# Homework 3 (due Tuesday, October 13)

Save this PDF as:

Size: px
Start display at page:

Download "Homework 3 (due Tuesday, October 13)"

## Transcription

1 Homework (due Tuesday, October 1 Problem 1. Consider an experiment that consists of determining the type of job either blue-collar or white-collar and the political affiliation Republican, Democratic, or Independent of the 1 members of an adult soccer team. How many outcomes are (a in the sample space; Each of the players has 2 possible job types and possible political affiliations, So a total of 6 possible choices for each of the 1 players. Thus a total of 6 1 possible outcomes for the soccer teams job and political composition. (b in the event that at least one of the team members is a blue-collar worker; If one wants to count the possibilities where at least one is a blue collar worker we count the ones in which no one is a blue collar worker from the total 6 1. If no one is to be a blue-collar worker then for each player there is only one choice of job type and three choices for political affiliation. Thus 1 total where no one is a blue-collar worker. Thus the number where there is at least one blue collar worker is (c in the event that none of the team members considers himself or herself as Independent? If no one considers themselves independent then there are only 2 possible choices for political choices and 2 job types. Thus for each player a possible of 4 different combos of job and political affiliations. Thus, there are 4 1 different outcomes of this type. Problem 2. probability of being dealt If it is assumed that all ( 2 poker hands are equally likely, what is the (a a full house? (A hand is said to be a full house if the hand has the values a, a, b, b, b. ( 2 The total number of hands one could be dealt without restriction is. This is the size of our sample space. Now choose from the 1 cards types the numbers for ( 1 the a s and the numbers for the b s from the remaining 12. Now for the a s 1 choose the 2 suits and suits for the b s. Thus the probability of getting 2 such a hand is: ( 2 =

2 (b two pairs? (This occurs when the cards have values a, a, b, b, c, where a, b, c are all distinct. To count the outcomes that look like this we choose the value that are to appear ( 1 ( in our hand. Then from those choose the 2 that are to be the pairs =. 2 Then from each of the 2 values that are to be the pairs you pick the 2 in the hand from the 4 possible cards of that value or = 6. Then for the card that appears 2 as the single you pick from the 4 possible cards of that value.thus the probability of getting such a hand is: ( = (c three of a kind? (This occurs when the cards have denominations a, a, a, b, c, where a, b, c are all distinct. ( 1 Again we choose the three values to appear in the hand. Then we choose from those the one to be the tripple ( choices. Now we have decided which is to be the triple and we choose the of the possible 4 cards of that denomination = 4. And for each of the values that appears as a single we choose from the four cards of that value or = 16. Consequently the probability is: 12 ( Problem. There are 0 physicists and 2 chemists attending a certain conference. Three of these people are randomly chosen to take part in a panel discussion. What is the probability that at least one chemist is chosen? 2

3 You could do this by cases considering the case when one chemist is chosen, where 2 chemists are chosen or where chemists are chosen. However I would rather consider the compliment of this event where no chemist is chosen and then subtract from one, the probability of this event happening. The total number of possible choices for a group of chosen from the total people is ( 0 where no chemist is chosen is (. The total number of groups. Thus the probability of no chemist is: = Thus the probability of getting at least one chemist is 1 minus the number above or: Problem 4. There are hotels in a certain town. If people check into hotels in a day, what is the probability that they each check into a different hotel? We could view this problem as the intersection of the following three events. E 1 = {event person 1 checks into a hotel}, E 2 = {event that person 2 doesn t check into the same hotel as person 1}, E = {event that person is not in hotels with either person one or person 2}. We would be interested in P (E 1 E 2 E = P (E 1 P (E 2 E 1 P (E E 1 E 2. The E 1 is the entire sample space since person 1 must check into SOME hotel, so P (E 1 = 1. Once person one has checked into a hotel if person 2 cant check into the same one then he has 4 choices out of where he or she can check in thus P (E 2 E 1 = 4. Once person one and person 2 are checked into different hotels then there is only choices left for where person can check in if he is not to be in the same hotel as person 1 or person 2. Thus P (E E 1 E 2 =. Consequently, the probability that we are after is equal to 4 = Problem. (a If n people, including A and B, are randomly arranged in a line, what is the probability that A and B are next to each other? Well the sample space has a total of n! possible orderings. Then we count the ways in which A and B can be next to each other. To do this we consider them as one person and order the n-1 people in (n 1! ways. Then we must order A and B among themselves in 2 possible ways. Thus the probability of selecting such an ordering would be 2 (n 1! n! = 2 n. (b What would the probability be if the people were randomly arranged in a circle?

