MB Qld- 8 Chapter Exercise A Informal description of chance a Double digits from 0 to 0 Probable b Only girls out of 30 Unlikely c No green marbles Impossible d Half the numbers are odd Fifty-fifty 2 a No negative numbers Impossible b All positive numbers Certain c Half the numbers are even Even chance d Half the cards are red Even chance e Most of the cards are numbered Probable f Only four of the cards are an ace Unlikely g No 30c pieces Impossible h Half the marbles are blue Even chance 3 Answers will vary. Possible answers include: a Certain: rolling a number with a b Probable: rolling a number less than with a c Even chance: rolling an even number with a d Unlikely: rolling a or 2 with a e Impossible: rolling a with a More school term days than school holidays. More likely to be during a school term. a More likely b Equally likely c Less likely d More likely e Less likely 6 There are three even numbers, two numbers less than three and four numbers greater than 2. Order is: Rolling a 6 Rolling a number less than 3 Rolling an even number Rolling a number greater than 2 Winning the raffle with tickets out of 30 Selecting a court card Drawing a green marble A number less than 3 on the die Landing a Head on a coin 8 Australia 9 Carl Bailey faster time and winner of the semifinal. Better past performance. 0 Five vowels out of 26 letters. The event would be unlikely. The answer is B. The option with the least total possibilities is selecting a diamond from a pack of cards, that is, one in four. The answer is B. 2 32 of the past 0 years is a favourable outcome. The answer is D. 3 Most of the 000 globes burned for more than 00 hours. It would be probable that the globe will burn for more than 00 hours. Well below half of the new cars sold had major mechanical problems. It would be unlikely for the car to have major mechanical problems. Half the people would vote for the government, therefore there would be a fifty-fifty chance they would vote for the government. Exercise B Single event probability S = {Heads, Tails}. One favourable outcome. 2 a S = {6} One favourable outcome. b S = {0,, 2} Three favourable outcomes. c S = {a, e, i, o, u} Five favourable outcomes. d S = {Saturday, Sunday} Two favourable outcomes. e S = {December, January, February} Three favourable outcomes. 3 a 26 red cards from the pack of b One horse from a -horse race c One ball from balls d Five tickets out of a possible 00 e Three balls from a total possible P(Tail) = 2 Total possible outcomes with a die is 6. a P(6) = 6 b P() = 6 c P(even) = 3 6 = 2 d P(prime) = 3 6 = 2 (Primes are 2, 3 and ) e P(less than ) = 6 = 2 3 f P(at least ) = 2 6 = 3 6 a P(23) = b P() = c P(even) = 22 d P(odd) = 23 e P(multiple of ) = 9 = f P(multiple of 3) = = 3 g P(less than 20) = 9 h P(greater than 3) = 0 = 2 9 6 i P(square) = = 2 There are cards in a standard deck. a P(Ace of diamonds) = b P(king) = = 3 c P(club) = 3 = d P(red) = 2 e P(picture) = 2 = 3 3
MB Qld- 9 f P(court) = 2 = 3 3 8 a P(yellow) = 2 b P(red) = 2 c P(orange) = 2 9 There are a total of digits. a P(2) = b P() = c P(even) = d P(odd) = 3 e P(divisible by 3) = 2 = 2 f P(prime) = 3 0 Ten pieces of fruit that are not pears. P(not pear) = 0 2 a P(even) = (ends in 2) b P(odd) = 3 c P(divisible by ) = (ends in ) (ends in 3, or 9) d P(less than 3000) = (starts with 2) e P(greater than 000) = 2 (starts with or 9) 2 Three favourable outcomes. The answer is C. 3 At least 3 means 3,, or 6 The answer is D. There are a total of 2 court cards from The answer is C. There are 0 spot cards from The answer is E. 6 a P(win) = 000 b Craig now has tickets left from 999 P(win) = 999 338 a P(win) = 60000 668 b P(win) = 60000 c 338 + 668 = 0 P(either) = 0 60000 8 a P(begin with 3) = b P(even) = (ends in 8) c P(odd) = (ends in, 3, or ) d P(divisible by ) = (ends in ) e P(greater than 30 000) = (starts with 3,, or 8) f P(less than 20 000) = (starts with ) 9 Answers will vary. Possible answers include: a Tossing a Head with a coin b Tossing 2 Heads when tossing two coins c Spinning a or a 2 on a five-sectioned spinner numbered, 2, 3, and (sections are evenly designed). 20 a There are two even digits that can occupy the last place (which determines whether or not a number is even). b The numbers less than 00 contain only 2 in the starting place. The numbers greater than 00 contain a or a in the starting place. Numbers greater than 00 are more likely to be formed. Exercise C Relative frequency Relative frequency = 3 0 = 0. 2 Relative frequency = 9 00 = 0.9 3 Relative frequency = 3 8 = 0.3 a Relative frequency = 2 60 = 0. b Relative frequency = 33 60 = 0. Relative frequency = 2 300 = 0.0 00% = % 0 6 a Relative frequency = 000 = 0.03 b Relative frequency = 20 000 = 0.9
