1 / 27

Chapter 2. Foundations of Probability Section 2.4. Counting Rules Useful in Probability

Chapter 2. Foundations of Probability Section 2.4. Counting Rules Useful in Probability. Jiaping Wang Department of Mathematical Science 01/23/2013, Wednesday. Outline. Fundamental Principle of Counting Permutations Combinations Partitions.

oliana
Download Presentation

Chapter 2. Foundations of Probability Section 2.4. Counting Rules Useful in Probability

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 2. Foundations of Probability Section 2.4. Counting Rules Useful in Probability Jiaping Wang Department of Mathematical Science 01/23/2013, Wednesday

  2. Outline Fundamental Principle of Counting Permutations Combinations Partitions

  3. Part 1. Fundamental Principle of Counting

  4. Examples A relative frequency definition of probability will work for any experiment that results a finite sample space with equally likely outcomes. So counting becomes a key step in obtaining the probability.

  5. Theorem 2.2 Fundamental Principle of Counting: If the first task of an experiment can result in n1 possible outcomes and for each such outcome, the second task can result in n2 possible outcomes, then there are n1n2 possible outcomes for the two tasks together. The principle can extend to more tasks in a sequence.

  6. Example 2.5 In connection with the national retail chain that plans to build two new stores, the eight possible combinations of locations are as shown in the figure. If all eight choices are equally likely (that is, if one of the pairs of cities is selected at random), find the probability that City E is selected. Solution: As E can appear in the combinations AE, BE, CE or DE, also there are total 8 possible combinations with equal chance to be selected, thus the probability of City E is selected is 1/8*4=1/2.

  7. Example 2.6 Five cans of paint (numbered 1 through 5) were delivered to a professional painter. Unknown to her, some of the cans (1 and 2) are satin finish and the remaining cans (3, 4 and 5) are glossy finish. Suppose she selects two cans at random for a particular job. Let A denote the event that the painter selects the two can of statin-finish paint and let B denote the event that the two cans have different finishes (one of satin and one of glossy). Find P(A) and P(B). Solution: Total we have 20 possible combinations. For event A, there are two possibilities: {1,2} or {2,1}, so P(A)=2/20; For event B, there are 12 possibilities: {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,1}, {3,2}, {4,1}, {4,2}, {5,1}, {5,2}, so P(B)=12/20.

  8. Part 2. Permutations

  9. With/Without Replacement Another Example: select telephone numbers. For example, given the area code (940), then how many possibilities can one choose from 0-9 numbers in the next 7 digits? Answer: the number of possible first digit is 10, the number of possible 2nd digit is also 10, and so for other digits. So total we have 10x10x10x10x10x10x10=107. From this example and Example 2.6, we can find two things affect the manner of counting: First is whether or not the order is important; Second is whether or not chosen with replacement or without replacement.

  10. Permutations Theorem 2.3 Permutations. The number of ordered arrangements or permutations Prn of r objects selected from n distinct objects (r≤n) is given by n! = n×(n-1) ×(n-2)ו••×3×2×1, 0!=1, n!=n(n-1)!.

  11. Example 2.7 A small company has 12 account managers. Three potential customers have been identified and each customer has quite different needs. The company’s director decides to send an account manager to visit each of the potential customers and considers the customer’s needs in making his selection. How many ways are there for him to assign three different account managers to make the contacts? Solution: P312 = 12!/(12-3)!=12!/9!=12 x 11 x 10 x 9!/9!=1320.

  12. Part 3. Combinations

  13. Order Is Not Important For example, playing bridge with dealing 13 cards, at this time, the order in which the cards are dealt does not affect the final hand. For this, the order is not important; Also, this selection is without replacement. For this kind of case, it is called combination. Another example in counting based on combination is drawing the numbers in lottery.

  14. Combinations Theorem 2.4 Combinations. The number of distinct subsets or combinations of size r that can be selected from n distinct objects (r ≤ n)is given by

  15. Example 2.8 Most states conduct lotteries as a means of raising revenue. In Florida’s lottery, a player select 6 numbers from 1 to 53. For each drawing, balls numbered from 1 to 53 are placed in a hopper. Six balls are drawn from the hopper at random and without replacement. To win the jackpot, all six of the player’s number must match those drawn in any order. How many wining numbers are possible? Solution: C653 = 53!/[6! (53 -6)!]=53x52x51x50x49x48/(6x5x4x3x2x1)=22957480 So the probability to win is about 1/22957480.

  16. Example 2.9 Twenty six states participate in the Powerball lottery. In this lottery, a player selects five number between 1 and 53 and a Powerball number between 1 and 42. For each drawing, five balls are drawn at random and without replacement from a hopper with 53 white balls numbered 1 to 53. A sixth ball is drawn from a second hopper with 42 red balls numbered 1 to 42. To win the jackpot, the five numbers selected by the player must match those of the five white balls drawn, and the player’s Powerball number must match the number on the red ball drawn from the hopper. How many possible wining numbers are there? Is there a greater probability of wining Florida’s lottery or the Powerball if one buys a single ticket? Solution: there are two tasks, in the first task, the number is C553 = 53x52x51x50x49/(5x4x3x2x1)=2869685; in the second task, there are C142 = 42. Based on the Fundamental Principle of Counting, there are total C553x C142 = 2869685 x 42 = 120,536,770, then we can say the probability of winning the Powerball lottery is 1/ 120,536,770, which is less than the probability of winning Florida’s lottery.

