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LEAVING CERTIFICATE. ARRANGEMENTS, SELECTIONS AND PROBABILITY 1 © MSS 2007 This presentation will emphasise: • The advantages of the counting method • The importance of learning by doing • Clear thinking over blind application of rules • The advantages of an experimental/investigative approaches • The suitability of PowerPoint as a teaching tool 2 © MSS 2007 Why Discrete Maths? • Short syllabus • Two Questions on Examination Paper (Possible third) • Least popular questions attempted by students: Q.6 - 68% Q.7- 57% • Important at Third Level Business - Science –Psychology – Medicine etc 3 © MSS 2007 Higher Syllabus 4 © MSS 2007 FUNDAMENTAL PRINCIPLE OF COUNTING Fundamental Principle of Counting: If one task can be accomplished in x different ways, and following this another task can be accomplished in y different ways, then the first task followed by the second can be accomplished in xy different ways. Example: A car manufacturer makes 3 models of cars, a mini, a saloon and an estate. These cars are all available in a choice of 5 colours; red, green, blue, black and orange. How many different cars are available ? 3 5 = 15 different versions 5 © MSS 2007 Example: A car manufacturer makes 3 models of cars, a mini, a saloon and an estate. These cars are all available in a choice of 5 colours; red, green, blue, black and orange. How many different cars are available ? A tree diagram can be used to illustrate this example 15 different versions 6 © MSS 2007 Example. In a restaurant you can have 6 types of muffin, 8 varieties of sandwich, and 5 drinks ( coffee, tea, coke, fanta, 7 up). Lunch is either a muffin and a hot drink or a sandwich and a cold drink. How many different choices of lunch are possible? 7 © MSS 2007 PERMUTATIONS (ARRANGEMENTS) In how many ways can we arrange the letters a,b,c,d,e taking the letters two at a time? Method 1. Write out all the arrangements (a,b) (a,c) (a,d) (a,e) (b,c) (b,d) (b,e) (c,d) (c,e) (d,e) Total = 20 (b,a) (c,a) (d,a) (e,a) (c,b) (d,b) (e,b) (d,c) (e,c) (e,d) Method 2. Box Method Possibilities 5 4 Total = 5x4 = 20 Method 3. Use the P notation This is called FACTORIAL 5 5! 5! 5.4.3.2.1 5 P2 5.4 20 (5 2)! 3! 3.2.1 The most common method : Box Method. 8 © MSS 2007 NOTES ON THE P NOTATION Formally n Pr (n 0)(n 1)(n 2).........[n (r 1)] n 6 Pr n(n 1)(n 2).........(n r 1) n! n Pr (n r )! 6! P2 65 (6 2)! 8 On calculator P4 8 7 6 5 1680 6 P 2 = Answer is 30 SPECIAL ONES 5 P5 5 4 3 2 1 120 This is called FACTORIAL 5 5! 4 P4 4! 4 3 2 1 24 9 © MSS 2007 n! n(n 1)(n 2)..........3 2 1 (n 3)! (n 3)(n 4)(n 5)...........3 2 1 Ex . 1 Simplify 3! (n 1)! n! 3.2.1.(n 1)(n)(n 1)...............3 2 1 n(n 1)(n 2).............3 2 1 6( n 1 ) 6n 6 10 © MSS 2007 Note 0! 1 Why? 5 ! 120 4 ! 24 3! 6 2! 2 1! 1 0! 1 120 5 24 24 4 6 632 2 2 1 1 1 1 11 © MSS 2007 Ex.1 Q.7(a) 2001 Paper 2 (i) How many different four letter arrangements can be made from the letters of the word FRIDAY if each letter is used no more than once in each arrangement ? (ii) How many of the above arrangements begin with the letter D and also end with a vowel ? 12 © MSS 2007 Ex.2 Q.6(a) 2000 Paper 2 (i) A Bank gives each of it's customers a four digit pin number which is formed from the digits 0 to 9. Examples are : 2475, 0865 and 3422. (i) How many different pin numbers can the bank use? (ii) If the bank decides not to use pin numbers that begin with 0, how many different numbers can it then use? The examples given are very important. NOTE : 0865 means that you can start with 0. NOTE : 3422 means that you can repeat the digits. 13 © MSS 2007 Ex. 3 Q.6(a) 1998 Paper 2 (i) In how many ways can the letters of the word IRELAND be arranged if each letter is used exactly once in each arrangment? (ii) In how many of these arrangments do the three vowels come together? (iii) In how many arrangments do the three vowels not come together? (i) Possibilities 7 6 5 4 3 2 1 7 6 5 4 3 2 1 7!ways 5040 14 © MSS 2007 ( ii ) POSSIBILITIES 5 4 3 VOWELS 2 1 Treat the vowels as one unit 5 ! . 3! 