4 If they are arranged in a circle, then as argued before, the number of ways to do that is (n 1!. The number of ways in which A and B are next to each other can be counted by fixing A at the 12 o clock position on the table and then placing B either to his left or right (2 choices and then ordering the remaining (n 2! possible ways. Thus this probability would be 2(n 2! = 2. (n 1! n 1 Problem 6. Find the simplest expression for the following events: (a (E F (E F c = E; (b (E F (E c F (E F c = F E; (c (E F (F G = F EG. Problem 7. In this problem we prove the following identity in two different ways. (2k! 2 k k! = (2k 1(2k (2k...1. (1 (a Prove the above identity using induction. To prove by induction we must do two things. First we must prove the base case when k=1. To do this we plug in 1 into both sides of the equation. Both sides give 1, thus the equation holds when k=1. Now we must assume that the formula holds for k = n and prove that this implies that the formula holds for k = n + 1. To do this we write down the left hand side of the equation replacing k with n + 1 giving: (2n + 2! 2 n+1 (n + 1! = (2n + 2(2n + 1(n! 2(n + 12 n n! = (2n + 1 [ (2n! 2 n n! Now the fraction in the square brakets in the equation above is exactly the LHS of the equation when k = n. We are assuming that the equation holds when k = n so we can replace this term in the square brackets with the right hand side of the equation when k = n. Thus we have that: ] (2 (2n + 2! = (2n+1(2n 1(2n (2kn...1 = (2(n+1 1(2(n n+1 (n + 1! ( Thus the equation holds for k = n + 1 if it is true for k = n. Consequently since it is true for the base case of k = 1 then it is true for all integers k by induction as desired. (b Prove the above identity combinatorially. To do this consider the way to divide 2k people up into pairs. 4

5 To prove combinatorially we count the number of ways to split 2k people into k pairs. One way of counting this is to first consider if the k pairs were distinct in some way, (this is like (a from homework #1. The number of ways of doing this ( 2k is Where there are k 2 s. But then we must divide by k! since we don t 2, 2, 2,...2 want to have the pairs ordered. Doing this gives the formula on the left hand side of the equation. Another way of counting the same thing is to envision the 2k people in a room. The first person then chooses a partner. They have (2k 1 for their partner. Now those two people are gone and we go to the next remaining person, and he chooses a partner, and he chooses a partner from the remaining 2k people. this continues. Thus the number of ways of forming groups is (2k 1(2k...1 or the right hand side of the equation. Since both sides of the equation count the same thing then they must be equal. Problem 8. Let A B, and P (A = 0., P (B = 0.4. Calculate the following probabilities: P (A B, P (A B c, P (B A, P (B A c, P (B c A c. Show your calculations in detail. Solution First note that on one hand P (A B = P (A + P (B P (AB. On the other hand since A B then P (A B = P (B. Putting these two statements together we have that P (AB = P (A =.. P (A B = P (AB P (B =..4 =.7 P (A B c = P (ABc P (B c = 0 1 P (B = 0 (since A B, then ABc = empty set P (B A = P (AB P (A =.. = 1 P (B A c = BAc P (A c = P (B P (AB 1 P (A =.1.7 = 1 7. (note that P (B = P (BAc + P (BA P (B c A c = P (Bc A c = P (Bc = 1.4 =.6 = 6. (Note that since A B, then P (A c P (A c B c A c and thus A c B c = B c. Problem 9. Suppose that an ordinary deck of 2 cards (which contains 4 aces is randomly divided into 4 hands of 1 cards each. We are interested in determining p, the probability that each hand has an ace. Let E i be the event that the ith hand has exactly one ace, and E = E 1 E 2 E E 4 be the event that each hand has exactly one ace. (a Find P (E without using the multiplication rule simply count the number of ways 2 cards can be divided in 4 groups of 1 cards in each group, if each group contains exactly one ace. Please specify clearly whether you think of the hands as distinct

6 ( East, West, North, South hand or not (this will change the intermediate results, but not the final answer. ( 2 If we consider the players as distinct then there are a total of ways to 1, 1, 1, 1 hand out the cards. Count the number of such hands such that each player has eexactly one ace would be counted by 4! ways to hand out the aces, then ( 48 12, 12, 12, 12 ways to hand out the remaining non aces. Thus a probability of: 4! 48! 12!12!12!12! 2! 1!1!1!1! = (b Determine P (E by using the multiplication rule. If I think of the event E i that person i has exactly one ace Then I need P (E 1 E 2 E E 4. By the multiplication rule we have: P (E 1 E 2 E E 4 = P (E 4 E 1 E 2 E P (E E 2 E 1 P (E 2 E 1 P (E 1 8 ( 2 If we calculate P (E 1 then there are 4 out of all possible hands that 12 1 have exactly one ace. Similarly, if E 1 has once ace, then if person 2 has once ace, ( 6 then there are remaining choices for the ace and for the other 12 cards out of 12 ( 9 the possible hands that they could have if the 1 cards that person 1 has are 1 out of the deck. Similarly we get that: ( 24 ( 6 8 P (E 1 E 2 E E 4 = ( ( 26 ( 9 ( (c Show that your answers to (a and (b are the same. Simple computations reveal that these are the same. Problem 10. A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns to be an ace. This ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace. 6

