Exam 1 Review 5.1-7.2.

Slides:



Advertisements
Similar presentations
Beginning Probability
Advertisements

22C:19 Discrete Math Counting Fall 2011 Sukumar Ghosh.
Counting. Counting = Determining the number of elements of a finite set.
Main Menu Main Menu (Click on the topics below) Combinations Example Theorem Click on the picture.
Counting Principles The Fundamental Counting Principle: If one event can occur m ways and another can occur n ways, then the number of ways the events.
Lecture 14: Oct 28 Inclusion-Exclusion Principle.
Chapter 8 Counting Techniques PASCAL’S TRIANGLE AND THE BINOMIAL THEOREM.
Mutually Exclusive: P(not A) = 1- P(A) Complement Rule: P(A and B) = 0 P(A or B) = P(A) + P(B) - P(A and B) General Addition Rule: Conditional Probability:
Consecutive Numbers Algebra I.
Exam 1 Review 5.1, , 8.1, 8.2.
Review of 5.1, 5.3 and new Section 5.5: Generalized Permutations and Combinations.
MAT 2720 Discrete Mathematics Section 6.1 Basic Counting Principles
HAWKES LEARNING Students Count. Success Matters. Copyright © 2015 by Hawkes Learning/Quant Systems, Inc. All rights reserved. Section 7.3 Using Counting.
Methods of Counting Outcomes BUSA 2100, Section 4.1.
Ch. 5: Counting. 5.1: The Basics of Counting Intro Example #1: If we have a class of 6 math majors and 7 CS majors (with no double majors) – A) In how.
A polynomial is an algebraic expression that includes addition, subtraction, multiplication, and whole number exponents, such as: 4x 3 – 3x 2 + 7x + 5.
Statistics 1: Elementary Statistics Section 4-7. Probability Chapter 3 –Section 2: Fundamentals –Section 3: Addition Rule –Section 4: Multiplication Rule.
Warm Up 2/1/11 1.What is the probability of drawing three queens in a row without replacement? (Set up only) 2.How many 3 letter permutations can be made.
10/23/ Combinations. 10/23/ Combinations Remember that Permutations told us how many different ways we could choose r items from a group.
Counting Methods Review General Guidelines. Fundamental Counting Principle Each category outcome is independent of any other category outcome OR Items.
Simple Arrangements & Selections. Combinations & Permutations A permutation of n distinct objects is an arrangement, or ordering, of the n objects. An.
DIVISIBILITY RULES.
Conditions : Perfect cube #’s ( 1, 8, 27, 64, 125, … ) Perfect cube exponents ( 3, 6, 9, 12,15, … ) Separated by a plus OR minus sign Factoring – Sum and.
Sullivan Algebra and Trigonometry: Section 14.2 Objectives of this Section Solve Counting Problems Using the Multiplication Principle Solve Counting Problems.
Zero elements. One element.Two elements.Three elements. { a } { b } { c } { a, b } { b, c } { a, c } { a, b, c } 8 subsets.
Starter Questions 1. It costs £6.40 for 4 theatre tickets.
AP Statistics Wednesday, 20 November 2015 OBJECTIVE TSW review for the test covering probability.
Normal Approximations to Binomial Distributions.  For a binomial distribution:  n = the number of independent trials  p = the probability of success.
Chapter 4 Probability, Randomness, and Uncertainty.
Section 9-3 Probability. Probability of an Event if E is an event in a sample space, S, of equally likely outcomes, then the probability of the event.
Factoring Polynomials. 1.Check for GCF 2.Find the GCF of all terms 3.Divide each term by GCF 4.The GCF out front 5.Remainder in parentheses Greatest Common.
6.7 Permutations & Combinations. Factorial: 4! = 4*3*2*1 On calculator: math ==> PRB ==> 4 7! = 5040 Try 12!
4-3 Addition Rule This section presents the addition rule as a device for finding probabilities that can be expressed as P(A or B), the probability that.
MATH 2311 Section 2.1. Counting Techniques Combinatorics is the study of the number of ways a set of objects can be arranged, combined, or chosen; or.
Copyright © Peter Cappello 2011 Simple Arrangements & Selections.
Section 5.3 Independence and the Multiplication Rule.
Chance of winning Unit 6 Probability. Multiplication Property of Counting  If one event can occur in m ways and another event can occur in n ways, then.

