Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Systematic Counting, Arrangements & Combinations, Probability, Complementary events & Trees, 2-Dice questions, Financial Expectation, For Year 11 & 12 NSW HSC General Maths 2 NSW HSC General Maths 2 Syllabus Prelim course RELATIVE FREQUENCY AND PROBABILITY oThe multiplication principle for counting no. of possible outcomes in multi-stage experiment oProbability definition oExperimental probability and relative frequency oComparing theoretical and experimental probabilities oComplementary events Now assumed knowledge: Language of probability, ordering events, sample space HSC course omit We use Counting to help calculate Probability. We count the number of different ways to achieve a desired result, to help find the probability of that desired result, from all the different results possible. We need to count Consider.. 1.How many different two-digit numbers can be formed using the digits 1, 2, 3, 4 and 5? 2.How many of these numbers are Odd? 3.What is the probability an Odd number is formed when 2 digits are selected at random? Go to answers Systematic counting. How many different results are possible? A Restaurant Menu has 3 courses: Entrée, Main Course & Dessert For an Entrée there are 2 choices: Soup or Pasta. For a Main Course there are 3 choices: Chicken Fish or Steak. For Dessert there are 2 choices: Apple Pie or Gelato. How many different meals are possible? Use a Tree Diagram Or use Boxes: How Many Different Meals? A Short Cut Using “ Boxes”. We use one Box for each Step in the Process (places to fill): 1st place (Entrée) 2nd place (Main) 3rd place (Dessert) In each Box write how many choices there are, for that step: 3 2 x x 2 Multiply the choices in each Box for the number of different meals = 12 Systematic counting Questions. Multiplication Principle. Multiply the number of possibilities at each stage of an event. 1.A menu has 3 entrées, 4 main courses and 2 deserts. How many different meals are possible. 2.Travelling from Sydney to Melbourne via Albury. 3 ways from Sydney to Albury, & 2 ways from Albury to Melbourne. How many different trips are possible? 3.How many different outfits are possible. If there are 2 different Shirts and 4 different pairs of pants. 4.How many different number plates are possible? Each number plate is 3 letters and 3 digits, Each letter is A to Z, and each digit is zero to nine. Repeats of letters & digits are OK 5.How many 3 digit numbers can be made, using the digits 1 to 6. Go to answers Counting the Occurrences of a desired Result. It helps to recognise the process for getting the results of an experiment or event. This make the counting of favourable results easier. 2 important ways that produce different results of an event are called “Arrangements” & “Combinations”. These are also called, Ordered selections & Unordered selections. Arrangements (also called “Permutations”) Use Boxes to find the number of different Arrangements (ordered selections). Always check if the question allows “repetition” of digits in numbers etc. 4 digit PIN numbers use the digits 0 - 9 to fill a place (position). How many different PINs are possible? Check if digits can be repeated or not. Where Repetition is NOT OK. 10 9 8 7 Different Numbers = 5040 possible Where Repetition is OK 10 10 10 10 Different Numbers = 10000 possible Arrangements ( “Permutations”) Count: the Positions to be filled (boxes). Count: Number of choices (possibilities) for each box First Check if digits can be repeated or not. 3 digit numbers are formed using any 3 of the digits 1, 3, 5, 7, 9. How many different 3 digit numbers are possible possible? Where Repetition is NOT OK. 5 4 3 Different Numbers = 60 possible Where Repetition is OK 5 5 5 Different Numbers = 125 possible Arrangements = Permutations = Ordered selections Consider This.. Using the digits 1, 2, 3, 4 and 5 (no repetitions) How many different numbers can be formed which are Three-digit, Odd numbers starting with 5? 