120.6 720 ways ( iii ) Number of ways in which the vowels do not come together The total arrangemen ts - The number of ways they are together 5040 720 4320 This last part is a negative statement. Total - Positive statement 15 © MSS 2007 ARRANGING THINGS IN A CIRCLE RULE : The number of arrangment s of n objects in a circle is (n - 1)! A circular permutation is not considered to be distinct from another unless one object is preceded or succeeded by a different object in both permutations Ex . In how many ways can we arrange six people around a circular table? Answer is (6 - 1)! 5! ways 16 © MSS 2007 B A A C B D D D C A C B B C D A Are these different arrangements? B A A C D D B D C C A C B B D A NO. ALL FOUR ARE REPRESENTATIONS OF THE SAME ARRANGEMENT 17 © MSS 2007 COMBINATIONS (SELECTIONS) In how many ways can we select the letters a,b,c,d,e two at a time. Method 1. Write out all the selections (a,b) (a,c) (a,d) (a,e) (b,c) (b,d) (b,e) (c,d) (c,e) (d,e) Total = 10 NOTE : (b,a) is the same selection as (a,b) Method 2. Use the C notation Formal Definition 5 2 n n! r !(n r )! r 5! 2!(5-2)! 5.4 1.2 5.4 10 2! 18 © MSS 2007 Origin of Formal Definition From n objects select r objects Arrange the chosen r objects From n objects select r objects r! From n objects select r objects r! From n objects select r objects We use this symbol n r From n objects arrange r objects pr n n! nr ! n! r ! nr ! n! r ! nr ! 19 © MSS 2007 NOTES ON THE C NOTATION 6 6.5.4.3.2.1 6.5 15 4 (4.3.2.1)( 2.1) 2.1 8 8.7.6.5 8.7.6.5 70 1.2.3.4 4! 4 n n(n1)(n2)(n3)...3.2.1 n(n1)(n2) 3.2.1.(n3)...3.2.1 3! 3 n n(n1)(n2)..........(nr1) r! r 20 © MSS 2007 TWO IMPORTANT RULES RULE 1 n 1 0 RULE 2 n n r n r Examples 5 1 0 5 5 Examples 3 2 20 1 0 20 20 18 2 21 © MSS 2007 Ex . 1 (i) In how many ways can a committee of four be choosen from nine people. (ii) If a certain person is always to be included, in how many ways can the committee be chosen. (i ) From 9 we are choosing 4. (ii ) 9 9.8.7.6 126 4 1.2.3.4 From 8 we are choosing 3, as one person is always included. 8 8.7.6 56 3 1.2.3 22 © MSS 2007 Ex .2 Q.7(a) 2000 Paper 2 The points a, b, c, d, e, and f lie on a cricle (i) If these points are used as as vertices, how many a different quadrilaterals can be formed? (ii) How many of these quadrilaterals will have ab as one side? (i ) Each 4-sided figure requires 4 points From 6 we are choosing 4. (ii) If ab is one of the sides we now have 4 points to choose 2. f b e c d 6 6 6.5 15 4 2 1.2 4 4.3 6 2 1.2 23 © MSS 2007 Ex .3 Q.6(a) 1999 Paper 2 (i) In how many ways can a group of five people be selected from four women and four men? (ii) In how many of these groups are there exactly three women? ( i ) We have a total of 8 people from whom we want to select 5. ( ii ) In order to have exactly 3 women, we can have the following. 3 women and 2 men 24 © MSS 2007 Ex .4 Q.6(a) 1996 Paper 2 (i) In how many ways can a group of five people be selected from ten people? (ii) How many groups can be selected if two particular people from the ten cannot be in the same group? ( i ) We have a total of 10 people from whom we want to select 5. 10 10.9.8.7.6 252 5 1.2.3.4.5 ( ii ) This is a negative statment " cannot" Cannot be included Total - When they are Included 8 8.7.6 252 56 3 1.2.3 252 - 56 196 25 © MSS 2007 Ex .5 Q.6(a) 2001 Paper 2 (i) How many different sets of three books or four books can be selected from six different books? (ii) How many of the above sets contain one particular book? ( i ) 3 BOOKS or 4 BOOKS 6 6 3 4 20 15 35 ways ( ii ) One book is always to be included we are choosing 2 books from 5 OR 3 books from 5. 5 5 10 10 20 ways 2 3 26 © MSS 2007 HEADS OR TAILS 27 © MSS 2007 Socks Problem Four distinguishable pairs of socks are in a drawer. If two socks are selected at random find the probability that a pair is chosen. 28 © MSS 2007 Fundamental Basis of Probability The number of things we want (favourable outcomes) P The total number of things available (total outcomes) Number of outcomes of interest P Number of possible outcomes Number of possible outcomes is also called the Sample Space 29 © MSS 2007 A bag contains five blue and six red discs. A disc is drawn at random from the bag. Find the probability that the disc is red. Favourable Outcomes : Total Outcomes : From 6 red choose 1 From 11 discs choose 1 6 Favourable 1 6 Probability = Total 11 11 1 30 © MSS 2007 A bag contains five blue and six red discs. Two discs are drawn at random from the bag. Find the probability