MB Qld- 80 a Relative frequency = 960 000 = 0.96 0 b Relative frequency = 000 = 0.0 8 Relative frequency = 36 0 = 0.2 The answer is A. 9 The highest relative frequency is option E. Relative frequency = 0 300 = 0. The answer is E. 0 a Relative frequency = 00 = 0. 8 b Relative frequency = = 0.3 c Relative frequency = = 0.03 a Relative frequency = 2 30 2 Percentage = 30 00% = 6.6% b Refunds = 2 30 200 = 80 200 2 a Relative frequency = 0000 = 0.02 b Total claims = 0.02 $20 000 = $00 3 Total repairs in the first year = (total repairs = 200) Relative frequency = 200 00% = 2% The assembly line will need upgrading. a Relative frequency = 200 00% = 2.% b Relative frequency = 03 200 00% =.% c Relative frequency = 6 200 00% = 82.% (00 82.) =. The relative frequency of a car not needing a mechanical repair in the first 3 years is.%. Number of kilometres Number of shock absorbers Relative frequency of shock absorbers not lasting Relative frequency of shock absorbers lasting 0 <20 000 0.00 0.99 20 000 <0 000 2 0.0 0.98 0 000 <60 000 6 0.2 0. 60 000 <80 000 6 0. 0. 80 000 <00 000 90 0 Therefore the maximum distance over which the manufacturer will guarantee the shock absorbers lasting with a relative frequency of 0.98 is 0000 km. 6 a Result Number b Win: Win Loss Draw Relative frequency = 0 = 0.3 Loss: Relative frequency = 0 = 0.3 Draw: Relative frequency = 0 = 0.2 Exercise D Modelling probability Note: Answers will vary in this exercise. Some advice is given below as guidance. to 3 Answers will vary. Students can use a graphics calculator to display results as is shown in worked examples and in chapter. Suggestion: Use the random number generator or a graphics calculator. Let represent heads and 2 represent tails. randint(, 2, 0) will display the results that simulate 0 coin tosses. The histogram may show that both bars get closer to the same level. The more trials may result in each outcome approaching the theoretical probability of 2. 6 a Generate 20 random numbers from to 6 twice. Pair the numbers in a table as follows: Game number 2 3... 20 First number Second number The numbers in these rows are obtained from the random number generator. b If the numbers and 2 appear at any game number, the player wins. Maximum loss of $20 is obtained if no wins result. Set out the results of the simulation in a similar manner to question 6. 8 Using the random number generator on the graphics calculator, for the first spinner, simulate 20 values between and and for the second spinner simulate 20 values between and.
MB Qld- 8 Exercise E Long-run proportion 2 Die roll Number of sixes 0 0 2 0 0 3 0.3 3 & 0. 0.2 6 0. 6 & 0.28 8 0. 9 0. 0 0. 0.09 09 && 2 2 0. 6 & 3 2 0.38 2 0.28 2 0.3 3 & 6 2 0. 2 0.6 8 2 0. 9 2 0.0 20 2 0. Innings Scoring shots Accumulated scoring shots Balls faced Accumulated balls faced 23 23 0.03082 2 26 9 0.29826 3 9 89 203 0.6308 9 03 9 222 0.6396396 23 26 6 268 0.09 6 3 60 2 30 0.08823 6 26 00 0 0.9090909 8 3 3 68 08 0.980396 9 8 2 3 39 0.028293 0 23 29 0 89 0.990 3 3 66 6 0.9683206 2 6 3 89 0.9869 3 a Long-run proportion = 28 = 0. b Day Sales Accumulated sales Houses visited Accumulated house visits 28 28 0. 2 9 0.2303 3 6 2 68 0. 8 93 0.268820 6 2 3 0.303030 6 2 3 29 6 0.323032 60 22 86 0.32806 8 2 2 3 2 0.33923 9 2 8 2 2 0.326229 0 3 9 26 20 0.3999
MB Qld- 82 The long-run proportion shows a steady increase from 0. (%) success to 0.36 (36%) success. Days Cleaners sold Accumulated cleaners sold Houses visited Accumulated house visited 20 20 0. 2 2 20 0 0. 3 20 60 0.8 3 & 6 20 80 0.2 3 9 20 00 0.9 9 00 = 0.9 The answer is D. More likely to be Statistician B, as in the long run, the experimental probability becomes close to the true probability. Therefore, the answer is D. 6 Long-run proportion = 600 = 0.08 Comment: One would expect the long-run proportion to be close to 0., so it seems the coin could be biased. Long-run proportion = 6 200 = 0.8 8 Long-run proportion = 9 = 0.82 Based on the long-run proportion, this student is more effective at answering mathematics questions. 9 0 Value Frequency 0.803 2 90 0.238 3 89 0.09 623 0.66 23 0.09223 6 0.0 3 Note: each proportion value is obtained by dividing the frequency of that value by the total frequency Comment: Theoretically each value should have a long-run proportion of 0.6. Values and 6 seem ok, but the others appear biased. Bread Shop 2 3 6 89 (0.) 6 99 (0.8) 26 309 (0.86) Bread Shop 3 seems more effective in long-run proportions, however, if you look at loaves wasted, Bread Shop is better with only 22 loaves wasted. Use the Excel files Coin tossing and Die rolling provided as support material with the Maths Quest text. 2 a Caught left handers: 6 = 0.29 Caught right handers: 2 6 = 0.29 Based on long-run proportions, there is no difference between being caught by a left-hander or a right-hander. b Bowled : 30 0 Caught : 30 0 c Not out (left) = 0.29 no difference Not out (right) 6 6 = 0. Higher value for not out left-handers than right-handers.