521 531 541 We can list them = 6 different 513 523 543 Or use the counting shortcut: Boxes Use 3 Boxes for 3 digit number (3 places to fill): Satisfy the rules Count the choices or ways to fill the boxes one by one 1st 1st Start=1 five 1 2nd 3 3rd 2 Leaves 3 choices 2nd = 6 different Odd=Only 2odds left Questions. Find the total number of different results. For these different arrangements (ordered selections). 1.There are five people in a race. How many different results are possible. 2.There are four different books to arrange on a shelf. How many different arrangements are possible? 3.There are five people in a race. How many different arrangements/results of 1st 2nd and 3rd. 4.How many different 3 digit numbers can be made from the digits 1 to 6. No repeats of digits allowed in a number. 5.For a 4 digit PIN number, using digits 0 to 9. How many PINS are possible if you (i) can (ii) can’t repeat a digit? Go to answers A different type of selection. “Combinations” There are 12 Cricketers to choose from: How many different.. ..batting orders of 11 players? ..teams of 11 players? The Order is Important.Arrangement Every selection of 11 with a different order is counted Order Not Important Combination Every selection of 11 with different contents is counted We have a Combination question when counting the number of different Teams, of a lesser number of people, which can be formed from the original bigger pool of people. Combinations are often called “Unordered selections”. This is because the internal order of the objects in the group does not matter. For Combinations you are looking for selected groups, with different contents. Combinations (Unordered selections) Choosing groups of 2 from the given 4 colours purple orange green yellow Colours to choose from Possible Combinations purple & orange Orange & green Purple & green Orange & yellow purple & yellow Green & yellow 6 different combinations Probability and Counting The 2 main ways that produce results of events are called “Arrangements” & “Combinations” We must recognise if results of an event are formed by an “Arrangement” or a “Combination”. We then use the related method, to count the possible results. This helps do Probability questions We will now compare the 2 ways & see how they are different. We will then see how counting is done for each type. Combinations or Arrangements : Know the Difference. 4 Students are used to form 3-person Debating teams: How many Different Teams of 3 are possible ? How many Different Speaking Orders: 1st, 2nd, 3rd are possible? Al, Bob, Cara & Dee: 4 students form 3-person Debating Teams: Different Speaking Orders Different Teams Possible = 4. Speaker 1st 2nd 3rd 4 3 2 = 24 ABC ABD ACD BCD. Arrangements & Combinations Arrangements: The Order matters. Combinations: Content matters, not Order. Different Speaking Orders Different Teams ABC ACB BAC BCA CAB CBA Al Bob & Cara ABD ADB BAD BDA DBA DAB Al Bob & Dee BCD BDC CBD CDB DBC DCB Bob Cara & Dee ACD ADC CAD CDA DCA DAC Al Cara & Dee There are 24 different speaking orders (6 different orders for each different team) There are 4 different teams (Unordered selections) The wording of the question must be carefully checked. This will convey if the question is a Combinations or Arrangements question. We have an Arrangement Question when counting different Line-ups. We have a Combination question when counting the number of Groups. Debating teams and speakers questions: For Arrangements (permutations): choose from 4, position in 3’s. For Teams (combinations): choose from 4, group in 3’s. Our calculator can help numerically: We can use 4P3 to find the number of different arrangements = 24 We can use 4C3 to find the number of different combinations(teams) = 4 12 Cricket players: How many different teams of 11? How many different batting orders? 12C 11 12P 11 Questions. Find the total number of different results. For these Combinations (Unordered selections). 1.The four winning numbers in a lottery are chosen from tiles numbered 1 to 20. If the selection order doesn’t matter, how many different results are possible. 