that both discs are red. Favourable Outcomes : Total Outcomes : From 6 red choose 2 From 11 discs choose 2 6 Favourable 2 3 Probability = Total 11 11 2 31 © MSS 2007 A bag contains five blue and six red discs. Two discs are drawn at random from the bag. Find the probability that exactly one red disc is drawn. Find the probability that one red and one blue disc are drawn Favourable Outcomes : } Note: Both are asking the same question From 6 red choose 1 and from 5 blue choose 1 6 5 Favourable 1 1 6 Probability = Total 11 11 2 32 © MSS 2007 A bag contains five blue and six red discs. Three discs are drawn at random from the bag. Find the probability that all three discs are red. Favourable Outcomes : Total Outcomes : From 6 red choose 3 From 11 discs choose 3 6 Favourable 3 4 Probability = Total 11 33 3 33 © MSS 2007 A bag contains five blue and six red discs. Three discs are drawn at random from the bag. Find the probability that exactly two of the discs are red. Note: Since 3 discs are drawn the third one must be considered and is blue Favourable Outcomes : From 6 red choose 2 and from 5 blue choose 1 6 5 Favourable 2 1 5 Probability = Total 22 11 3 34 © MSS 2007 A bag contains five blue and six red discs. Three discs are drawn at random from the bag. Find the probability that at least one red disc is drawn. Note: This is the negative of drawing three blue discs. Unfavourable Outcomes : Total Outcomes : Favourable Outcomes : From 5 blue choose 3 From 11 discs choose 3 5 3 11 3 11 5 3 3 11 5 Favourable 3 3 31 Probability (at least one red) = 11 Total 33 3 35 © MSS 2007 Another method A bag contains five blue and six red discs. Three discs are drawn at random from the bag. Find the probability that at least one red disc is drawn. Note: We can also do this by counting all of the cases that will yield a favourable outcome. Favourable Outcomes : From 6 red choose 3 or From 6 red choose 2 and from 5 blue choose 1 or From 6 red choose 1 and from 5 blue choose 2 Probability (at least one red)= 6 20 3 6 5 75 2 1 6 5 60 1 2 Favourable 20 75 60 31 Total 33 11 3 36 © MSS 2007 RELATIVE FREQUENCY number of successful trials RELATVE FREQUENCY= total number of trials 37 © MSS 2007 Example 1. A drawing-pin can land “point up” or “point down” when dropped. Pat drops a drawing-pin 100 times and it lands “point up”35 times. Estimate the probability of the drawing pin landing “point up”. Solution: Probability = number of successful trials total number of trials 35 100 7 20 38 © MSS 2007 Example 2 A coin was tossed 1000 times and the number of heads were recorded. It was found that a head occurred 495 times. Estimate the probability of getting a head. Solution: Probability = number of successful trials total number of trials 495 1000 0 495 Note: As the number of trials increases the relative frequency tends to stabilise about a constant value. This constant value is called the probability of the event occurring. Therefore the probability of getting a head is 0.5 or 50% 39 © MSS 2007 Example 3 A five sided spinner, with sides marked A, B, C, D and E, was spun 10,000 times. The frequency of occurrence of each side is shown in the following frequency table. Side A B C D E Frequency 1988 2023 2009 1981 Calculate, correct to three decimal places, the relative frequency of 1999 occurrence of each side and hence estimate the probability of occurrence of each side. SOLUTION: Side A Frequency 1988 Relative 0.20 Frequency B 2023 0.202 C D E 2009 1981 1999 0.201 0.198 0.20 Probability of occurrence of each side = 20% 40 © MSS 2007 PROBABILITY 1. P(event ) Number of favourable outcomes Total number of outcomes Number of outcomes of interest Number of possible outcomes 2. 0 P(event ) 1, 0 1 3. P(event not occuring)1P(event occuring) This can be useful in certain questions but is not required 41 © MSS 2007 SINGLE EVENT PROBLEMS Ex . 1 A card is picked at random from a pack of 52 cards. What is the probability that it is (i) an ace (ii) a spade. Favourable 4 1 (i) P(ace) Total 52 13 (ii) P(spade) Favourable 13 1 Total 52 4 Ex . 