7 Hint: Condition on whether or not the interchanged card is selected. As the hint suggests we condition on whether or not the interchanged card is selected. Define the events A = { An Ace is selected on the second draw} and let I = {The interchanged card is selected}. Now we can write: P (A = P (IP (A I + P (I C P (A I C Since there are 27 cards in the pile after the interchanged card is placed in them, P (I = 1/27 and P (I C = 26/27. Now if we select the interchanged card the probability that I get an Ace is 1 since the interchanged card is an ace, thus P (A I = 1. Finally, if I select another card other then the interchanged card it should be equally likely to be any of the other 1 cards in the deck, of which are aces so we have that P (A I C = /1. Thus we have, P (A = (1/27(1 + (26/27(/1 = Problem 11. Suppose that an insurance company classifies people into one of the following three classes: low risk, average risk, and high risk. Their records indicate that the probabilities that a low, average, and high risk persons will be involved in an accident over a 1-year period are 0.0, 0.1, and 0.0, respectively. Let 0% of the population be low-risk, 0% be average-risk, and 20% be high-risk. (a What proportion on people have accidents in a fixed year? Let A be the event that someone has an accident in a fixed year. Let L, V, and H be the events thats a certain person is Low, Average, or High risk respectively. In this question we are seeking to find P (A P (A = P (HP (A H+P (LP (A L+P (V P (A V = (.2(.+(.(.0+(.(.1 =.1 (b If Ms. Barry had no accidents in 200, what is the probability that she is a low-risk policy-holder? Average-risk? What we are seeking here is P (L A c and P (V A c P (L A c = P (LP (Ac L P (A c P (V A c = P (V P (Ac V P (A c = (.(.9 1 P (A = = (.(.8.8 Problem 12. A true-false question is to be posed to a husband and wife team on a quiz 7 =.

8 show. Both the husband and the wife will, independently, give the correct answer with probability p. Which of the following is a better strategy for this couple? (Give your calculations in detail for both strategies. Choose one of them and let that person answer the question; or In this situation we break the problem down be defining the following events. C is the event that they submit the correct answer. W is the event that the wife is chosen to answer. H is the probability that the husband is chosen to answer. (a P (C = P (CW + P (CH = P (W P (C W + P (HP (C H = p p1 2 = p. Here we are assuming that the husband and wife are chosen to answer the question equally likely. (b Have them both consider the question and then either give the common answer if they agree or, if they disagree, flip a coin to determine which answer to give? In this case we break up the problem in the following disjoint events, CC is the event that both answer correctly, II is the event that they both answer incorrectly. IC is the event that the husband answers incorrectly and the wife correctly, and CI is the event that the husband answered correctly and the wife incorrectly. Thus: P (C = P (CCP (C CC + P (IIP (C II + P (ICP (C IC + P (CIP (C CI. Now P (C CC = 1, P (C II = 0 and P (C IC = P (C CI = 1 2. Then substituting we get: P (C = p 2 (1 + (1 p 2 (0 + (1 p(p (1 p(p 1 2 = p Thus either strategy will be equally as good. 8

### 6.3 Conditional Probability and Independence

222 CHAPTER 6. PROBABILITY 6.3 Conditional Probability and Independence Conditional Probability Two cubical dice each have a triangle painted on one side, a circle painted on two sides and a square painted

More information

### Exam 1 Review Math 118 All Sections

Exam Review Math 8 All Sections This exam will cover sections.-.6 and 2.-2.3 of the textbook. No books, notes, calculators or other aids are allowed on this exam. There is no time limit. It will consist

More information

### Statistics 100A Homework 1 Solutions

Chapter 1 tatistics 100A Homework 1 olutions Ryan Rosario 1. (a) How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers? The first two

More information

### Jan 31 Homework Solutions Math 151, Winter 2012. Chapter 3 Problems (pages 102-110)

Jan 31 Homework Solutions Math 151, Winter 01 Chapter 3 Problems (pages 10-110) Problem 61 Genes relating to albinism are denoted by A and a. Only those people who receive the a gene from both parents

More information

### Module 6: Basic Counting

Module 6: Basic Counting Theme 1: Basic Counting Principle We start with two basic counting principles, namely, the sum rule and the multiplication rule. The Sum Rule: If there are n 1 different objects

More information

### Combinatorial Proofs

Combinatorial Proofs Two Counting Principles Some proofs concerning finite sets involve counting the number of elements of the sets, so we will look at the basics of counting. Addition Principle: If A

More information

### 16. Let S denote the set of positive integers 18. Thus S = {1, 2,..., 18}. How many subsets of S have sum greater than 85? (You may use symbols such

æ. Simplify 2 + 3 + 4. 2. A quart of liquid contains 0% alcohol, and another 3-quart bottle full of liquid contains 30% alcohol. They are mixed together. What is the percentage of alcohol in the mixture?