2.A committee of three students is to be chosen from a class of 30. How many different committees are possible. 3.A meal with 2 vegetables is made from carrots, peas, potatoes and pumpkin; how many different combinations of 2 vegetables are possible. 4.You can choose 2 languages from five which are offered. How many different 2-language combinations are possible. 5.You select 3 books from twelve in a bag; how many different sets of 3 books are possible. Go to answers Experimental Probability: Relative Frequency By counting results we can estimate future probability. The distance travelled by car tyres is measured before they wear out. Use the results in the table to estimate the % that last from 50k to 100 k Kilometres 0 to 20 000 20 001 to 50 000 50 001 to 100 000 100 001 to 150 000 150 001 to …… Total Tyres Tyres 4 9 15 22 30 = 80 What is the relative frequency for tyres 50 001 to 100 000 kms? Relative Frequency = 15 80 = 0.1875 Estimate that 18.75% of tyres last between 50 000 & 100 000 kms Single choice probability. P = Number of ways to get what’s wanted Total number of results possible We need to count 1.Choosing 1 number from the digits one to eight. (i) Chance it is even; (ii) a multiple of three; 2.Rolling a die once. (i) Chance of getting greater than a 2; (ii) less than or equal to 4; (iii) a 2 or 3. 3.The spinner has 6 equal-sized sections. 2 are red, one green, 3 blue. Spin once. (i) Probability of red; (ii) not red. Go to answers Probability: Use Arrangements for counting. Then find the probability You get one attempt at guessing a forgotten PIN number (uses 0 to 9). If the PIN has 4 digits, what is the Probability you get it right 1st go. Need the number of different Pins possible 1st 2nd 3rd 4th position Ways to fill each place using 0 - 9 If you Can’t Repeat a digit once used If you can Repeat digits you have 1 chance out of 5040 you have 1 chance out of 10000 P = 1/5040 P = 1/10000 Questions. Find the probability 1.In a race of 5 people. What is the chance you win (assume equal ability) 2.In making a 3 digit number from digits 1-6. What is the chance of making the number 123 randomly. 3.In the lottery, 4 nos. are chosen (order doesn’t matter), from numbers 1-20; what is the probability you win if you have one ticket. 4.For the student committee, 3 students chosen from 30 randomly; what is the chance you are chosen with your two best friends. 5.For the vegetables carrots, peas, potatoes & pumpkin You pick 2 randomly; what is the chance you get peas and carrots. Go to answers Complementary Events are Opposites You find Probability of one from the other’s Probability There is a connection between events that are opposites. We can get the probability of an Opposite from its partner 1.A gun hits a target. The gun misses the target 2.A student is late on any day. The student is not late on any day Complementary events 3.A man is colorblind A man is not colourblind If something has a 10% chance of happening, then it’s 90% that it won’t happen. Probability of Complement of A = 1 – probability of A Complementary Events are Opposites: You find one’s Probability from the other’s Probability Lights at Queen street You get a Green Light 4 days out of 5 at Queen St, thus …. The Prob of NOT getting a Green is 1 out of 5 at Queen st. P(green) = 4/5….. P(not green) = 1/5… at Queen St You also have lights at King St. At King St you get a Green 3 days out of 5….. P = 3/5 and hence you get a Not Green light 2 days out of 5…..P = 2/5 Questions. Complementary events What is the chance? 1.An arrow is fired at a target. The Chance it hits the target is ¼. Find the chance it misses. 