MATH 1324 (Finite Mathematics or Business Math I) Lecture Notes Author / Copyright: Kevin Pinegar

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1 MATH 1324 Module 4 Notes: Sets, Counting and Probability 4.2 Basic Counting Techniques: Addition and Multiplication Principles What is probability? In layman s terms it is the act of assigning numerical value to a particular event. The event might be Winning a lottery, a raffle or even a multimillion dollar powerball jackpot (don t get excited) Guessing the correct answer on a multiple choice or true/false test question Selecting an Ace from a deck of cards, or getting a full house (poker hand) Rolling a sum of 7 or 11 on a pair of dice or flipping heads or tails on a coin Having a boy or girl (given that you are pregnant of course) That a light bulb or battery will last more than 1000 hours. That you will wait in line more than 15 minutes at a fast food restaurant. That an electronic device will have a defect within first year of use That a person contracts a disease or loses their before they turn 40 That a person will file an auto insurance claim during a time period These are just a few examples of the many applications of probability. How do you determine the probability of an event? To put it simply, you take the number of ways that your event can occur divided by the number of total possibilities. (ASSUMING each event is equally likely to occur) On events that are not equally likely we must take a different approach. Counting the number of elements in a set is a very important tool for probability. That is why these counting techniques are so important. Example: Suppose you toss two coins. What is the probability that both coins are heads? There are 4 possible outcomes {(h,h),(h,t),(t,h),(t,t)} and only one way to get the desired event {(h,h)}. Thus, the probability is 1/4 if we assume each output is equally likely. In other examples it might not be so easy to count the total number of possibilities (or the number of ways to get the event). Suppose you buy a lotto ticket and want to know the probability that your numbers will be selected. Could you actually write out all possible scenarios? I doubt you would want to try. What if you are dealt a 5-card hand of cards, and you want to know your chances of getting a full house (3 of one kind and two of another). Would you want to have to write down every 5-card hand that results in a full house, so you could count them? Again, these examples assume each event is equally likely to occur. Counting Techniques: Counting techniques give us a way to count how many elements are in a set, without having to actually count them. Caution: During the first couple of sections of this chapter, we will only be determining how many ways an event can occur. We will not be calculating probabilities until later. I say this because students commonly make the following error my probability test. When a question asks how many possible outcomes there are for an event, students sometimes give the probability for the event. And, when I ask for the probability, students will give the number of possible outcomes. Read the questions carefully. Counting Techniques Discussed I. Addition Principle II. Multiplication Principle III. Permutations IV. Combinations Before we can talk about how many elements are in a set, we must first recall some terminology about sets we discussed earlier. 1

2 Set Notations Definition i. U is the universal set (not to be confused with the union notation) The universal set contains all of the elements. ii. A B is read A union B. An element is in the set A B if the element lies in either set A or set B. Note: Even if an element occurs in both sets, you only list it once. iii. A B is read A intersect B. An element is in the set A B if the element lies (simultaneously) in both set A and set B. iv. A or A is read A complement or not A. An element is in the set A if the element lies outside of set A. Notice for each set A, B, C I simply listed elements from U that are not in those sets. v. The capital letter notation A defines the set A while n(a) is the notation used to give the number of elements in set A. vi. Represents the empty set. {a set with no elements, like, the number of pregnant men in class} vii. Lower case letters can represent elements of a set. a A and so on. Example Suppose we selected a card from a standard deck of cards. U would be the set containing all 52 cards. But f we selected a 3-card hand from the deck, U would be the set containing all 3 card-hands. Let U be the following common shapes. U={circle, triangle, square, pentagon, hexagon, heptagon, octagon, diamond, heart, star} Suppose sets A, B and C represents the following sets of common shapes: A={circle, triangle, square, hexagon} B={octagon, square, diamond, triangle} C={pentagon, heart, diamond} Then, A B ={circle, triangle, square, hexagon, octagon, diamond} A C ={circle, triangle, square, hexagon, pentagon, heart, diamond} Using the same sets above, A B={triangle, square} B C={diamond} Using the same sets above, A ={pentagon, heptagon, octagon, diamond, heart, star} B ={circle, pentagon, hexagon, heptagon, heart, star} C ={circle, triangle, square, hexagon, heptagon, octagon, star} Using the same sets above, A={circle, triangle, square, hexagon} But, n(a)=4 A B={triangle, square} But, n(a B)=2 A ={pentagon, heptagon, octagon, diamond, heart, star} But, n(a )=6 Notice that if I define the set A C, that set is empty. A C= and n(a C)=0 and 0 are related but not same meaning. For set A we can say circle A and star A. 2