More information

### 10 k + pm pm. 10 n p q = 2n 5 n p 2 a 5 b q = p

Week 7 Summary Lecture 13 Suppose that p and q are integers with gcd(p, q) = 1 (so that the fraction p/q is in its lowest terms) and 0 < p < q (so that 0 < p/q < 1), and suppose that q is not divisible

More information

### A fairly quick tempo of solutions discussions can be kept during the arithmetic problems.

Distributivity and related number tricks Notes: No calculators are to be used Each group of exercises is preceded by a short discussion of the concepts involved and one or two examples to be worked out

More information

### 94 Counting Solutions for Chapter 3. Section 3.2

94 Counting 3.11 Solutions for Chapter 3 Section 3.2 1. Consider lists made from the letters T, H, E, O, R, Y, with repetition allowed. (a How many length-4 lists are there? Answer: 6 6 6 6 = 1296. (b

More information

### If a question asks you to find all or list all and you think there are none, write None.

If a question asks you to find all or list all and you think there are none, write None 1 Simplify 1/( 1 3 1 4 ) 2 The price of an item increases by 10% and then by another 10% What is the overall price

More information

### Hoover High School Math League. Counting and Probability

Hoover High School Math League Counting and Probability Problems. At a sandwich shop there are 2 kinds of bread, 5 kinds of cold cuts, 3 kinds of cheese, and 2 kinds of dressing. How many different sandwiches

More information

### 4. An isosceles triangle has two sides of length 10 and one of length 12. What is its area?

1 1 2 + 1 3 + 1 5 = 2 The sum of three numbers is 17 The first is 2 times the second The third is 5 more than the second What is the value of the largest of the three numbers? 3 A chemist has 100 cc of

More information

### Jan 17 Homework Solutions Math 151, Winter 2012. Chapter 2 Problems (pages 50-54)

Jan 17 Homework Solutions Math 11, Winter 01 Chapter Problems (pages 0- Problem In an experiment, a die is rolled continually until a 6 appears, at which point the experiment stops. What is the sample

More information

### Combinatorics 3 poker hands and Some general probability

Combinatorics 3 poker hands and Some general probability Play cards 13 ranks Heart 4 Suits Spade Diamond Club Total: 4X13=52 cards You pick one card from a shuffled deck. What is the probability that it

More information

### STUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS

STUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS The intermediate algebra skills illustrated here will be used extensively and regularly throughout the semester Thus, mastering these skills is an

More information

### Hawkes Learning Systems: College Algebra

Hawkes Learning Systems: College Algebra Section 1.2: The Arithmetic of Algebraic Expressions Objectives o Components and terminology of algebraic expressions. o The field properties and their use in algebra.

More information

### An approach to Calculus of Probabilities through real situations

MaMaEuSch Management Mathematics for European Schools http://www.mathematik.unikl.de/ mamaeusch An approach to Calculus of Probabilities through real situations Paula Lagares Barreiro Federico Perea Rojas-Marcos

More information

### Chapter 3: The basic concepts of probability

Chapter 3: The basic concepts of probability Experiment: a measurement process that produces quantifiable results (e.g. throwing two dice, dealing cards, at poker, measuring heights of people, recording

More information

### 4. How many integers between 2004 and 4002 are perfect squares?

5 is 0% of what number? What is the value of + 3 4 + 99 00? (alternating signs) 3 A frog is at the bottom of a well 0 feet deep It climbs up 3 feet every day, but slides back feet each night If it started

More information

### HMMT February Saturday 21 February Combinatorics

HMMT February 015 Saturday 1 February 015 1. Evan s analog clock displays the time 1:13; the number of seconds is not shown. After 10 seconds elapse, it is still 1:13. What is the expected number of seconds

More information

### Right Triangles and Quadrilaterals

CHATER. RIGHT TRIANGLE AND UADRILATERAL 18 1 5 11 Choose always the way that seems the best, however rough it may be; custom will soon render it easy and agreeable. ythagoras CHATER Right Triangles and

More information

### Warm-Up 16 Solutions. Peter S. Simon. Homework: January 26, 2005

Warm-Up 16 Solutions Peter S. Simon Homework: January 26, 2005 Problem 1 On a number line, point M is the midpoint of segment AB. The coordinates of A and M are 2 and 7, respectively. What is the coordinate

More information

### Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note 11

CS 70 Discrete Mathematics and Probability Theory Fall 2009 Satish Rao,David Tse Note Conditional Probability A pharmaceutical company is marketing a new test for a certain medical condition. According

More information

### Interlude: Practice Midterm 1

CONDITIONAL PROBABILITY AND INDEPENDENCE 38 Interlude: Practice Midterm 1 This practice exam covers the material from the first four chapters Give yourself 50 minutes to solve the four problems, which

More information

### Section P.9 Notes Page 1 P.9 Linear Inequalities and Absolute Value Inequalities

Section P.9 Notes Page P.9 Linear Inequalities and Absolute Value Inequalities Sometimes the answer to certain math problems is not just a single answer. Sometimes a range of answers might be the answer.

