Mathematics-advanced-probability-and-discrete-distributions



Mathematics AdvancedMA-S1 Probability and Discrete Probability DistributionsTable of contents TOC \o "1-3" \h \z \u Mathematics Advanced PAGEREF _Toc48198926 \h 1MA-S1 Probability and Discrete Probability Distributions PAGEREF _Toc48198927 \h 1Table of contents PAGEREF _Toc48198928 \h 2Syllabus outcomes PAGEREF _Toc48198929 \h 4Outcomes PAGEREF _Toc48198930 \h 4Content PAGEREF _Toc48198931 \h 4Supplementary resources PAGEREF _Toc48198932 \h 6Department of Education resources PAGEREF _Toc48198933 \h 6NESA resources PAGEREF _Toc48198934 \h 6WOOTUBE resources PAGEREF _Toc48198935 \h 6Examination-style questions PAGEREF _Toc48198936 \h 7Sample question 1 PAGEREF _Toc48198937 \h 7Sample question 2 PAGEREF _Toc48198938 \h 7Sample question 3 PAGEREF _Toc48198939 \h 8Sample question 4 PAGEREF _Toc48198940 \h 8Sample question 5 PAGEREF _Toc48198941 \h 9Sample question 6 PAGEREF _Toc48198942 \h 9Sample question 7 PAGEREF _Toc48198943 \h 10Sample question 8 PAGEREF _Toc48198944 \h 10Sample question 9 PAGEREF _Toc48198945 \h 11Sample question 10 PAGEREF _Toc48198946 \h 11Sample question 11 PAGEREF _Toc48198947 \h 12Sample question 12 PAGEREF _Toc48198948 \h 12Sample question 13 PAGEREF _Toc48198949 \h 13Sample question 14 PAGEREF _Toc48198950 \h 13Sample question 15 PAGEREF _Toc48198951 \h 14Sample question 16 PAGEREF _Toc48198952 \h 14Sample question 17 PAGEREF _Toc48198953 \h 15Sample question 18 PAGEREF _Toc48198954 \h 15Sample question 19 PAGEREF _Toc48198955 \h 16Sample question 20 PAGEREF _Toc48198956 \h 17Sample question 21 PAGEREF _Toc48198957 \h 18Sample question 22 PAGEREF _Toc48198958 \h 19Solutions PAGEREF _Toc48198959 \h 20Sample question 1 PAGEREF _Toc48198960 \h 20Sample question 2 PAGEREF _Toc48198961 \h 20Sample question 3 PAGEREF _Toc48198962 \h 20Sample question 4 PAGEREF _Toc48198963 \h 21Sample question 5 PAGEREF _Toc48198964 \h 21Sample question 6 PAGEREF _Toc48198965 \h 22Sample question 7 PAGEREF _Toc48198966 \h 22Sample question 8 PAGEREF _Toc48198967 \h 22Sample question 9 PAGEREF _Toc48198968 \h 23Sample question 10 PAGEREF _Toc48198969 \h 23Sample question 11 PAGEREF _Toc48198970 \h 23Sample question 12 PAGEREF _Toc48198971 \h 24Sample question 13 PAGEREF _Toc48198972 \h 25Sample question 14 PAGEREF _Toc48198973 \h 26Sample question 15 PAGEREF _Toc48198974 \h 27Sample question 16 PAGEREF _Toc48198975 \h 28Sample question 17 PAGEREF _Toc48198976 \h 29Sample question 18 PAGEREF _Toc48198977 \h 30Sample question 19 PAGEREF _Toc48198978 \h 31Sample question 20 PAGEREF _Toc48198979 \h 32Sample question 21 PAGEREF _Toc48198980 \h 33Sample question 22 PAGEREF _Toc48198981 \h 34DisclaimerThis document is to be used to supplement the support teachers are offering students undertaking HSC Mathematics courses. Questions can be printed off for students individually, with or without solutions, or as an entire booklet. Questions have been sourced from various states across Australia and the source of each question has been referenced. Permission to use these resources was provided in June 2020. Solutions for each of the questions can be found at the end of the document.OutcomesAll outcomes referred to in this booklet are from Mathematics Advanced Syllabus ? 2017 NSW Education Standards Authority (NESA) for and on behalf of the Crown in right of the State of New South Wales. Syllabus outcomesThe examination-style questions presented in this document refer to the following outcomes and syllabus content.OutcomesA student:uses concepts and techniques from probability to present and interpret data and solve problems in a variety of contexts, including the use of probability distributions MA11-7provides reasoning to support conclusions which are appropriate to the context MA11-9ContentS1.1: Probability and Venn diagramsestablish and use the rules:? P(A)=1-P(A)?and? P(A∪B)=P(A)+P(B)-PA∩B(ACMMM054) AAM understand the notion of?conditional probability?and recognise and use language that indicates conditionality(ACMMM056)use the notation?P(A|B)?and the formula?P(A|B)=P(A∩B)p(B), P(B)≠0?for conditional probability(ACMMM057) AAMunderstand the notion of