3 Addition Principle For any two sets A and B, n(a B)=n(A)+n(B) n(a B) This is true because if you simply add the number in each set together, you add their intersection both times. So the subtraction corrects the duplication. If A and B are disjoint (their intersection is ), then n(a B)=n(A)+n(B) 0 = n(a)+n(b) Example from our sets above. 1) Find the number of elements in A B without counting them. n(a B)=n(A)+n(B) n(a B)=4+4-2=6 Check: A B ={circle, triangle, square, hexagon, octagon, diamond} (6 elements) 2) Find the number of elements in A C without counting them. Since A and C are disjoint, n(a C)=n(A)+n(C)= 4+3=7 Check: A C ={circle, triangle, square, hexagon, pentagon, heart, diamond} (7 elements) In other sets it will be too difficult to check the answers. That is why we need this principle. Playing Card Example Consider a standard deck of 52 cards. U= {the set of all 52 cards} Let A represent the set of all red cards in a standard deck. {hearts,diamonds} A={Ah,2h,3h,..., Ad,2d,3d,...} n(a)=26 Let B represent all face cards in a standard deck. (jacks, queens, kings} B= {Jh.Qh,Kh,Jd,Qd,Kd,Js,Qs,Ks,Jc,Qc,Kc} n(b)=12 Let C represent the set of all two s in the deck. C={2h,2d,2s,2c} n(c)=4 1) Find the n(a B) Answer: A B={Jh,Qh,Kh,Jd,Qd,Kd} n(a B)=6 2) Find the n(a B) Answer: n(a B)= =32 n(a B)=32 3) Find the n(b C) Answer: B C = ; n(b C)=0 4) Find the n(b C) Answer: n(b C)=12+4=16 n(b C)=16 Let s fill out a Venn Diagram for sets A and B. n(u)=52 (all the cards) n(a)=26 (all red cards) n(b)=12 (all face cards) n(a B)=6 (red AND face cards) n(a B )=20 (red but not face cards) n(a B)= 6 (not red but are face cards) n(a B )= 20 (not red nor face cards) Once this is completed you can answer other questions. How many are not red?(6+20=26) How many are not face cards? (20+20=40) U (52) A (26) B (12) A B 26-6 =20 A B (6) A B (52-32=20) A B 12-6 =6 3

4 More Addition Principle Examples 1. In a class of 35 students, 19 are married and 20 are blondes. Given that there are 7 students that are both married and blonde, answer the following questions. {See diagram} a) How many are married, but not blonde? a) Answer: 12 b) How many are blonde but not married? b) Answer: 13 c) How many are blonde or married? c) Answer: =32 d) How many are neither blonde nor married? d) Answer: 35-32=3 e) How many are not blonde? e) Answer: 12+3=15 Venn Diagram: You could have used the addition principle formula for part c: n(a B)= =32 Let A represent the married students and B represent the blonde students and then fill in the diagram. U=35 A(19) B(20) In a survey of 500 businesses it was found that 250 had copiers and 300 had fax machines. It was also determined that 100 businesses had both copiers and fax machines. {Draw your own diagram} a) How many had either a copier or a fax machine? a) Answer: 450 b) How many had neither a copier nor a fax machine? b) Answer: 50 c) How many had a copier, but no fax machine? c) Answer: 150 d) How many had a fax machine, but no copier? d) Answer: 200 {Just the fax ma am, just the fax!} e) How many had no fax machines? e) Answer: 200 U=500 A(250) B(300) Given: n(u)=110, n(a)=50, n(b)=75, n(a B)=105, find the following: a) n(a B) b) n(a B) c) n(a B ) d) n[(a B) ] e) n(b ) From the addition principle, n(a B)=n(A)+n(B) - n(a B), thus 105= n(a B) solve the equation to get n(a B)=20 b) 55, c) 5, d) 5, e)

5 Practice 1: Suppose your class has 50 students. There are 23 students majoring in psychology and 16 majoring in sociology. Seven students are seeking a double major in psychology and sociology. Fill out a two-set Venn diagram and answer the questions. a) How many are majoring in psychology alone? b) How many are majoring in sociology alone? c) How many are majoring in EITHER psychology or sociology? d) How many are majoring in NEITHER psychology nor sociology? e) How many are NOT majoring in psychology Answers: a) 23-7=16 b) 16-7=9 c) =32 d) 50-32=18 e) 9+18=27 Practice 2 (Don t Peek): A survey of 100 employees who exercise regularly found the following: 45 jog; 30 swim; 20 cycle; 6 jog and swim; 1 jog and cycle; 5 swim and cycle; 1 does all three. Fill out a 3-set Venn Diagram to answer the following. a) How many jog and swim but do not cycle? b) How many jog or swim? c) How many jog or swim or cycle? d) How many do only one of the three exercises? e) How many do none of the three exercises? Answers (see diagram below) a) 5 b) 6 c) 84 d) =74 e) =16 Notes: n(u)=100 n(j)=45 n(s)=30 n(c)=20 J 39 S 5 0 C