More information

### High School Math Contest

High School Math Contest University of South Carolina January 31st, 015 Problem 1. The figure below depicts a rectangle divided into two perfect squares and a smaller rectangle. If the dimensions of this

More information

### 1. By how much does 1 3 of 5 2 exceed 1 2 of 1 3? 2. What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3.

1 By how much does 1 3 of 5 exceed 1 of 1 3? What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3 A ticket fee was \$10, but then it was reduced The number of customers

More information

### Pythagorean Triples. Chapter 2. a 2 + b 2 = c 2

Chapter Pythagorean Triples The Pythagorean Theorem, that beloved formula of all high school geometry students, says that the sum of the squares of the sides of a right triangle equals the square of the

More information

### Matrices: 2.3 The Inverse of Matrices

September 4 Goals Define inverse of a matrix. Point out that not every matrix A has an inverse. Discuss uniqueness of inverse of a matrix A. Discuss methods of computing inverses, particularly by row operations.

More information

### Graduate Management Admission Test (GMAT) Quantitative Section

Graduate Management Admission Test (GMAT) Quantitative Section In the math section, you will have 75 minutes to answer 37 questions: A of these question are experimental and would not be counted toward

More information

### Discrete Mathematics: Homework 6 Due:

Discrete Mathematics: Homework 6 Due: 2011.05.20 1. (3%) How many bit strings are there of length six or less? We use the sum rule, adding the number of bit strings of each length to 6. If we include the

More information

### Assigning Probabilities

What is a Probability? Probabilities are numbers between 0 and 1 that indicate the likelihood of an event. Generally, the statement that the probability of hitting a target- that is being fired at- is

More information

### Math 421: Probability and Statistics I Note Set 2

Math 421: Probability and Statistics I Note Set 2 Marcus Pendergrass September 13, 2013 4 Discrete Probability Discrete probability is concerned with situations in which you can essentially list all the

More information

### Example: If we roll a dice and flip a coin, how many outcomes are possible?

12.5 Tree Diagrams Sample space- Sample point- Counting principle- Example: If we roll a dice and flip a coin, how many outcomes are possible? TREE DIAGRAM EXAMPLE: Use a tree diagram to show all the possible

More information

### a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.

Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given

More information

### The Advantage Testing Foundation Solutions

The Advantage Testing Foundation 013 Problem 1 The figure below shows two equilateral triangles each with area 1. The intersection of the two triangles is a regular hexagon. What is the area of the union

More information

### Grade 7 & 8 Math Circles. Mathematical Games - Solutions

Faculty of Mathematics Waterloo, Ontario N2L 3G1 Grade 7 & 8 Math Circles November 19/20/21, 2013 Mathematical Games - Solutions 1. Classic Nim and Variations (a) If the Nim Game started with 40 matchsticks,

More information

### Induction Problems. Tom Davis November 7, 2005

Induction Problems Tom Davis tomrdavis@earthlin.net http://www.geometer.org/mathcircles November 7, 2005 All of the following problems should be proved by mathematical induction. The problems are not necessarily

More information

### 6.2 Unions and Intersections

22 CHAPTER 6. PROBABILITY 6.2 Unions and Intersections The probability of a union of events Exercise 6.2- If you roll two dice, what is the probability of an even sum or a sum of 8 or more? Exercise 6.2-2

More information

### Introductory Problems

Introductory Problems Today we will solve problems that involve counting and probability. Below are problems which introduce some of the concepts we will discuss.. At one of George Washington s parties,

More information

### Statistics 100A Homework 2 Solutions

Statistics Homework Solutions Ryan Rosario Chapter 9. retail establishment accepts either the merican Express or the VIS credit card. total of percent of its customers carry an merican Express card, 6

More information

### Notes from February 11

Notes from February 11 Math 130 Course web site: www.courses.fas.harvard.edu/5811 Two lemmas Before proving the theorem which was stated at the end of class on February 8, we begin with two lemmas. The

More information

### Partial Fractions Examples

Partial Fractions Examples Partial fractions is the name given to a technique of integration that may be used to integrate any ratio of polynomials. A ratio of polynomials is called a rational function.