2.The Chance a student is late on any day is 1/8. Find the chance she is not late. 3.The Chance any man is colour-blind = 10%. Find the chance he is not colour-blind. Go to answers 2+ Step Probability To Draw a tree to see the possible results: Draw a tree diagram to show all the possible meals you can have Drink? Cookie? Possible results Wm & cc Wm & om Cm & cc Cm & om A Tree Diagram shows the choices at each step. Follow branches->results 1st step choices, followed by 2nd step choices, etc. A Restaurant Menu has 3 courses: so the tree has 3 steps across. There are 2 Entrées, so step 1 shows these, the soup and pasta. There are 2 Mains, so step 2 shows these, chicken, fish & steak. There are 2 Desserts, so step 3 shows these, apple pie & gelato Dessert Main Entree c f s ap s c ap g scg ap s f ap g sfg ap s g ap c p s s ap ssg p c ap g pcg ap p f ap g pfg f s Possible meals ap p s ap psg g Tossing a coin twice. Probability tree How do you calculate the overall probabilities? You multiply probabilities along the branches. You add probabilities down columns Probability Trees: Where 2 or more steps follow each other. We do a Tree & show the Chances on the Branches. Then multiply along the Branches to the result(s) you want. What is the chance of 2 green lights , if at Queen St, the chance of a Green is 4/5, and at King St the chance of a Green is 3/5? 3/5 4/5 G 2/5 3/5 1/5 QUEEN ST G R G R 2/5 R KING ST GG 12/25 2 greens Chance is 12/25. GR 8/25 RG 3/25 RR 2/25 Using R for “not green Not replacement of 1st choices There are 5 blue fish and 4 red fish in a pond. All fish have an equal chance of being caught The probability of a blue fish being caught 1st is 5/9 If This fish is not replaced after it’s caught Then, the probability of catching a 2nd blue fish is 4/8 4 blue fish of 8 fish left, after 1st not replaced TREE DIAGRAM FOR PROBABILITIES 5 blue fish and 4 red fish in a pond Two fish are taken without replacement What does “without replacement “ mean? B BB 4/8 Pr(2 blue fish) = 54 9 8 BB Pr(2 red fish) = 43 9 8 RR B R 4/8 5/9 BR BR+RB B 4/9 R 5/8 3/8 1stFish RB Pr( one red & one blue) = 5 4 4 5 9 8 9 8 RR Pr(at least 1 blue) = 1 - P(2 red) R 2ndFish BB results = 1 - 12 / 72 = 60/72 Probability tree exploration. Geogebratube link below screenshot http://www.geogebratube.org/student/m55049 Questions. Do a Probability Tree and answer the questions Find the Probability for.. 1.Toss of a coin three times. Chance of two heads and 1 tail? 2.A bag has 2 red & 3 blue marbles. Select 2, with replacement. Chance they are all the same colour? 3.There are 3 men and 4 women. Select 2, without replacement. Chance of at least 1 man? Harder questions involve: all are..; none are..; at least one; etc. Go to answers Tree diagram blank Make a question for this Tree: about firing shots at a target, etc Tree diagram blank Make a question for this Tree: about choosing coloured balls from a bag, etc BETTING TO WIN The Sum game with 2 dice. Which bet would you take? Bet 1 or Bet 2? You roll the 2 dice, 36 times & pay $1 to play each roll. Bet 1: You are paid $12 each time you get a sum of 2. Bet 2: you are paid $ 6 each time you get a sum of 7. Use the table showing the possible results of rolling 2 dice . See next slide 2 DICE TABLE. Roll a red & a black dice together The table shows what you can get. Black Die You roll the 2 dice, 36 times & pay $1 to play each roll: pay total = $36 Bet 1: a sum of 2 is expected once from 36 rolls. P = 1/36 Bet 2: a sum of 7 is expected 6 times from 36 rolls. P = 6/36 Bet 1: Money Expected = Get $12 once in 36 rolls Bet 2: Money Expected = Get $6 six times in 36 rolls Bet 2 is better 2 Dice Difference Game. Roll a red & a black dice together The table shows the Differences (subtract) for the 2 dice . Red Die Black Die Diff. 1 2 1 0 1 2 1 0 3 2 1 4 3 2 5 4 3 6 5 4 3 4 5 6 2 3 4 5 1 2 3 4 0 1 2 3 1 0 1 2 2 1 0 1 3 2 1 0 Is this a fair game? You win if the Dice-Difference= 0,1 or 2. Opponent wins if Dice-Difference= 3,4 or 5 Not a Fair Game Diff = 0,1 or 2 24 out of 36 results Diff = 3,4 or 5 12 out of 36 results Expectation • When you do an action….. different results are possible • If you repeat the action … How many times would one special result be expected? • To find the expected number, What do you need to know? The number of repeats & the probability of the result Expected occurrences of special result = Prob. x No. of repeats Expectation and Probability If a game