Example_class_1_slides

Example_class_1_slides - Example class 1 STAT1301...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Example class 1 STAT1301 Probability and Statistics I Chan Chi Ho, Chan Tsz Hin & Shi Yun September 14, 2011 Information about tutors Tutors: Chan Chi Ho (MW515), Chan Tsz Hin (MW518) and Shi Yun (MW518) Emails: [email protected], [email protected], [email protected], respectively. Assignment: Put assignments in assignment box outside MW515. Deadline of Assignment 1: Sep. 26th Information about tutors Tutors: Chan Chi Ho (MW515), Chan Tsz Hin (MW518) and Shi Yun (MW518) Emails: [email protected], [email protected], [email protected], respectively. Assignment: Put assignments in assignment box outside MW515. Deadline of Assignment 1: Sep. 26th Information about tutors Tutors: Chan Chi Ho (MW515), Chan Tsz Hin (MW518) and Shi Yun (MW518) Emails: [email protected], [email protected], [email protected], respectively. Assignment: Put assignments in assignment box outside MW515. Deadline of Assignment 1: Sep. 26th Review: Combinatorial Analysis Multiplication Principle: If an operation consists of a sequence of k separate steps of which the first can be performed in n1 ways, followed by the second n2 ways, and so on until the k th can be performed in nk ways, then the operation consisting of k steps can be performed by k n1 × n2 × ... × nk = ∏ ni i =1 ways. Review: Combinatorial Analysis (cont’d) I Selection of r from n distinct objects: 1 Ordered selection with replacement: Consider drawing r balls from n distinct balls. The total number of ways is n × n × ... × n = nr . r 2 times Ordered selection without replacement: Consider drawing r balls from n distinct balls without replacement. The total number of ways is n × (n − 1) × (n − 2) × ... × (n − r + 1) = n! . (n − r )! Review: Combinatorial Analysis (cont’d) II 3 Unordered selection with replacement: Consider drawing r balls from n distinct balls with replacement and without regard to order. This is the same as placing r identical objects into n distinct containers (why?). The problem is the same as arranging the “0” and “|” in, say, 00|000||00|0|||00.....|0||0 r "0"s and n−1 "|"’s The total number of ways is r +n−1 r 4 = (r + n − 1)! . r !(n − 1)! Unordered selection without replacement: Consider drawing r balls from n distinct balls without replacement and without regard to order. The total number of ways is n r = n! . r !(n − r )! Review: Combinatorial Analysis (cont’d) III Arrangement of n objects, with r distinct types. In other words, there are n1 objects of type 1, n2 objects of type 2, ..., nr objects of type r . (e.g. number of different letter arrangements can be formed using the letters STATISTICS.) Total number of ways is n! . n1 ! × n2 ! × · · · × nr ! Partition of n distinct objects into r distinct groups with specified size n1 , n2 , · · · , nr . (e.g. number of ways in dividing a class of 40 into groups of 10, 10 and 20.) Toal number of ways is n n1 , n2 , · · · , nr = n! . n1 ! × n2 ! × · · · × nr ! Review: Combinatorial Analysis (cont’d) IV Partition of n indistinguishable objects into r distinct groups (i.e. we only concern the number of objects in each group). The problem is the same as arranging the “0” and “|” in, say, 00|000||00|0|||00.....|0||0 n "0"s and r −1 "|"’s Total number of ways is n+r −1 n = (n + r − 1)! . n!(r − 1)! Problem 1 Problem 1. A product code of manufacturer consists of 2 letters followed by 3 digits. The manufacturer only uses the letters A, B, C and D. For example, AA032 and BD369 are valid product codes. If a product is selected at random, what is the probability that 1 2 3 4 it begins with the letter A? the first two letters are identical? the letters are identical and the digits are equal? the letters are formed by B and D, and the 3-digit number is even? Solution to Problem 1 Solution. 1 2 3 4 Since the four letters are equally likely to be the first letter, P(beginning with A) = 1 . 4 4 P(two letters are identical) = 4×4 = 1 . 4 1 P(the digits are equal) = 10×10×10 = 100 . 10 1 1 P(letters are identical and digits are equal) = 100 × 1 = 400 . 4 P(the letters are formed by B and D) = P({B,D} or {D,B}) = 2 1 4×4 = 8 . 5 P(the 3-digit number is even) = 10 = 1 . 2 P(the letters are formed by B and D, and the 3-digit number is even) 1 1 1 = 8 × 2 = 16 . Problem 2 Problem 2. Three boys and four girls line up randomly in a row. What is the probability that 1 2 3 4 all boys are on one side? boys and girls are in alternate positions? no 2 boys sit next to each other? the boy named John sits on the right of the girl named Mary (assuming all boys and girls have different names)? Solution to Problem 2 Solution. 