More information

### Solution, Assignment #6, CSE 191 Fall, 2014

Solution, Assignment #6, CSE 191 Fall, 2014 General Guidelines: This assignment will NOT be collected, nor graded. However, you should carefully complete it as if it were to be graded. There will be a

More information

### CHAPTER 8: ACUTE TRIANGLE TRIGONOMETRY

CHAPTER 8: ACUTE TRIANGLE TRIGONOMETRY Specific Expectations Addressed in the Chapter Explore the development of the sine law within acute triangles (e.g., use dynamic geometry software to determine that

More information

### Algebra Revision Sheet Questions 2 and 3 of Paper 1

Algebra Revision Sheet Questions and of Paper Simple Equations Step Get rid of brackets or fractions Step Take the x s to one side of the equals sign and the numbers to the other (remember to change the

More information

### Cognitive Abilities Test Practice Activities. Teacher Guide. Form 7. Quantitative Tests. Level. Cog

Cognitive Abilities Test Practice Activities Teacher Guide Form 7 Quantitative Tests Level 9 Cog Test 4: Number Analogies, Level 9 Part 1: Overview of Number Analogies An analogy draws parallels between

More information

### 2.1. Inductive Reasoning EXAMPLE A

CONDENSED LESSON 2.1 Inductive Reasoning In this lesson you will Learn how inductive reasoning is used in science and mathematics Use inductive reasoning to make conjectures about sequences of numbers

More information

### Math 3C Homework 3 Solutions

Math 3C Homework 3 s Ilhwan Jo and Akemi Kashiwada ilhwanjo@math.ucla.edu, akashiwada@ucla.edu Assignment: Section 2.3 Problems 2, 7, 8, 9,, 3, 5, 8, 2, 22, 29, 3, 32 2. You draw three cards from a standard

More information

### Alg2 Notes 7.4.notebook February 15, Two Way Tables

7 4 Two Way Tables Skills we've learned 1. Find the probability of rolling a number greater than 2 and then rolling a multiple of 3 when a number cube is rolled twice. 2. A drawer contains 8 blue socks,

More information

### Probability. A random sample is selected in such a way that every different sample of size n has an equal chance of selection.

1 3.1 Sample Spaces and Tree Diagrams Probability This section introduces terminology and some techniques which will eventually lead us to the basic concept of the probability of an event. The Rare Event

More information

### UNIT 2 MATRICES - I 2.0 INTRODUCTION. Structure

UNIT 2 MATRICES - I Matrices - I Structure 2.0 Introduction 2.1 Objectives 2.2 Matrices 2.3 Operation on Matrices 2.4 Invertible Matrices 2.5 Systems of Linear Equations 2.6 Answers to Check Your Progress

More information

### Clock Arithmetic and Modular Systems Clock Arithmetic The introduction to Chapter 4 described a mathematical system

CHAPTER Number Theory FIGURE FIGURE FIGURE Plus hours Plus hours Plus hours + = + = + = FIGURE. Clock Arithmetic and Modular Systems Clock Arithmetic The introduction to Chapter described a mathematical

More information

### SAT Math Facts & Formulas Review Quiz

Test your knowledge of SAT math facts, formulas, and vocabulary with the following quiz. Some questions are more challenging, just like a few of the questions that you ll encounter on the SAT; these questions

More information

### INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS

INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28

More information

### MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 1. SYSTEMS OF EQUATIONS AND MATRICES 1.1. Representation of a linear system. The general system of m equations in n unknowns can be written a 11 x 1 + a 12 x 2 +

More information

### Discrete probability and the laws of chance

Chapter 8 Discrete probability and the laws of chance 8.1 Introduction In this chapter we lay the groundwork for calculations and rules governing simple discrete probabilities. These steps will be essential

More information

### Arithmetic and Geometric Sequences

Arithmetic and Geometric Sequences Felix Lazebnik This collection of problems is for those who wish to learn about arithmetic and geometric sequences, or to those who wish to improve their understanding

More information

### P (A B) = P (AB)/P (B).

1 Lecture 8 Conditional Probability Define the conditional probability of A given B by P (A B) = P (AB) P (B. If we roll two dice in a row the probability that the sum is 9 is 1/9 as there are four combinations

More information

### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

More information

### Grade 7/8 Math Circles Greek Constructions - Solutions October 6/7, 2015

Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Grade 7/8 Math Circles Greek Constructions - Solutions October 6/7, 2015 Mathematics Without Numbers The

More information

### CS 173: Discrete Structures, Fall 2010 Homework 6 Solutions

CS 173: Discrete Structures, Fall 010 Homework 6 Solutions This homework was worth a total of 5 points. 1. Recursive definition [13 points] Give a simple closed-form definition for each of the following

More information

### Lecture 1 Introduction Properties of Probability Methods of Enumeration Asrat Temesgen Stockholm University

Lecture 1 Introduction Properties of Probability Methods of Enumeration Asrat Temesgen Stockholm University 1 Chapter 1 Probability 1.1 Basic Concepts In the study of statistics, we consider experiments

