a series of advisors and some will not. It randomly decides which of the students will have this requirement and which will not. In the data, 516 of the 885 students that had this requirement...

The average age of students at the University of Houston – Downtown is about 28 years...

The average age of students at the University of Houston – Downtown is about 28 years old. The president of the university believes that the age of full-time students attending classes during the day is lower that the average of all students. Test this hypothesis. Be sure to show all five steps. Show your Excel file ONCE per group with all the names of your team members included. Age 14-21: 23% 22-24: 23% 25-30: 27% 31-35: 12% 36-50: 13% Over...

2. A researcher is concerned about the level of knowledge possessed by university students in the...

2. A researcher is concerned about the level of knowledge possessed by university students in the field of algebra. Students completed a high school senior level standardized algebra exam. Major for students was also recorded. Data in terms of percent correct is recorded below for 32 students. Use the Microsoft Excel "Anova Single-Factor" Data Analysis tool to conduct a 1-way ANOVA test for the data in the following table: Education Business/Management Behavioral/Social Science Fine Arts 62 82 42 30 81...

A large university is interested in learning about the average time it takes students to drive...

A large university is interested in learning about the average time it takes students to drive to campus. The university sampled 238 students and asked each to provide the amount of time they spent traveling to campus, the average and standard deviation were 21.5 and 4.32 respectively. This variable, travel time, was then used conduct a test of hypothesis. The goal was to determine if the average travel time of all the university's students differed from 20 minutes. Use a...

The director of admissions at a large university advises parents of incoming students about the cost...

The director of admissions at a large university advises parents of incoming students about the cost of textbooks during a typical semester. He selected a sample of 100 students and recorded their textbook expenses for the semester. He then calculated a sample mean of $675.60 and a sample standard deviation of $45.20. You may assume that the distribution of textbook expenses is approximately normally distributed. (a) Is there sufficient evidence that the population mean textbook expense per semester is above...

A recent survey asked 64 CMU students, “In the past year, about how many books did...

A recent survey asked 64 CMU students, “In the past year, about how many books did your read?” The results indicated a mean of 16.8 books with a S.D. of 6,7 books. a. What is the Standard Error of the estimate of the population mean? b. What is the Margin of Error at a C.L. of 97%? c. What is the right boundary of the C.I. at a C.L. of 97%? d. At a 99% C.L., if the researcher wants...