1 2 3 4 2 7 =. 3 35 1 P(boys and girls are in alternate positions) = 35 . The number of ways of arranging in this way is 5 = 10 ( Set the girls 3 as GGGG, put the three boys in to the interspace of the girls). Thus P(no two boys sit next to each other) = 10 = 2 35 7 6 × 5! 1 P(John sits on the right of Mary)= =. 7! 7 P(all boys on one side)= 2/ Problem 3 Problem 3. There are 4 red balls, 3 yellow balls, 2 blue balls and 1 white balls. In how many distinct ways can the balls be arranged in a circle? Solution. First we consider the case of arranging the balls in a row. It is a problem of arrangement of n objects (10 balls), with r distinguishable types (4 colors). 10 10! = 4! × 3! × 2! × 1! 4, 3, 2, 1 = 12600 We then understand that for each arrangement of balls in a circle, we can derive 10 distinct arrangements in a row, each by choosing a different ball to be the first in a row. The number of ways the balls can be arranged in a row is 10 times of that they can be arranged in a circle. Thus, the answer is: 12600 10 = 1260 Problem 4 Problem 4. (Exam Fall 2008) Three candidates, A, B and C, participate in an election in which eight voters will cast their votes. The candidate who receives the absolute majority, that is at least five, of the votes will win the election. All voters must vote for one and only one candidate and cannot abstain. Suppose the voting is done by a secret ballot such that the voter’s identities are kept confidential. 1 2 3 Show that there are 45 distinct ways to distribute the eight votes among the three candidates. Show that there are 10 distinct ways to distribute the eight votes among the three candidates so that A will win the election. How many distinct ways are there to distribute the eight votes among the three candidates so that nobody will win the election? Solution to Problem 4 I Solution. 1 This is a problem of partitioning n indistinguishable objects (8 votes) into r distinguishable groups (3 candidates). 8+3−1 8 = = 2 10! 8! × 2! 45 As we have fixed the result to be A winning the election, we first allocate 5 votes to A and consider the number of ways to distribute the remaining 3 votes. The problem becomes a partition of n indistinguishable objects (3 votes) into r distinguishable groups (3 candidates). 3+3−1 3 = = 5! 3! × 2! 10 Solution to Problem 4 II 3 We can simply deduct the number of ways to distribute the votes for candidate A, B and C to win respectively from the total number of ways to distribute the eight votes. This is based on the fact that the events {A winning}, {B winning}, {C winning} and {nobody winning} are mutually exclusive and exhaustive. 45 − 10 − 10 − 10 = 15 Problem 5 Problem 5. A total of r identical robots are being lifted from the ground floor to the n upper floors of a building. 1 2 Find the total number of distinct patterns. A student claims that if each robot is independently and equally probably allocated to each of the n floors, then the probability that all robots end up on the first floor is given by Number of ways of sending all robots to 1/F = Number of distinct exit patterns Comment on the above statement. Is it correct? n+r −1 r −1 . Solution to Problem 5 Solution. 1 This is a problem of allocating r identical objects into n distinct containers. Thus the total number of ways is r +n−1 r 2 . The statement is wrong because, if each robot is independently and equally probably allocated to each of the n floors, the distinct patterns are not equally likely (Why? For example, if there are 3 floors and 2 robots, what is the probability of all robots going to the 1st floor? What is the probability of one robot going to the 1st floor and the other robot going to the 2nd floor?). The correct probability should be 1 . nr Problem 6 Problem 6. In an experiment k balls are drawn randomly without replacement from an urn containing 1 red ball, 2 blue balls and 17 yellow balls. All the balls are identical apart from their colours. 1 Suppose 0 ≤ k ≤ 17. Show how you may obtain the number of distinct combinations of the k balls by considering the identity (1 + x )(1 + x + x 2 )(1 − x )−1 ≡ (1 − x 2 )(1 − x 3 )(1 − x )−3 . 2 Show that there are six distinct ways of drawing three balls randomly without replacement from the urn. Solution to Problem 6 Solution. 