MB Qld- 83 d The relatively small number of left-handed observations means comparisons are not very accurate. However, there seems little difference between left- and right-handed effectiveness. 3 Let the population of trout in lake (including tagged trout) be x. Therefore, population (excluding the tagged trout) = x 00 Sample of 00 fish, were tagged So, are tagged 00 Use this to estimate the total population in the lake. 00 represents 00 tagged fish in x the lake of x fish. = 00 00 x x = 0000 x = 29 (rounded to the nearest whole number) There are 29 trout in the lake, of which 00 are tagged. So there are 329 untagged trout in the lake. Chapter review More options to receive a higher card than a lower card. 2 a Most numbers are less than 6 Probable b There is no of diamonds Impossible c Half the options is Tails Even chance d One of the options is 2 (2 to 2) Unlikely e Hard to win the lottery Unlikely 3 Answers will vary. Possible answers include: a Certain: rolling a number with a b Impossible: rolling a with a Hot weather in Alice Springs in January. Rolling a die and getting a number greater than Selecting a picture card Selecting the blue marble Winning the lottery 6 Mark has won 6 out of the past 9. He is most likely to win their next match. S = {, 2, 3,, } Favourable outcomes = {3,, } 8 a P(3) = b P() = c P(odd) = 3 d P(square) = = 9 e P(prime) = (Primes are 2, 3,,,, 3,, 9 and 23) f P(double-digit) = 6 9 There are cards in a standard deck. a P(2 of clubs) = b P(any 2) = = 3 c P(any club) = 3 = d P(black) = 2 e P(court) = 2 = 3 3 f P(spot) = 0 = 0 3 0 The total DVDs is 0. a P(comedy) = 0 = 20 b P(horror) = 0 = 0 c P(not romance) = 30 0 a P(89) = b P(odd) = 3 2 = 3 c P(divisible by ) = (ends in, or 9) d P(greater than 000) = 3 (starts with, 8 or 9) (ends in ) e P(less than 8000) = 2 (starts with or ) = 2 2 a P(win) = = 00 b P(win) = 999 2 3 Relative frequency = 00 = 0.02 a Relative frequency = 0 000 = 0. b Premium = $000 0. = $0 Answers will vary. There are 36 total possibilities when rolling two dice together, five which sum to 6. 6 Die roll Number of sixes (3 decimal places) 0 0 2 0 0 3 0 0 0 0 0.2 6 0.6 0. 8 0. 9 2 0.222 0 2 0.2 3 0.22 2 3 0. 3 3 0.23 3 0.2 3 0.2 6 0. 0.23 8 0.222 9 0.2 20 0.2 The long-run proportion for twos is 0.2. Probability = 20 = 0.
MB Qld- 8 Modelling and problem solving All elements in the sample space are not equally likely. There are a number of factors which influence this. One factor that can be considered is the financial status of Joanna and her friends or acquaintances. Joanna, and most of the people she knows, have similar finances so they will use similar airlines at similar times. Also, they will tend to have holidays at similar times and tend to choose the same locations. (She is less likely to meet at Reykjavik Airport.) So the chance of accidently meeting someone she knows at JFK terminal is considerably higher than the in 0 000 000 chance that might be expected. 2 P(6) = 6, P() = 6 and P(2, 3, or ) = 6 Expected win = 6 $6 + 6 $ + 6 $ = $ + $ 6 $ 6 = $ 2 The player would expect to win on average 0 cents per game.