independence of an event?A?from an event?B, as defined by P(A|B)=P(A)(ACMMM058)use the multiplication law P(A∩B)=P(A)P(B)?for?independent events?A?and?B?and recognise the symmetry of independence in simple probability situations (ACMMM059)S1.2: Discrete probability distributionsdefine and categorise random variablesknow that a random variable describes some aspect in a?population?from which samples can be drawnknow the difference between a?discrete random variable?and a?continuous random variableuse discrete random variables and associated probabilities to solve practical problems(ACMMM142) AAMuse relative frequencies obtained from data to obtain point estimates of probabilities associated with a discrete random variable(ACMMM137)recognise uniform discrete random variables and use them to model random phenomena with equally likely outcomes(ACMMM138)examine simple examples of non-uniform discrete random variables, and recognise that for any random variable,?X, the sum of the probabilities is 1 (ACMMM139)recognise the mean or?expected value,?E(X)=μ, of a discrete random variable?X?as a measure of centre, and evaluate it in simple cases(ACMMM140)recognise the?variance,Var(X), and?standard deviation?(σ) of a discrete random variable as?measures of spread, and evaluate them in simple cases(ACMMM141)use?Var(X)=E(X-μ2)=E(X2)-μ2?for a random variable and?Var(x)=σ2?for a datasetunderstand that a sample mean,?x, is an estimate of the associated population mean?μ, and that the sample standard deviation,?s, is an estimate of the associated population standard deviation,?σ, and that these estimates get better as the sample size increases and when we have independent observations Supplementary resourcesDepartment of Education resourcesUnits of workMA S1 Probability and discrete probabilityHSC Hub videosConditional probabilityDiscrete probability distributions Q19 NESA sample examinationNESA resourcesMathematics Advanced – Sample examination materials (2020)WOOTUBE resourcesMA S1 Probability and Discrete probability distributionsExamination-style questionsSample question 1Question 7The discrete random variable X has the following probability distribution:X0123P(X=x)a3a5a7aThe mean of X is116135161782Source: ? VCAA 2019 Mathematical Methods: Written examination 2Sample question 2Question 11A and B are events from a sample space such that P(A) = p, where p>0, PBA)=m and P(B|A’) = n.A and B are independent events whenm = nm = 1 – pm + n = 1m = pm + n = 1 - pSource: ? VCAA 2019 Mathematical Methods: Written examination 2 Sample question 3Question 12The discrete random variable X has the following probability distribution:X01236Pr(X=x)14920110120320Let μ be the mean of XP(X< μ) is1214172045710Source: ? VCAA 2018 Mathematical Methods: Written examination 2Sample question 4Question 14Two events, A and B, are independent, where P(B)=2P(A) and P(A∪B)=0.52P(A) is equal to0.10.20.30.40.5Source: ? VCAA 2018 Mathematical Methods: Written examination 2Sample question 5Question 14The random variable X has the following probability distribution, where 0< p <13.X-101P(X=x)p2p1-3pThe variance of X is2p(1 - 3p)1 – 4p(1 – 3p)26p – 16p2p(5 – 9p)Source: ? VCAA 2017 Mathematical Methods: Written examination 2Sample question 6Question 7The number of pets, X, owned by each student in a large school is a random variable with the following discrete probability distribution.X0123P(X=x)0.50.250.20.05If two students are selected at random, the probability that they own the same number of pets is:0.30.3050.3550.4050.8Source: ? VCAA 2016 Mathematical Methods: Written examination 2Sample question 7Question 19Consider the discrete probability distribution with random variable X shown in the table below.X-10b2b4P(X=x)abb2b0.2The smallest and largest possible values of E(X) are respectively-0.8 and 1-0.8 and 1.60 and 2.40.2125 and 10 and 1Source: ? VCAA 2016 Mathematical Methods: Written examination 2Sample question 8Question 14Consider the following discrete probability distribution for the random variable, X.X12345P(X=x)p2p3p4p5pThe mean of this distribution is23721134Source: ? VCAA 2015 Mathematical Methods (CAS): Written examination 2Sample question 9Question 