  17. Example 2.10 A department in a company has 12 members: 8 males and 4 females. To gain greater insight into the employees’ view of various benefits, the human resources office plans to form a focus group from members of this department. Five members will be selected at random from the department’s members. What is the probability that the focus group will only have males? What is the probability that the focus group will have two males and three females? Solution: It is selection without replacement and order is not important. So the total number of ways is 12!/(5!7!)=792; Q1: The number of ways to select only males is that select 5 males from 8 which is 8!/(5!3!)=56, so the probability is 56/792. Q2: The number of ways to select 2 males is 8!/(2!6!)=28 and the number of ways to select 3 females is 4!/(3!1!)=4, then the probability is 28x4/792.

  18. Part 4. Partitions

  19. Theorem 2.5 Partitions Consider a case: If we roll a die for 12 times, how many possible ways to have 2 1’s, 2 2’s, 3 3’s, 2 4’s, 2 5’s and 1 6’s? Solution: First, choose 2 1’s from 12 which gives 12!/(2!10!), second, since there are two positions are filled by 1’s, the next choice appears in the left 10 positions, so there are 10!/(8!2!) ways, and so similar for next other selections which provides the final result is 12!/(2!10!)x10!/(2!8!)x8!/(3!5!)x5!/(2!3!)x3!/(2!1!)x1!/(1!0!) =12!/(2!x2!x3!x2!x2!x1!) Theorem 2.5 Partitions. The number of partitioning n distinct objects into k groups containing n1, n2,•••, nk objects, respectively, is

  20. Examples 2.11, 2.12 2.11 Suppose 10 employees are to be divided among three job assignments, with 3 employees going to job I, 4 to job II, 3 to job III. In how many ways can the job assignments be made? Solution: 10!/(3!4!3!)=4200. 2.12 Suppose that three employees of a certain ethnic group all get assigned to job I. Assumes that they are the only employees among the 10 under consideration who belong to this ethnic group, what is the probability of this happening under a random assignment of employees to jobs? Solution: As the job I is filled by these three employees with the same ethnic, we only consider the left employees. At this time, we have total 7 employees with two different groups (job II and job III), so the number of ways is 7!/(4!3!)=35, then we can find the probability is 35/4200 = 1/120.

  21. Determining whether or order is important Is critical to counting Suppose a fair coin is flipped three times and the number of heads observed. From here, we can find the probabilities are same for ordered or unordered results, but the sample space is changed. A key to determining whether or not order is important is an understanding of the random process giving rise to the outcomes.

  22. Example 2.13 The birthday problem. Suppose n people are in a room. Assume no one has birthday at 2/29 and there are total 365 days per year. Q1: What is the probability that no two of them have the same birthday? Solution: Each person can have 365 possible birthdays, so there are total 365n ways. If for the first person, there are 365 possible birthdays, then the second one has only 364 possible birthdays (as the first one filled one of them) and so on for the next Persons), so total there are 365(364)••(365-n+1) ways such that no two have the same Birthday. Then we have the probability 365(364)••(365-n+1)/ 365n. If n>365, based on the Pigeon Hole theorem, at least two of them have the same birthday, so the probability that no two of them have the same birthday is 0. Q2: How many people must be in the room for the probability that at least two of the n people have the same birthday to be greater than 1/2? Solution: Consider the opposite event: no two of them have the same birthday, so the probability is 365(364)••(365-n+1)/ 365n<1/2, by solving this inequality to obtain The maximum n=22, so the opposite event should need 23 people.

  23. Example 2.14 A poker hand consist of five cards. If all of them are from the same suit but not in consecutive order, we say that the hand is a flush. For instance, if we have five clubs that are not in consecutive order (say, 2, 4, 5, 6, 10), then we have a flush. What is the probability of a flush but not straight flush? Solution: There are total 52!/(47!5!) ways to choose 5 from 52 cards. Now to choose The flush, first there are 4 different suits, second given a suit, there are 13!/(5!8!) flush including the straight flush. Now we know there are 10 different straight flush from (ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, ace). So the probability would be 4x(13!/(5!8!)-10)/(52!/(47!5!).

  24. Homework #2 Page 45: 2.36, 2.38, 2.40, 2.42, 2.43, Page 46: 2.46, 2.48, 2.50 Due Wed., 03/02/2013

  25. Part 5. More Counting Rules

  26. Order and Replacement Theorem 2.6 The number of ways of making r selections from n objects when selection is made with replacement and order is not important is

  27. Example 2.16 In a lottery, a player selects 6 numbers between 1 and 44. The same number may be chosen more than once. For each drawing, 6 balls are drawn at random with replacement from a hopper with 44 white balls numbered 1 to 44. Sufficient time is allowed between selections of a ball for the previously selected ball to be mixed with the others. To win the jackpot, all six numbers of the player must match those drawn in any order. How many wining numbers are possible? Solution: C644+6-1 = 49!/(6!43!)

More Related