More information

### Year 9 set 1 Mathematics notes, to accompany the 9H book.

Part 1: Year 9 set 1 Mathematics notes, to accompany the 9H book. equations 1. (p.1), 1.6 (p. 44), 4.6 (p.196) sequences 3. (p.115) Pupils use the Elmwood Press Essential Maths book by David Raymer (9H

More information

### EXPONENTS. To the applicant: KEY WORDS AND CONVERTING WORDS TO EQUATIONS

To the applicant: The following information will help you review math that is included in the Paraprofessional written examination for the Conejo Valley Unified School District. The Education Code requires

More information

### Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00

18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list

More information

### A set is a Many that allows itself to be thought of as a One. (Georg Cantor)

Chapter 4 Set Theory A set is a Many that allows itself to be thought of as a One. (Georg Cantor) In the previous chapters, we have often encountered sets, for example, prime numbers form a set, domains

More information

### Strong Induction. Catalan Numbers.

.. The game starts with a stack of n coins. In each move, you divide one stack into two nonempty stacks. + + + + + + + + + + If the new stacks have height a and b, then you score ab points for the move.

More information

### M243. Fall Homework 2. Solutions.

M43. Fall 011. Homework. s. H.1 Given a cube ABCDA 1 B 1 C 1 D 1, with sides AA 1, BB 1, CC 1 and DD 1 being parallel (can think of them as vertical ). (i) Find the angle between diagonal AC 1 of a cube

More information

### Solution to Homework 2

Solution to Homework 2 Olena Bormashenko September 23, 2011 Section 1.4: 1(a)(b)(i)(k), 4, 5, 14; Section 1.5: 1(a)(b)(c)(d)(e)(n), 2(a)(c), 13, 16, 17, 18, 27 Section 1.4 1. Compute the following, if

More information

### 3(vi) B. Answer: False. 3(vii) B. Answer: True

Mathematics 0N1 Solutions 1 1. Write the following sets in list form. 1(i) The set of letters in the word banana. {a, b, n}. 1(ii) {x : x 2 + 3x 10 = 0}. 3(iv) C A. True 3(v) B = {e, e, f, c}. True 3(vi)

More information

### MODULAR ARITHMETIC KEITH CONRAD

MODULAR ARITHMETIC KEITH CONRAD. Introduction We will define the notion of congruent integers (with respect to a modulus) and develop some basic ideas of modular arithmetic. Applications of modular arithmetic

More information

### 0 ( x) 2 = ( x)( x) = (( 1)x)(( 1)x) = ((( 1)x))( 1))x = ((( 1)(x( 1)))x = ((( 1)( 1))x)x = (1x)x = xx = x 2.

SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5. Problem 8. Prove that if x and y are real numbers, then xy x + y. Proof. First we prove that if x is a real number, then x 0. The product of two positive

More information

### Arithmetic Operations. The real numbers have the following properties: In particular, putting a 1 in the Distributive Law, we get

Review of Algebra REVIEW OF ALGEBRA Review of Algebra Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus. Arithmetic Operations The real

More information

### Section 2.1. Tree Diagrams

Section 2.1 Tree Diagrams Example 2.1 Problem For the resistors of Example 1.16, we used A to denote the event that a randomly chosen resistor is within 50 Ω of the nominal value. This could mean acceptable.

More information

### Axiom A.1. Lines, planes and space are sets of points. Space contains all points.

73 Appendix A.1 Basic Notions We take the terms point, line, plane, and space as undefined. We also use the concept of a set and a subset, belongs to or is an element of a set. In a formal axiomatic approach

More information

### Pick s Theorem. Tom Davis Oct 27, 2003

Part I Examples Pick s Theorem Tom Davis tomrdavis@earthlink.net Oct 27, 2003 Pick s Theorem provides a method to calculate the area of simple polygons whose vertices lie on lattice points points with

More information

### Homework #1-16 (Write your answer as a percent rounded to 1 decimal place.)

Homework #1-16 (Write your answer as a percent rounded to 1 decimal place.) 1) A class consists of 19 girls and 15 boys. If 5 of the students are to be selected at random, determine the probability they

More information

### Solving Equations of Degree 1 (Linear Equations):

Solving Equations of Degree 1 (Linear Equations): Definition: The equation a b0 is called an equation of degree 1. Eample: Solve for : 0. 0, Or. Dividing both sides by : giving. Easy! Check the answer!