14If X is a random variable such that P(X>5) = a and P(X>8) = b, then PX<5 X<8) isaba-b1-b1-b1-aab1-ba-1b-1Source: ? VCAA 2014 Mathematical Methods (CAS): Written examination 2Sample question 10Question 17A and B are events of a sample space.Given that PA B) = p, P(B) = p2 and P(A) = p13, PB A) is equal topp43p73 p83p3Source: ? VCAA 2013 Mathematical Methods (CAS): Written examination 2Sample question 11Question 10For events A and B, PA∩B=p,PA'∩B=p-18 and PA∩B'= 3p5If A and B are independent, then the value of p is014381235 Source: ? VCAA 2013 Mathematical Methods (CAS): Written examination 2Sample question 12Question 6 (4 marks)Two boxes each contain four stones that differ only in colour.Box 1 contains four black stones.Box 2 contains two black stones and two white stones.A box is chosen randomly and one stone is drawn randomly from it.Each box is equally likely to be chosen, as is each stone.What is the probability that the randomly drawn stone is black? (2 marks)It is not known from which box the stone has been drawn.Given that the stone that is drawn is black, what is the probability that it was drawn from Box 1? (2 marks)Source: ? VCAA 2018 Mathematical Methods: Written examination 1Sample question 13Question 8 (5 marks)For events A and B from a sample space, P(A | B) = 15 and P(B | A) = 14. Let P(A∩B) = pFind P(A) in terms of p (1 mark)Find PA'∩B'in terms of p (2 marks)Given that PA∪B≤15, state the largest possible interval for p (2 marks)Source: ? VCAA 2017 Mathematical Methods: Written examination 1Sample question 14Question 7 (3 marks)A company produces motors for refrigerators. There are two assembly lines, Line A and Line B. 5% of the motors assembled on Line A are faulty and 8% of the motors assembled on Line B are faulty. In one hour, 40 motors are produced from Line A and 50 motors are produced from Line B. At the end of an hour, one motor is selected at random from all the motors that have been produced during that hour.What is the probability that the selected motor is faulty? Express your answer in the form 1b, where b is a positive integer. (2 marks)The selected motor is found to be faulty.What is the probability that it was assembled on Line A? Express your answer in the form 1c, where c is a positive integer (1 mark)Source: ? VCAA 2016 Mathematical Methods: Written examination 1Sample question 15Question 8 (3 marks)For events A and B from a sample space, P(A|B) = 34 and P(B) = 13.Calculate P?(A∩B) (1 mark)Calculate P?(A'∩B), where A’ denotes the complement of A (1 mark)If events A and B are independent, calculate P?(A∪B) (1 mark)Source: ? VCAA 2015 Mathematical Methods: Written examination 2Sample question 16Question 9 (4 marks)An egg marketing company buys its eggs from farm A and farm B. Let p be the proportion of eggs that the company buys from farm A. The rest of the company’s eggs come from farm B. Each day, the eggs from both farms are taken to the company’s warehouse.Assume that 35 of all eggs from farm A have white eggshells and 15 of all eggs from farm B have white eggshells.An egg is selected at random from the set of all eggs at the warehouse.Find, in terms of p, the probability that the egg has a white eggshell. (1 mark)Another egg is selected at random from the set of all eggs at the warehouse.Given that the egg has a white eggshell, find, in terms of p, the probability that it came from farm B. (2 marks)If the probability that this egg came from farm B is 0.3, find the value of p.(1 mark)Source: ? VCAA 2015 Mathematical Methods: Written examination 1Sample question 17Question 9 (6 marks)Sally aims to walk her dog, Mack, most mornings. If the weather is pleasant, the probability that she will walk Mack is 34, and if the weather is unpleasant, the probability that she will walk Mack is 13Assume that pleasant weather on any morning is independent of pleasant weather on any other morning.In a particular week, the weather was pleasant on Monday morning and unpleasant on Tuesday morning. Find the probability that Sally walked Mack on at least one of these two mornings. (2 marks)In the month of April, the probability of pleasant weather in the morning was 58Find the probability that on a particular morning in April, Sally walked Mack. (2 marks)Using your answer from part b.i., or otherwise, find the probability that on a particular morning in April, the weather was pleasant, given that Sally walked Mack that morning. (2 marks)Source: ? VCAA 2014 Mathematical Methods: Written examination 1Sample question 18Question 7 (6 marks)The probability distribution of a discrete random variable, X, is given by the table below.X01234P(X=x)0.20.6p20.11-p0.1Show that p=23 or p=1 (3 marks)Let p=23Calculate E(x) (2 marks)Find P?