2 A bag contains 8 blue marbles and 6 white marbles A marble is drawn at random from the bag. What is the the probability that the marble is white. P(white) Favourable 6 3 Total 14 7 42 © MSS 2007 MULTIPLE EVENTS PROBLEMS. F.P.C Ex.1 A die is thrown and a coin is tossed. What is the probability of getting a head and a 5? Method 1. Sample Space H 1 1,H 2 2,H 3 4 3,H 4,H 5 5,H 6 6,H T 1,T 2,T 3,T 5,T 6,T 4,T 43 © MSS 2007 MULTIPLE EVENT PROBLEMS. Ex. 1 An unbiased die is thrown twice. Find the probability of getting two equal scores or a total of 10 Method 1. Grid method F.P.C 1 1 1,1 2 2,1 3 3,1 4 5 4,1 5,1 6 6,1 2 3 1,2 1,3 2,2 2,3 3,2 3,3 4,2 5,2 4,3 5,3 6,2 6,3 4 1,4 2,4 3,4 4,4 5,4 6,4 5 1,5 2,5 3,5 4,5 5,5 6,5 6 1,6 2,6 3,6 4,6 5,6 6,6 We have 36 outcomes and 8 favourable outcomes 8 P ( equal or total 10) 36 44 © MSS 2007 1999 Q. 9 (a) An unbiased die is thrown twice. Find the probability of getting a total less than four. 1 1 1,1 2 2,1 3 4 5 3,1 4,1 5,1 6 6,1 2 3 1,2 1,3 2,2 2,3 3,2 4,2 5,2 3,3 4,3 5,3 6,2 6,3 4 1,4 2,4 3,4 4,4 5,4 6,4 5 1,5 2,5 3,5 4,5 5,5 6,5 6 1,6 2,6 3,6 4,6 5,6 6,6 P(event) Favourable 3 1 Total 36 12 45 © MSS 2007 REPLACEMENT AND NO REPLACEMENT Ex. 1 A card is picked at random from a pack of 52 cards. The card is replaced and again a card is then picked. What is the probability that the two cards picked are clubs. Note: Replacement does not occur unless explicitly stated 13 13 Favourable 1 1 13 13 1 52 52 Total 52 52 16 1 1 Ex. 2 Two cards are picked at random from a pack of 52 cards. What is the probability that the two cards picked are clubs. 13 Favourable 2 13 12 12 1 Total 52 52 51 204 17 2 46 © MSS 2007 ORDER DOES NOT MATTER Ex.1 Three cards are drawn at random, without replacement, from from a pack of 52. Find the probability that the three cards drawn are, the Jack of hearts, the Queen of diamonds and the King of clubs. No. of favourable outcomes = Total no. of outcomes = P( ) 52 3 33 3 3 1 52 22100 3 47 © MSS 2007 2005 Discrete maths questions 48 © MSS 2007 2005 1 2 Outcomes of Interest: (Favourable Outcomes) : Total Outcomes : 3 4 5 6 7 8 9 From remaining cards choose 3. Therefore we have 8 cards left to choose 3 = We have 9 cards to choose 3 = 8 Favourable 3 56 2 Probability = Total 9 84 3 3 8 56 3 9 84 3 49 © MSS 2007 1 2 Outcomes of Interest: (Favourable Outcomes) 3 4 From remaining cards choose 3 Probability = 5 6 7 8 9 Therefore we have 5 cards left to choose 3 = 5 10 3 Favourable 10 5 Total 84 42 50 © MSS 2007 1 7 + 8 2 + 9 3 24 4 > 5 1 6 + 2 7 + 8 3 + 4 9 + 5 + 6 21 6 + 8 + 9 23 > 1 + 2 + 3 + 4 + 5 + 7 22 5 + 8 + 9 22 < 1 + 2 + 3 + 4 + 6 + 7 23 Only Two Favourable Outcomes Probability = Favourable 2 1 Total 84 42 51 © MSS 2007 What the Chief Examiner had to say about Q6 -2005 • (c) Parts (i) and (ii) involved relatively standard counting techniques in a familiar context. Nonetheless they required some clarity of thought, • Part (iii) was poorly answered. Candidates seemed unable to identify favourable outcomes and many did not even attempt the question. It would appear that, other than with certain well-rehearsed question types, candidates lack the capacity to systematically list and/or systematically count outcomes satisfying particular criteria. Even the basic generic skills for this topic, such as exploring the situation by looking at examples of outcomes that satisfy or do not satisfy relevant criteria, were lacking. – 52 © MSS 2007 2005 Outcomes of Interest: Any 4 blue discs Favourable Outcomes : Total Outcomes : We have 16 discs to choose 4 = Probability = 5 5 4 Therefore we have 5 blue to choose 4 = 16 1820 4 Favourable 5 1 Total 1820 364 53 © MSS 2007 Outcomes of Interest: Favourable Outcomes : 5 blue choose 4 or 6 red choose 4: 5 4 OR 6 4 5 + 15 Probability = = 20 Favourable 20 1 Total 1820 91 54 © MSS 2007 Outcomes of Interest: 5 Blue choose 1 and 3 Green choose 1 and 6 Red choose 1 and 2 Yellow choose 1 Favourable Outcomes : 5 1 AND 3 1 AND 6 1 AND 2 1 5 x 3 x 6 x 2 Probability = = 180 Favourable 180 9 Total 1820 91 55 © MSS 2007 Outcomes of Interest: Favourable Outcomes : 5 blue choose 2 and other discs, 11 choose 2 5 2 AND 11 2 10 x 55 Probability = = 550 Favourable 550 55 Total 1820 182 56 © MSS 2007 What the Chief Examiner had to say about Q7 2005 • (b) Parts (i) and (ii) were very well answered by most candidates. Parts (iii) and (iv), however, caused much more difficulty than ought to be expected at this level. It is worth noting that the majority of candidates