is played many times, we can predict the number of wins, if we know the probability of a win in a single game. When we roll a die, the Chance of a 6 is 1 out of 6. How many 6’s would you expect if you roll the die 60 times? Solution: Number of 6’s expected = Prob. x No. of rolls = 1 x 60 6 = 10 times (a 6 is expected) Expected Value The expected value of an event gives the theoretical mean of the outcomes if the experiment is conducted a very large number of times. Forty people are given Drug X, thirty people are give Drug Y and twenty people are given Drug Z for the treatment of a disease. The success rates are shown in the table below. Drug Success Rate X 80% Y 70% Z 50% How many patients can you expect to be treated successfully? Expected Value Forty people are given Drug X, thirty people are give Drug Y and twenty people are given Drug Z for the treatment of a disease. The success rates are shown in the table below. Drug Success Rate X 80% Y 70% Z 50% How many patients can you expect to be treated successfully? Expected Value Data collected in a certain town suggests that the probabilities of there being 0, 1, 2, 3, 4 or 5 car thefts in one day are 0.10, 0.35, 0.30, 0.08, 0.15 and 0.02, respectively. What is the expected number of car thefts on any one particular day? If it makes it easier draw a table! Financial expectation • An event has different possible results. • Each result has different probability and different $ gains or $ losses. • Financial Expectation = the average $ result to expect, in the long run. Financial expectation Financial Expectation = the average $ result to expect, in the long run. Businesses use Financial expectation to plan. It’s a useful tool to find if the risk in an investment is justified. Gamblers or Casino operators use this to calculate the losses, or winnings, they can expect from a particular game where money is involved. Financial Expectation Would you play this Game?? The Chance of a Win is 40%, while the Chance of a Loss is 60%, When you Lose you pay $1, but when you Win you get $2. Solution: If you played 100 games you should win 40 games win $80 & you’d lose 60 games lose $60. Overall your profit is $20 in 100 games, Average profit of $0.20 per game is expected Financial Expectation (E): Formula Short Cut Overall, do you expect to WIN or LOSE ; & How Much? Use the rule E = P1 x $1 + P2 x $2 + ... The Chance for 1st Result Note any LOSS Results are Negative E = 40% x $2 + 60% x (- $1) …$1 is negative since it’s a Loss. E = $0.20 per game. You expect to WIN on the average $0.20 per game What is the Financial Expectation for this game? Result Throw dice Dice = 6 Dice not 6 Probability 1/6 Win Lose $10 5/6 $2 E = P1 x $1 + P2 x $2 Financial Expectation = 1/6 x $10 + 5/6 x -$2 = $0 This means it’s fair: don’t expect to win or lose in long run. Financial Expectation Financial Expectation Financial Expectation If the result is negative it is a loss. Financial Expectation Text Exercise 7J – 4, 5 7 7K – 2 d , 5 a, 8 a, 9 a, c, 10, 11, 12 a, 13, 15 ANSWERS Introductory question slide4 BACK TO SLIDE 3 1= 20 numbers; 2= 12 odds; 3=12/20 or 3/5 BACK TO SLIDE 6 Systematic counting slide 7.. 1=24 meals; 2=6 trips; 3=8 outfits; 4=17576000 number plates; 5=120 numbers Arrangements slide 11.. 1=120 race results; 2=24 book arrangements; 3=60 1st 3 placings; 4=120 numbers; 5(i)=10000 Pins (ii) = 5040 Pins Combinations slide 18.. 1=4845 results; 2=1015 committees; 3=6 2-veg combo’s; 4= 10 2-language pairs; 5= 220 3-book sets BACK TO SLIDE 11 BACK TO SLIDE 18 ANSWERS continued Single choice prob. Slide 20.. 1=(i)1/2; (ii)¼; 2= (i)2/3; (ii)2/3; (iii)1/3; 3=(i)1/3 (ii)2/3 Find prob. Slide 22.. 1. P(win)=1/5; 2.P(1,2,3)=1/120; 3. P(win) = 1/4845; 4.P(you+friends)=1/4060; 5.P(peas+carrots)=1/6 BACK TO SLIDE 20 BACK TO SLIDE 22 Complementary events. Slide 25.. 1.P(misses) = ¾; 2.P(not-late)=7/8; 3.P(not colourblind)=90% BACK TO SLIDE 25 BACK TO SLIDE 33 Prob trees & questions. Slide 33.. 1.P(2H +1T) = 3/8; 2.P(2-same colour)=13/25; 3. P(at least 1M)=1-12/42 = 30/42