6 MULTIPLICATION PRINCIPLE 1. If 2 operations O1 and O2 are performed in order, with N1 possible outcomes for O1 and N2 possible outcomes for O2, then there are N1*N2 possible combined outcomes of the first operation followed by the second operation. 2. This principle can be generalized for any number of operations. If there are O1,O2,..On operations performed with N1,N2,.Nn possible outcomes for each respectively; then there are N1*N2*N3*..*Nn possible combined outcomes of the operations performed in a given order. Example 1: Suppose you are getting ready for work and notice you have only 2 shirts and 3 pants that are clean. How many different outfits can you put together with one shirt and one pant? O1 is selecting the shirt and N1 is equal to 2. O2 is selecting the pant and N2 is equal to 3. Answer: You have N1*N2=2*3=6 possible outfits. Example 1: Suppose you are getting ready for work and notice you have only 2 shirts, 3 pants that are clean with 4 pair of socks, 2 belts and 5 pair of shoes. (We won t ask about your underwear) Now, how many different outfits can you put together with one shirt and one pant? O1 is selecting the shirt and N1 is equal to 2. O2 is selecting the pant and N2 is equal to 3. O3 is selecting the socks and N2 is equal to 4. O4 is selecting the belt and N2 is equal to 2. O5 is selecting the shoes and N2 is equal to 5. Answer: You have 2*3*4*2*5=240 possible outfits!!! See you have enough outfits for 8 months. Of course, we are assuming the fashion police are not on the prowl. Note: There was a trait in the above problems that is not always true. There was no repetition. In other words, there was no chance of you selecting the same shirt twice, same pants twice etc The next example is an example that does allow repetition. Example: Suppose you take a 5 question multiple choice quiz and decide to randomly select each answer. a) How many different answer sheets can be submitted if each question has 4 responses? b) How many different answer sheets can be submitted if each question has 5 responses? Solution a) In this example could your answers be repeated? For example: {a,a,b,b,b}. Of course they could. Even the sheet {a,a,a,a,a} is a possible selection. This problem allows repetition and each of the five questions represent an operation. Each operation has 4 choices that can be repeated. Answer: 4*4*4*4*4=4 5 =1024 possible answer sheets. Solution b) This is just like part a except that each operation has five choices. Answer: 5*5*5*5*5=5 5 =3125 possible answer sheets. You can see that the possibilities build dramatically as we add more questions and more choices. If this were 10 questions the answers would be 4 10 for a and 5 10 for b. 6

7 A Note About Tree Diagrams A tree diagram is a visual representation of the choices. The first set of branches represent the choices for operation 1, the second set of branches represent the choices for operation 2 and so on. On the right is a tree diagram for the shirts and pants in the earlier example. You can see there are 6 paths to follow. {Shirt 1, Pant 1} {Shirt 1, Pant 2} {Shirt 1, Pant 3} {Shirt 2, Pant 1} {Shirt 2, Pant 3} {Shirt 2, Pant 3} Tree Diagram 2 choices followed by 3 choices P3 If we added the 2 belts we would get two branches coming out of each P branch (B1,B2) to give 24 outfits. You can imagine how big this diagram would get if we added 4 th and 5 th tiered branches for socks and shoes. On the earlier question regarding multiple choice questions, the branches would grow exponentially. Suppose we just assume 4 possible choices per question. Just for two questions there are 16 paths on the tree diagram. For three questions, we would have 64 paths, four questions yield 256 paths and for the fifth question we would have 1024 paths. Tree Diagram Two multiple choice questions With 4 choices each. (4x4=16) P3 a b c d a b c d a b c d a b c d 7