approached these parts using multiplicative laws for probability, which are not on the core course. Such candidates (unless they have developed their understanding substantially through studying the material on the option) tend to favour a blind application of rules over clear thinking, and generally suffer the consequences in all but the most basic of situations. In this case, they made errors related to ordering. As might be expected, the candidates who stuck to the basic principle for dealing with all situations involving equally likely outcomes (i.e., count the outcomes and put favourable over possible) fared much better than those attempting to use the more complex rules. 57 © MSS 2007 LC 2003 Paper 2 Question 6 (a) (i) (a) Eight people, including Kieran and Anne, are available to form a committee. Five people must be chosen for the committee (i) In how many ways can a committee be formed if both Kieran and Anne must be on the chosen ? How do you get pupils to decide between combinations and permutations ? 58 © MSS 2007 LC 2003 Paper 2 Question 6 (a) (i) (a) Eight people, including Kieran and Anne, are available to form a committee. Five people must be chosen for the committee (i) In how many ways can a committee be formed if both Kieran and Anne must be on the chosen ? How do you get pupils to decide between combinations and permutations ? 59 © MSS 2007 LC 2003 Paper 2 Question 6 (a) (i) (a) Eight people, including Kieran and Anne, are available to form a committee. Five people must be chosen for the committee (i) In how many ways can a committee be formed if both Kieran and Anne must be on the chosen ? How do you get pupils to decide between combinations and permutations ? If combinations is being used what are the two important questions ? 1) How many are available ? n 2) How many am I choosing/picking/selecting ? r n r 60 © MSS 2007 (i) In how many ways can a committee be formed if both Kieran and Anne must be on the chosen ? Kieran Anne Committee (of 5) ? ? ? 1) How many are available ? n 2) How many am I choosing/picking/selecting ? r 61 6 3 © MSS 2007 (ii) In how many ways can a committee be formed if neither Kieran nor Anne can be chosen Kieran Anne Committee (of 5) ? ? ? ? ? 1) How many are available ? n 2) How many am I choosing/picking/selecting ? r 62 6 5 © MSS 2007 LC 2003 Paper 2 Question 6 (c) Ten discs, each marked with a different whole number from 1 to 10, are placed in a box. Three of the discs are drawn at random (without replacement) from the box. (i) What is the probability that the disc with the number 7 is drawn ? (ii) What is the probability that the three numbers on the discs drawn are odd ? (iii) What is the probability that the product of the three numbers on the discs drawn is even ? (iv) What is the probability that the smallest number on the discs drawn is 4 ? 63 © MSS 2007 (i) What is the probability that the disc with the number 7 is drawn ? Method F.P.C The number of things we want (favourable outcomes) P= The total number of things available (total outcomes) 1 2 3 4 5 6 Favourable Outcomes : Total Outcomes : 7 8 From the 10 discs choose any 3 9 10 P A seven and two other discs A seven and from 9 choose any 2 discs From 10 choose any 3 discs 1 9 1 2 3 P 10 10 3 64 © MSS 2007 (ii) What is the probability that the three numbers on the discs are odd ? Method F.P.C The number of things we want (favourable outcomes) P= The total number of things available (total outcomes) Favourable Outcomes : Total Outcomes : 1 2 Three odd discs From the 10 discs choose any 3 3 P= 4 7 10 5 8 6 9 From the 5 odd discs choose any 3 From 10 choose any 3 discs 5 3 1 P 10 12 3 65 © MSS 2007 (iii) What is the probability that the product of the three numbers on the discs is even ? Note: If any or all the numbers are even then the product will be even. Unfavourable Outcomes : Three odd discs Total Outcomes : From the 10 discs choose any 3 Favourable Outcomes = Total – unfavourable = 1 2 3 4 5 6 7 8 9 10 10 5 3 3 10 5 3 3 11 P 10 12 3 66 © MSS 2007 (iv) What is the probability that the smallest number on the discs drawn is 4 ? P= The number of things we want (favourable outcomes) The total number of things available (total outcomes) Favourable Outcomes : Total Outcomes : 1 2 From the 10 discs choose any 3 3 4 5 6 7 8 9 10 A four and two other discs larger than 4 A four and from (5, 6, 7, 8, 9, 10) any 2 discs P= From 10 any 3 discs 1 6 1 2 1 P 8 10 3 67 © MSS 2007 LCH 2003 Paper 2 Question 7 (a) (i) (a) Five cars enter a car park. There are exactly five vacant spaces in the car park. (i) In how many different ways can the five cars park in the vacant spaces ? 