More information

### In a triangle with a right angle, there are 2 legs and the hypotenuse of a triangle.

PROBLEM STATEMENT In a triangle with a right angle, there are legs and the hypotenuse of a triangle. The hypotenuse of a triangle is the side of a right triangle that is opposite the 90 angle. The legs

More information

### Homework 6 (due November 4, 2009)

Homework 6 (due November 4, 2009 Problem 1. On average, how many independent games of poker are required until a preassigned player is dealt a straight? Here we define a straight to be cards of consecutive

More information

### MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS

MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS Systems of Equations and Matrices Representation of a linear system The general system of m equations in n unknowns can be written a x + a 2 x 2 + + a n x n b a

More information

### Chapter 6. 1. What is the probability that a card chosen from an ordinary deck of 52 cards is an ace? Ans: 4/52.

Chapter 6 1. What is the probability that a card chosen from an ordinary deck of 52 cards is an ace? 4/52. 2. What is the probability that a randomly selected integer chosen from the first 100 positive

More information

### Partial Fractions. (x 1)(x 2 + 1)

Partial Fractions Adding rational functions involves finding a common denominator, rewriting each fraction so that it has that denominator, then adding. For example, 3x x 1 3x(x 1) (x + 1)(x 1) + 1(x +

More information

### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Wednesday, January 29, :15 a.m. SAMPLE RESPONSE SET

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, January 29, 2014 9:15 a.m. SAMPLE RESPONSE SET Table of Contents Question 29................... 2 Question 30...................

More information

### Mathematical Procedures

CHAPTER 6 Mathematical Procedures 168 CHAPTER 6 Mathematical Procedures The multidisciplinary approach to medicine has incorporated a wide variety of mathematical procedures from the fields of physics,

More information

### ECE302 Spring 2006 HW1 Solutions January 16, 2006 1

ECE302 Spring 2006 HW1 Solutions January 16, 2006 1 Solutions to HW1 Note: These solutions were generated by R. D. Yates and D. J. Goodman, the authors of our textbook. I have added comments in italics

More information

### Mathematical Induction

Mathematical Induction Victor Adamchik Fall of 2005 Lecture 2 (out of three) Plan 1. Strong Induction 2. Faulty Inductions 3. Induction and the Least Element Principal Strong Induction Fibonacci Numbers

More information

### Mathematical Induction with MS

Mathematical Induction 2008-14 with MS 1a. [4 marks] Using the definition of a derivative as, show that the derivative of. 1b. [9 marks] Prove by induction that the derivative of is. 2a. [3 marks] Consider

More information

### 0.7 Quadratic Equations

0.7 Quadratic Equations 8 0.7 Quadratic Equations In Section 0..1, we reviewed how to solve basic non-linear equations by factoring. The astute reader should have noticed that all of the equations in that

More information

### Math 421, Homework #5 Solutions

Math 421, Homework #5 Solutions (1) (8.3.6) Suppose that E R n and C is a subset of E. (a) Prove that if E is closed, then C is relatively closed in E if and only if C is a closed set (as defined in Definition

More information

### 4.3. Addition and Multiplication Laws of Probability. Introduction. Prerequisites. Learning Outcomes. Learning Style

Addition and Multiplication Laws of Probability 4.3 Introduction When we require the probability of two events occurring simultaneously or the probability of one or the other or both of two events occurring

More information

### Oral Solutions. Harvard-MIT Math Tournament February 27, Problem O1 [25 points]

Oral Solutions Harvard-MIT Math Tournament February 7, 1999 Problem O1 [5 points] Start with an angle of 60 and bisect it, then bisect the lower 30 angle, then the upper 15 angle, and so on, always alternating

More information

### Pigeonhole Principle Solutions

Pigeonhole Principle Solutions 1. Show that if we take n + 1 numbers from the set {1, 2,..., 2n}, then some pair of numbers will have no factors in common. Solution: Note that consecutive numbers (such

More information

### Solutions for Practice problems on proofs

Solutions for Practice problems on proofs Definition: (even) An integer n Z is even if and only if n = 2m for some number m Z. Definition: (odd) An integer n Z is odd if and only if n = 2m + 1 for some

More information

### Answer Key for California State Standards: Algebra I

Algebra I: Symbolic reasoning and calculations with symbols are central in algebra. Through the study of algebra, a student develops an understanding of the symbolic language of mathematics and the sciences.

More information

### 3. Conditional probability & independence

3. Conditional probability & independence Conditional Probabilities Question: How should we modify P(E) if we learn that event F has occurred? Derivation: Suppose we repeat the experiment n times. Let

More information

### Polynomials and Vieta s Formulas

Polynomials and Vieta s Formulas Misha Lavrov ARML Practice 2/9/2014 Review problems 1 If a 0 = 0 and a n = 3a n 1 + 2, find a 100. 2 If b 0 = 0 and b n = n 2 b n 1, find b 100. Review problems 1 If a

More information

### STAB47S:2003 Midterm Name: Student Number: Tutorial Time: Tutor:

STAB47S:200 Midterm Name: Student Number: Tutorial Time: Tutor: Time: 2hours Aids: The exam is open book Students may use any notes, books and calculators in writing this exam Instructions: Show your reasoning

More information