8 Multiplication Principle Examples 1. If a married couple is planning to have two children, how many possible outcomes are possible if the order of the births are taken into account? O1= result of first child {B or G} N1=2 possible outcomes O2= result of second child {B or G} N2=2 possible outcomes Answer: 2*2=4 possible outcomes Note: Having multiple children, tossing multiple coins and answering multiple T/F questions follow a similar pattern. 1a) How many outcomes are possible if THREE coins are tossed. Answer: 2*2*2=8 1b) How many outcomes are possible if FIVE coins are tossed. Answer: 2*2*2*2*2=32 1c) How many outcomes are possible if THREE T/F are answered. Answer: 2*2*2=8 1d) How many outcomes are possible if FIVE T/F are answered. Answer: 2*2*2*2*2=32 Binary Code. Here is an interesting example: We know computer technology is based on binary numbers (numbers consisting of only 0 s or 1 s). You don t need to know binary arithmetic to understand this example. A bit is a basic unit of information in computing and digital devices. A bit can have only one of two values (0 or 1). When someone discusses 8 bits, 128 bits or 256 bits; they are referring to a binary number consisting of only 0 s or 1 s. For example, some 8 bit numbers might be: { , , , } How many 8-bit numbers exist? You can easily determine the answer. There would be 8 operations with 2 choices per operation (repetition is allowed). (2)(2)(2)(2)(2)(2)(2)(2)=2 8 =256 possibilities. Computer experts use numbers that are very large in length as an aide in protecting our accounts, , passwords and so on. This is digital security or encryption. The topic is too broad to explain here, but let me give you an analogy. We know a car door might have a 5-digit code of the numbers 0-9. Suppose for a moment, that the code consisted of only numbers 0 or 1. Then there would be 2 5 =32 possible codes which is not very secures. A thief could try all 32 codes until he got one that works. BUT, suppose there 128 buttons on the car door. How many codes would be possible? Answer: = x possibilities. You have a much greater chance at buying the winning lottery ticket than breaking into the car during this decade. Computers can use 64-bit, 128-bit and 256-bit encryption. That s what keeps your credit card number secure over cyberspace. 2. You go to your closet and find that you only have 2 clean shirts, 3 clean pants, 3 decent pairs of shoes, and 4 belts. How many outfits can you put together assuming all other garments are predetermined? {shoes must match} O1= shirt,n1=2; O2= pants,n2=3; O3= shoes,n3=3; O4=belt,N4=4 Answer: 2*3*3*4=72 outfits 4. How many possible 10-digit phone numbers are possible under the following conditions? a) No restrictions (each number can be 0-9) =10 10 =10,000,000,00 b) The first number cannot be a 0 or = =8,000,000,000 Phone companies keep track so they do not run out of phone numbers to issue. Of course there are even less numbers that can actually be used because you can t begin a number with 911, 411, etc. 8

9 3. Suppose a debit card has a PIN consisting of 4 digits. a) How many possible codes can be selected if there are no restrictions on the selection? Solution: Each number can be {0,1,2,3,4,5,6,7,8,9} and you must select 4 numbers. Answer: 10*10*10*10=10000 options b) How many possible codes can be selected if digits cannot be repeated? Solution: There are 10 choices for first number, 9 choices for second number and so on. Answer: 10*9*8*7=5040 c) How many possible codes can be selected if adjacent digits must be different? Solution: The first number can be any of the 10 numbers; the second number cannot be the same as the first number (9 choices); the third number can t be same as second number but can be any other number (9 choices); the fourth number can t be same as third number but can be any other number (9 choices). Answer: 10*9*9*9=7290 Dice and Coins 4) If you toss 4 coins, how many ways can they land? 2*2*2*2=16 5) If you toss two 6-sided dice, how many ways can they land? 6*6=36 6) If you toss three 6-sided dice, how many ways can they land? 6*6*6=216 7) If you toss a 4-sided, 5-sided and 6-sided die, how many ways can they land? 4*5*6=120 8a) How many different license plates can be manufactured if they must consist of exactly 6 characters {letters or digits}? There are 36 characters (26 letters,10 digits). Solution: Each space can be any of 36 characters. Answer: 36*36*36*36*36*36=36 6 8b) What if we could have up to and including 6 characters? Solution: Ask how many one character plates can be assigned. {36} Ask how many two character plates can be assigned. {36*36=36 2 } Ask how many one character plates can be assigned. {36*36*36=36 3 } Up to how many six character plates can be assigned. {36 6 } Adding up all the possibilities gives: = 2,238,976,116 Do you think 2 billion is enough? If not, we can squeeze one more plate,36 0 =1. This represent the plate with nothing on it 8. How many ways can you answer a five question multiple choice test if the test consists of; a) 5 True/False questions. Answer: 2*2*2*2*2=2 5 =32 b) 5 multiple choice questions with 4 possible responses each. Answer: 4*4*4*4*4=4 5 =1024 c) 3 T-F questions followed by 2 of the multiple choice questions. Answer: 2*2*2*4*4=2 3 *4 2 =128 9) Pizza-Pizza: Suppose you can order a pizza that has 4 options for sauce, 10 options for toppings, 3 options for crust in 2 different sizes. How many different pizzas can be ordered if you will only choose one sauce, one topping, one crust and one size? Answer: 4*10*3*2=240 10) How many different 3 letter code words can be created from the first 9 letters of the alphabet under the given conditions. a) No restrictions on the code words. 9*9*9=729 b) No letter can be repeated. 9*8*7=504 Next we will have fun with permutations and combinations!!! 9