5 4 3 2 1 5.4.3.2.1 = 120 ways 68 © MSS 2007 LCH 2003 Paper 2 Question 7 (a) (ii) Two of the cars leave the car park without parking. In how many different ways can the remaining three cars park in the five vacant spaces ? There are 5 different spaces this car could choose. However once a space is picked there is only 4 spaces left for the next car. Similarly, only 3 spaces are left for the final car. No. of ways 5 . 4 . 3 = 60 69 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) L and k are distinct parallel lines. a, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices? L . . . . b a K c d . . . x y z Two identified methods here. (1) Traditional C Notation (2) Inspection 70 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) L and k are distinct parallel lines. a, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Choose 1 point from line L and b c d a L two points from line K . . . . K . . . x y 4 3 1 2 OR Choose 1 point from line K and two points from line L z 4 3 2 1 Two identified methods here. (1) Traditional C Notation 4.3 + 6.3 = 12+18 = 30 triangles 71 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) Land k are distinct parallel lines. A, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Using just point a b c d a L . . . . 1 K . . . x y z Two identified methods here. (2) Inspection 72 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) Land k are distinct parallel lines. A, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Using just point a b c d a L . . . . 2 K . . . x y z Two identified methods here. (2) Inspection 73 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) Land k are distinct parallel lines. A, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Using just point a b c d a L 3 triangles from a . . . . K . . . 3 x y z 4 points (a,b,c,d) 4 x 3 = 12 triangles Two identified methods here. (2) Inspection 74 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) Land k are distinct parallel lines. A, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Using just point x b c d a L 1 . . . . K . . . 3 x y z Two identified methods here. (2) Inspection 75 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) Land k are distinct parallel lines. A, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Using just point x b c d a L 2 . . . . K . . . x y z Two identified methods here. (2) Inspection 76 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) Land k are distinct parallel lines. A, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Using just point x b c d a L 3 . . . . K . . . x y z Two identified methods here. (2) Inspection 77 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) Land k are distinct parallel lines. A, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Using just point x b c d a L 4 . . . . K . . . x y z Two identified methods here. (2) Inspection 78 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) Land k are distinct parallel lines. A, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Using just point x b c d a L 5 . . . . K . . . x y z Two identified methods here. (2) Inspection 79 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (i) Land k are distinct parallel lines. A, b,c and d are points on L such that |ab|=|bc|=|cd|=1cm. x,y and z are points on K such that |xy|=|yz|=1cm. (i)How many different triangles can be constructed using three of the named points as vertices. Using just point x b c d a L 6 triangles from x 6 3 points (x,y,z) . . . . K . . . x y z Two identified methods here. 3 x 6 = 18 triangles 12 red + 18 blue = 30 triangles (2) Inspection 80 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (ii) (ii) How many different quadrilaterals can be constructed ? Again two methods could be used here. (1) Traditional C Notation (2) Inspection a L . . . . K b c d . . . x y z 81 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (ii) (ii) How many different