Copyright © 2010, 2007, 2004 Pearson Education, Inc. Lecture Slides Elementary Statistics Eleventh Edition and the Triola Statistics Series by.
1-2 Simplifying Expressions. Free powerpoints at
Discrete Probability Distributions
MAT 2720 Discrete Mathematics
Lecture Slides Elementary Statistics Twelfth Edition
Counting Principles Ex. Eight pieces of paper are numbered 1 to 8 and placed in a box. One piece of paper is drawn from the box, its number is written.
L14: Permutations, Combinations and Some Review
1-2 Simplifying Expressions.
Consecutive Numbers Algebra I.
Warm Up Identify the slope and y-intercept of each equation. Then graph. 1. Y = -5X X + 5Y = X = Y = 12.
Ch. 8 – Sequences, Series, and Probability
CS 2210 Discrete Structures Counting
Lecture 08 Counting Profs. Koike and Yukita
Lial/Hungerford/Holcomb: Mathematics with Applications 10e
Mathematics for Computer Science MIT 6.042J/18.062J
A Brief Summary for Exam 2
Algebraic Expressions, Equations, and Symbols
Algebraic Expressions
Combinatorics.
Lesson 7.4 Solving by Multiplication
Counting Elements of Disjoint Sets: The Addition Rule
Kinder Campus Math Bee School Year.
Subtracting Real Numbers
Lial/Hungerford/Holcomb: Mathematics with Applications 10e Finite Mathematics with Applications 10e Copyright ©2011 Pearson Education, Inc. All right.
Inclusion-Exclusion Principle
Kinder Math Bee Practice Power point School Year.
Counting Elements of Disjoint Sets: The Addition Rule
Kinder Campus Math Bee Skill: Counting School Year.
Kinder Campus Math Bee School Year.
{ a } { a, b } { a, b, c } { a, c } { b } 8 subsets. { b, c } { c }
Factoring Polynomials
MATH 2311 Section 2.1.
Presentation transcript:

Exam 1 Review 5.1-7.2

Basic Counting Rules- ch. 5 SUM rule (for +) PRODUCT rule (for *) INCLUSION/EXCLUSION COMPLEMENT rule number= total – opposite Ex: number with at least 2 vowels = total – (number with 0 or 1 vowels)

5.1, 5.3, 5.5 Order matters, repetition allowed Multiplication Rule Ex: Social Security numbers Order matters, repetition NOT allowed Permutations: P(n,r)= n!/(n-r)! Ex: number of ways to pick 1st, 2nd, 3rd from 30 Order DOESN’T matter, repetition allowed section 5.5 (stars and bars; objects and dividers) n categories, n-1 dividers, r objects C(n-1+r, r) = C(n-1+r, n-1) Order DOESN’T matter, repetition NOT allowed Combinations: C(n,r)= n!/ [(n-r)!*r!] Ex: number of ways to pick a committee of 3 from 30

Binomial P(X=k)= nCk * p k q n-k  = np σ = (npq)

Basic probability rules P(E)=|E|/|S| 0<= P(E) <= 1 P(E ‘ ) = 1 – P(E) For Bayes Thm: do tree diagram Expected value: E(X)=  = for binomial  = np

Sample Ch. 5 and 6 problems Sample problems: 1. Passwords can be comprised of letters or digits. (uses sum, multiplication, complement rules) How many of them are: a) 4-6 characters b) 4-5 characters, with exactly 1 digit c) 4-5 characters, with exactly 2 digits d) 4-5 characters, with at least 2 digits  

…Probability 2. Which "type" of counting problems are these? (case 1,2,3,4, or 5?) a)An ice cream parlor has 28 different flavors, 8 different kinds of sauce, and 12 toppings. i)In how many different ways can a dish of three scoops of ice cream be made where each flavor can be used more than once and the order of the scoops does not matter? ii)How many different kinds of small sundaes are there if a small sundae contains one scoop of ice cream, a sauce, and a topping?