1 The number of ways of drawing k balls from the urn is equal to the coefficient of x k in the expression (why? Note that (1 + x )(1 + x + x 2 )(1 − x )−1 = (1 + x )(1 + x + x 2 )(1 + x + x 2 + x 3 + ...). 2 The number of ways of drawing k balls from the urn is equal to the coefficient of x k in the above expression.) Using the expression on the RHS, we get (1 − x 2 )(1 − x 3 )(1 − x )−3 = (1 − x 2 − x 3 + x 5 )(1 + 3x + 6x 2 + 10x 3 + ...). From above, we see that the coefficient of x 3 is −1 − 3 + 10 = 6. Problem 7 Problem 7. (Class test Fall 2008) A lady has three rings and eight distinguishable fingers (excluding the two thumbs). 1 Suppose the three rings are identical. 1 2 3 2 How many distinct ways are there to wear the three rings on the same finger? How many distinct ways are there to wear two rings on the same finger and the third ring on a different finger? How many distinct ways are there to wear the three rings on three separate fingers? Now suppose the three rings are distinguishable. Using part (a) or otherwise, find the total number of distinct ways to wear the three rings on the lady’s eight fingers without any restriction. Solution to Problem 7 Solution. 1 Since the rings are identical, 1 2 3 2 Number of distinct ways to wear the three rings on the same finger = number of distinct fingers = 8. First we choose a finger to wear 2 rings, then we choose another finger to wear 1 ring. Thus the number of ways = 8 × 7 = 56. The number of ways to wear the three rings on three separate fingers is 8 = 56. 3 Since the rings are distinguishable, the total number of distinct ways is 8 × 3! + 56 × 3! + 56 × 3! = 720. Problem 8 Problem 8. Mary has 5 dresses, 3 skirts, 4 blouses, 3 pairs of shoes, and 2 hats. She always wears shoes and either a dress or a blouse and a skirt. She may or may not wear a hat. 1 2 3 How many different combinations can she wear? Suppose Mary can afford to buy either an extra dress or an extra hat (but not both). Which should she buy if she decides to maximize the number of different combinations that she can wear? Suppose Mary’s brown shoes do not match her pink or blue dress, and that the blue hat does not match her yellow blouse. How many matching combinations can she wear? Solution to Problem 8 Solution. 1 2 3 Total number of combinations = 3 × (2 + 1) × (5 + 3 × 4) = 153. Extra dress: total number of combinations = 3 × (2 + 1) × (6 + 3 × 4) = 162. Extra hat: total number of combinations = 3 × (3 + 1) × (5 + 3 × 4) = 204. Therefore she should buy an extra hat. Total number of combinations with brown shoes and pink or blue dress = 1 × 3 × 2 = 6. Total number of combinations with blue hat and yellow blouse = 3 × 1 × (1 × 3) = 9. Since the two events “brown shoes and pink or blue dress” and “blue hat and yellow blouse” are mutually exclusive, there is no double count in the above two countings. Therefore the matching combinations she can wear is 153 − 6 − 9 = 138. Problem 9 Problem 9 (a more difficult question). Suppose there are N boys and M girls sitting around a table, and the chairs are all different. 1 2 How many different ways to sit them? How many ways to sit them such that there are k boys with a girl sitting to his right (assume M ≥ k )? Solution to Problem 9 I Solution 1 2 (M + N )!. We can count the total number of combinations using the following steps: 1 2 3 4 Suppose there is a boy called John. Line up the boys with John in front. The number of ways to line up the N − 1 boys is (N − 1)!. Choose k boys from the N boys and mark them. The number of ways N is . k Line up the M girls. The number of ways is M !. Divide the M girls (which are already all lined up) into k groups M −1 without disturbing the order. The number of ways is (why? k −1 Consider for example there are 4 girls lined up in a row G1 G2 G3 G4 , how many ways to separate them into 3 groups without disturbing the order?). Put the first group on the right of the first marked boy, and put the 2nd group on the right of the 2nd marked boy, etc. Solution to Problem 9 II 5 Choose a sit for John. People queuing behind John occupy seats to John’s right one by one anticlockwise. There are M + N ways. Total number of ways = the product of the number of ways in (i), (ii), (iii), (iv), (v) =(N − 1)! =N !M ! N k N −1 k −1 M! M −1 k −1 M −1 k −1 (M + N ) M +N k ...
View Full Document

This note was uploaded on 01/16/2012 for the course STAT 1301 taught by Professor Smslee during the Fall '08 term at HKU.

Ask a homework question - tutors are online