quadrilaterals can be constructed ? Again two methods could be used here. (1) Traditional C Notation L K Quadrilateral: . . . . b a c d . . . x y z Choosing 2 points from line L 4 2 Choosing 2 points from line K 3 2 6 x 3 = 18 different quadrilaterals 82 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (ii) (ii) How many different quadrilaterals can be constructed ? Again two methods could be used here. (2) Inspection L Using just [ab] . . . . b a c d 1 K . . . x y z 83 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (ii) (ii) How many different quadrilaterals can be constructed ? Again two methods could be used here. (2) Inspection L Using just [ab] . . . . b a c d 2 K . . . x y z 84 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (ii) (ii) How many different quadrilaterals can be constructed ? Again two methods could be used here. (2) Inspection L . . . . b a c d There are 6 possible line segments 3 K . . . x Using just [ab] There are just 3 quadrilaterals which can be formed using [ab] y z ([ab],[ac],[ad],[bc],[bd],[cd]) 6 x 3 = 18 different quadrilaterals 85 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (iii) (iii)how many different parallelograms can be constructed using four of the named points as vertices ? . . . . b a c d [ab] 1 . . . x y z 86 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (iii) (iii)how many different parallelograms can be constructed using four of the named points as vertices ? . . . . b a c d [ab] 2 . . . x y z 87 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (iii) (iii)how many different parallelograms can be constructed using four of the named points as vertices ? . . . . b a c d [ac] 3 . . . x y z 88 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (iii) (iii)how many different parallelograms can be constructed using four of the named points as vertices ? . . . . b a c d [bc] 4 . . . x y z 89 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (iii) (iii)how many different parallelograms can be constructed using four of the named points as vertices ? . . . . b a c d [bc] 5 . . . x y z 90 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (iii) (iii)how many different parallelograms can be constructed using four of the named points as vertices ? . . . . b a c d [bd] 6 . . . x y z 91 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (iii) (iii)how many different parallelograms can be constructed using four of the named points as vertices ? . . . . b a c d [cd] 7 . . . x y z 92 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (iii) (iii)how many different parallelograms can be constructed using four of the named points as vertices ? . . . . b a c d [cd] 8 . . . x y 8 parallelograms z 93 © MSS 2007 LCH 2003 Paper 2 Question 7 (b) (iv) (iv) If one quadrilateral is constructed at random, what is the probability that it is not a parallelogram ? Again, keep pupils on alert for total – the opposite of what we are looking for. Quadrilateral which is not a parallelogram = 18-8 P( Quadrilateral which is not a parallelogram) = or 18 8 5 18 9 P(Success) = 1- P(Failure) Quadrilateral which is not a parallelogram 8 P=1 18 Parallelogram 10 5 P= = 18 9 94 © MSS 2007 2004 Question 6 (c) 1 2 3 4 5 6 77 78 95 © MSS 2007 1 2 3 4 5 6 77 78 We want to pick 4 cards all different in colour 7 Number of Favourable Outcomes Probability = = Favourable Outcomes = Total Outcomes Total Number Outcomes = Number of Favourable Outcomes Probability = Total Number Outcomes 2 2 2 2 = 16 1 1 1 1 8 = 70 4 16 3 70 35 96 © MSS 2007 1 2 3 4 5 77 6 78 We want to pick 4 cards 2 odd and 2 even Colour does not matter here. O O E Favourable Outcomes = From 4 odd pick 2 and from 4 even pick 2 Number of Favourable Outcomes Probability = Total Number Outcomes E 4 4 2 2 36 18 70 35 = 36 97 © MSS 2007 1 Example 2 O 3 4 O 5 E O O O O E E E 6 77 78 E E 4 If we pick any the odd cards first there are ways of doing this. 2 From the cards that are left we have one option: ie. to pick the even blue and the even yellow 4 Number of Favourable Outcomes 6 3 Probability = 82 98 © MSS 2007 Total Number Outcomes 70 35 4 2004 Question 7 (i) The first runner can go in any one of 8 lanes. The second runner can now occupy any of the 7 remaining lanes. And so on until the eight runner has only one lane