…Probability b) How many ways are there to choose a dozen donuts from 20 varieties: i)if there are no two donuts of the same variety? ii)if all donuts are of the same variety? iii)if there are no restrictions? iv)if there are at least two varieties? v)if there must be at least six blueberry-filled donuts? vi)if there can be no more than six blueberry-filled donuts?

…Probability c) A professor writes 20 multiple choice questions, each with possible answer a,b,c, or d, for a test. If the number of questions with a,b,c, and d as their answer is 8,3,4, and 5, respectively, how many different answer keys are possible, if the questions can be placed in any order? d) How many ways are there to assign 24 students to five faculty advisors? e) A witness to a hit and run accident tells the police that the license plate of the car in the accident, which contains three letters followed by three digits, starts with the letters AS and contains both the digits 1 and 2. How many different license plates can fit this description?

… f) There are 7 types of bagels at the store. i) How many different ways could you pick 12 of them and bring them to a meeting? ii) How many different ways could you choose to select bagels to each on 12 consecutive days? g)How many ways could we rearrange 13 books on a bookshelf: i)if all are different? ii)if 4 are identical chemistry books, 6 physics, and 3 math?

… h)How many ways could I there be to select 6 students out of 20 to receive A's? i)How many ways could I guess who in this class will get the best, second, and third score on the exam? j) How many ways can I select 3 women and 3 men from a Math Team (of 20 female mathletes and 25 male mathletes) to go to the National Math Tournament?

… 3. Number of solutions a) How many nonnegative solutions are there to x1 + x2 + x3 = 30, where x1>1, x2>4, x3>2?

… 7. how many bit strings of length 8: i) have at least 6 zero‘s? ii) start with 10 and end with 010?

… 8. a) How many one-to-one functions exist from a set with 3 elements to a set with 7 elements (section 5.1)?

Ch. 6: Probability Basic Def P(E’) Bayes

7.1- Recurrence relations example Prove: an=n! is a solution to an=n*an-1, a0=1 Find a solution to an=n*an-1, a0=1

7.2– Thm. 1 Thm. 1: Let c1, c2 be elements of the real numbers. Suppose r2-c1r –c2=0 has two distinct roots r1 and r2, Then the sequence {a n} is a solution of the recurrence relation an = ____________ iff an= __________ for n=0, 1, 2… where______

Thm. 2 Thm. 1: Let c1, c2 be elements of the real numbers. Suppose r2-c1r –c2=0 has ____root , Then the sequence {a n} is a solution of the recurrence relation an = ____________ iff an= __________ for n=0, 1, 2… where______

Summation formula Given: =

Solving 2nd degree LHRR-K For degree 2: the characteristic equation is r2-c1r –c2=0 (roots are used to find explicit formula) Basic Solution: an=α1r1n+ α2r2n where r1 and r2 are roots of the characteristic equation

Thm. 1 for two roots Theorem 1: Let c1, c2 be elements of the real numbers. Suppose r2-c1r –c2=0 has two distinct roots r1 and r2, Then the sequence {a n} is a solution of the recurrence relation an = c1an-1 + c2 an-2 iff an=α1r1n+ α2r2n for n=0, 1, 2… where α1 and α2 are constants.

Thm. 2 for one root Theorem 2: Let c1, c2 be elements of the real numbers. Suppose r2-c1r –c2=0 has only one root r0 , Then the sequence {a n} is a solution of the recurrence relation an = c1an-1 + c2 an-2 iff an=α1r0n+ α2 n r0n for n=0, 1, 2… where α1 and α2 are constants.

Ex: 6. an =8an-1 -16an-2 for n≥2; a0=2 and a1=20. Find characteristic equation Find solution

Ex: 6. an =8an-1 -16an-2 for n≥2; a0=2 and a1=20. Prove the solution you just found is a solution