left 99 Possibilities = 8 7 6 5 4 3 2 1 = 8! © MSS 2007 2004 Question 7 (ii) The first runner can go in any one of 8 lanes. The second runner can now occupy any of the 7 remaining lanes. And so on until the fifth runner has 4 lanes left Possibilities =87654= 8 100 © MSS 2007 P5 8.7.6.5.4 Q.7 These students do not study Biology = 21 Total number of students = 56 P (Not Biology) = 21 56 101 © MSS 2007 This gives a total of 26 who study at least two subjects These 4 do not study Biology P= 4 26 102 © MSS 2007 Total of 28 study Physics Number of Favourable Outcomes Probability = Favourable = Total = Total Number Outcomes 28 Choose 2 28 2 56 Choose 2 562 Probability = 282 378 27 56 1540 110 2 103 © MSS 2007 104 © MSS 2007 2006 Question 6 (a) (i) n 11 3 165 2) How many am I choosing/picking/selecting ? r 1) How many are available ? (ii) Number of Favourable Outcomes Probability = Total Number Outcomes 5 3 11 3 105 10 2 165 33 © MSS 2007 2006 Question 6 (c) No. of favourble outcomes = No. of possible outcomes = 301 301 301 301 301 301 301 (i) P 30 1 1 1 1 1 1 1 1 1 1 1 1 1 No. of favourble outcomes 30 1 1 1 1 1 1 30 1 7 6 No. of possible outcomes 30 30 30 30 30 30 30 30 30 106 © MSS 2007 2006 Question 6 (c) No. of favourble outcomes = No. of possible outcomes = (ii) P 30 29 28 27 26 25 24 1 1 1 1 1 1 1 30 30 30 30 30 30 30 1 1 1 1 1 1 1 No. of favourble outcomes 30 29 28 27 26 25 24 2639 0 469 No. of possible outcomes 30 30 30 30 30 30 30 5625 107 © MSS 2007 2006 Question 6 (c) No. of unfavourble outcomes = No. of possible outcomes = No. of favourble outcomes = 30 29 28 27 26 25 24 1 1 1 1 1 1 1 307 307 30 29 28 27 26 25 24 1 1 1 1 1 1 1 No. of favourble outcomes 30 30 29 28 27 26 25 24 2986 P 0 53 7 No. of possible outcomes 5625 30 7 (iii) 0 5 or Probability = 1- P( all seven have the same birthday) 1 30 29 28 27 26 25 24 2639 1 0 5309 0 5 7 30 5625 108 © MSS 2007 2006 Question 7 (a) (i) (ii) 10 1 10 10 10 10 10 10 10 5 105 1 103 5 109 © MSS 2007 2006 Question 7 (b) 35 (i) 324632 5 5 30 (ii) 150 4 1 (iii) 5 30 4350 3 2 (iv) P(matching at least three numbers) = P( match three or match four or match five) 43501501 4501 0138649 0014 423632 423632 110 © MSS 2007 The Odds Real Lotto Source www.lotto.ie Odds on winning 111 © MSS 2007 Before the Lotto Draw From 45 numbers you choose 6 numbers Lotto Draw B A C E D X Bonus F 38 balls remain 6 numbers 112 © MSS 2007 Jackpot – Match 6 B A C E D X Bonus F 38 balls remain All 6 match N/A N/A 6 No. of favourable = 1 6 total possible 45 8145060 6 6 No. of favourble 6 1 P 45 total possible 8145060 6 113 © MSS 2007 Match 5 + Bonus B A C E D X Bonus F 38 balls remain From 6 you match 5 Yes N/A 6 1 favourable = 6 5 1 6 5 1 favourble 6 1 P total possible 45 8145060 1357510 6 114 © MSS 2007 Match 5 B A C E D X Bonus F 38 balls remain From 6 you match 5 N/A From 38 choose 1 38 1 38 6 5 = 6 6 38 favourable = 228 5 1 favourble 228 1 1 1 P 8145060 total possible 8145060 35723.94 35724 228 115 Trick © MSS 2007 Match 4 + Bonus B A C E D X Bonus F 38 balls remain From 6 you match 4 Yes 6 4 15 1 1 1 From 38 choose 1 38 1 38 favourable = 15 1 38 570 favourble 570 1 1 1 P 8145060 total possible 8145060 14289.57 14290 570 116 © MSS 2007 Match 4 B A C E D X Bonus F 38 balls remain From 6 you match 4 N/A From 38 choose 2 38 2 703 6 4 15 favourable = 15 703 10545 favourble 10545 1 1 1 P 8145060 total possible 8145060 772.409 772 10545 117 © MSS 2007 Match 3 + Bonus B A C E D X Bonus F 38 balls remain From 6 you match 3 6 3 20 Yes 1 1 1 From 38 choose 2 38 2 703 favourable = 20 1 703 14060 favourble 14060 1 1 1 P 8145060 total possible 8145060 579.30 579 14060 118 © MSS 2007 And finally back to the Socks Problem Four distinguishable pairs of socks are in a drawer. If two socks are selected at random find the probability that a pair is chosen. No of favourable outcomes = (There are 4 pairs in the drawer) Total Number of Outcomes = (From 8 objects choose 2) Number of Favourable Outcomes Probability = Total Number Outcomes 4 4 1 8 28 2 4 1 28 7 119 © MSS 2007 1 1. 2004 Ordinary Level Q6 (c) 2 2. 2003 Higher Level Q6 (c) 120 © MSS 2007 3 3. 2006 Ordinary Level Q6 (c) 4 4. 2003 Foundation Level Q6 (c) 121 © MSS 2007 5 5. 2005 Higher Level Q6 (c) 6 6. 2006 Foundation Level Q6 (c) 122 © MSS 2007 123 © MSS 2007