(X≥E(X) (1 mark)Source: ? VCAA 2013 Mathematical Methods (CAS): Written examination 1Sample question 19Question 10 (7 marks)A group of researchers conducted a study into the number of siblings of adult Australian citizens. They surveyed a total of 200 participants and recorded the number of siblings, X, of each participant. A few days later the lead researcher discovered that the survey data had been misplaced. Fortunately, one of the research assistants had been doing some rough calculations on a whiteboard and the lead researcher was able to recover the following information about the probability distribution for X and the mean μ. X0123P(X=x)0.2ab0.1μ = 1.3The letters a and b have been used to denote unknown probabilities. i.Write two independent equations for a and b. (2 marks) ii.Hence solve for the unknown probabilities. (2 marks) Later that day the research assistant found the complete probability distribution in their records and discovered that they had made an error in their original calculation of the mean. The correct probability distribution is given in the table below. X0123P(X=x)0.20.30.40.1i.Given that there were 200 participants in the study, complete the table below to show the number of participants N with 0, 1, 2 and 3 siblings. (1 mark) X0123P(X=x)0.20.30.40.1N40ii.Determine the correct mean and standard deviation of the number of siblings X. (2 marks)Source: ? WA SCSA 2019 Mathematics Methods – Section 2: Calculator assumedSample question 20Question 4 (4 marks)Ten shop owners in a coastal resort were asked how many extra staff they intended to hire for the next holiday season. Their responses are shown below: 3, 0, 2, 1, 2, 1, 1, 0, 2, 1 If N = number of additional staff, complete the probability distribution of N below. (2 marks)n0123Pr(N=n)what is the mean number of staff the shop owners intend to hire? (2 marks)Source: ? WA SCSA 2018 Mathematics Methods – Section 1: Calculator freeSample question 21Question 13 (9 marks)Ravi runs a dice game in which a player throws two standard six-sided dice and the sum of the uppermost faces is calculated. If the sum is less than five, the player wins $20. If the sum is greater than eight, the player wins $10. Otherwise the player receives no money. Complete the table below. (2 marks) Amount wonProbabilityWhat is the expected amount of money won by a player each time they play? (2 marks) Liu Yang decides to play the game. If Ravi charges her $5 to roll two dice, who is likely to be better off in the long-term? Explain. (3 marks) If Ravi wants to make a long-term profit per game of 20% of what he charges, what should he charge a player to roll the two dice? (2 marks)Source: ? WA SCSA 2017 Mathematics Methods – Section 2: Calculator assumedSample question 22Question 15 (6 marks)A tetrahedral die has the numbers 1 to 4 on each face. When thrown, each side is equally likely to land facedown. Let X be defined as the sum of the numbers on the facedown side when the die is thrown plete the following table (1 mark)Roll twoSum of two rolls1234Roll One11 + 1 = 2323354i.Hence, or otherwise, complete the probability distribution of X, which is given by the following table. (1 mark)X2345678P(X=x)116ii.Calculate the probability of obtaining a sum of five or less (2 marks)iii.Determine the mean and standard deviation for X (2 marks)Source: ? WA SCSA 2016 Mathematics Methods – Section 2: Calculator assumedSolutionsSample question 1Question 7a + 3a + 5a + 7a = 1a = 116E(x) = 0 x 116+ 3 x 116+ 5 x 116+ 7 x 116E(x) = 178Correct answer = DSample question 2Question 11If A and B are independentP(B|A) = P(B) and P(B|A’) = P(B)So, P(B|A) = P(B|A’)m = nCorrect answer = ASample question 3Question 12μ=1×920+2×110+3×120+6×320μ=1.7P(X<1.7) = P(0) + P(1) = 710Correct answer = ESample question 4Question 14PA∪B=PA+PB-P(A∩B)0.52 = PA+ 2PA-P(A∩B)PA∩B=PA×PB for independent events0.52 = PA+ 2PA-P(A)×2P(A)2x2-3x+0.52=0Solving with quadratic formula gives: PA= 1.3 or 0.2Correct = B (disregard 1.3 as probability cannot be greater than 1)Sample question 5Question 14μ=-1×p+0×2p+1×(1-3p)μ=-p+1-3pμ=-4p+1E(X2) = p+1-3pE(X2) = -2p+1VarX= EX2- μ2Var(X) = -2p + 1 – -4p + 12Var (X) = -16p2 + 6pCorrect answer= DSample question 6Question 7P(both students own the same number of pets) = 0.52 + 0.252 + 0.22 + 0.052P(both students own the same number of pets) = 0.355Correct answer = CSample question 7Question 19a+4b+0.2=1a=0.8-4b…1EX= -a+5b2+0.8…2Substitute1 into (2)EX=5b2+4b, 0≤b≤0.2, maximum value of b occurs when a is zero4b+0.2=1, b=0.2The smallest value for E(X) is 0 and the largest value is 1.Correct answer = ESource: ? VCAA 2016 Mathematical Methods: Examination 2 ReportSample question 8Question 14p + 2p + 3p + 4p + 5p = 1p = 115E(X) = 1×115+2×215+3×315+4×415+5×515E(X) = 113Correct = DSource: ? VCAA 2015 Mathematical Methods: Written examination 2 reportSample question 9Question 14If X is a continuous random variable, thenPX<5X<8= P?(X<5∩X<8)P?(X<8)PX<5X<8= P?(X<5)P?(X<8)=1-a1-b=a-1b-1Correct = ESource: ? VCAA 2014 Mathematical Methods (CAS): Written examination 2 reportSample question 10Question 17PA|B= P(A∩B)P?(B)=p,PA∩B=p3PB|A=P?(A∩B)P?(A)=p3p13=p83Correct = DSource: ? VCAA 2013 Mathematical Methods (CAS): Written examination 2 reportSample question 11Question 10If A and B are independent eventsP(A)× P(B) = PA∩B=pp+p-18×p+3p5=p, p=38Correct = CSource: ? VCAA 2013 Mathematical Methods (CAS): Written examination 2 reportSample question 12Question 6aSeveral approaches were possible using a tree diagram or a counting argument.P(Black) = 12×1+12×12=341-P(White) = 1-14=34Since choosing either box is equally likely and choosing any stone is equally likely and there are 8 stones, 6 of which are black, P(Black) = 68=34This question was generally well answered. Many students showed their reasoning via a tree diagram or some written explanation. Some students overworked the problem by trying to use the binomial distributionQuestion 6bP(Box 1|Black) = 12÷34=23Students generally recognised the conditional probability (reduced sample space). Some students incorrectly worked P(Black | Box 1), resulting in a probability greater than 1, which is not feasible.Source: ? VCAA 2018 Mathematical Methods: Written examination 1 reportSample question 13Question 8aPA= PA∩BPB|APA= p14PA=4pThis questions was generally answered well. The most common errors included solving for P(B), and incorrectly transposing pP?(A) to yield 14Question 8bAA’Bp4p5pB’3p1 – 8p1 – 5p4p1-4p1Or, P?(A∪B)=4p+5p-p=8pAnd A∪B'=A'∩B'PA'∩B'=1-8pStudents who scored highly usually used a table or a Venn diagram to arrive at their answer. There were various misconceptions of the connection between conditional probabilities and PA'∩B'. Many students assumed that events A and B were independent, hence incorrectly used PA'∩B'=PA'×P?(B')Question 8cPA∪B=8pIf 8p≤15p≤140 , thus 0<p≤140Most students identified that PA∪B=8p. Only a few students identified the correct interval because students did not consider that in this case p≠0. Common incorrect answers included p=140 or p≤140 (allowing negative probabilities) and 0≤p≤140Source: ? VCAA 2017 Mathematical Methods: Written examination 1 reportSample question 14Question 7aPF= 49×120+59×225PF= 115While a tree was not required to answer the question, it may have assisted some students to determine the two required cases. Many students stated probabilities greater than 1.Question 7bPA|F= P?(A∩F)P?(F)PA|F= 145115=13In general the conditional probability was recognised but not the reduced sample space. Often the instruction regarding the form of the final answer was overlooked.Source: ? VCAA 2016 Mathematical Methods: Written examination 1 reportSample question 15Question 8aPA∩B=PA|B×PB= 34×13=14Question 8bPA'∩B=PB-PA∩B=13-14=112 Question 8cIf A and B are independent, then P(A\B) = P(A) = 34PA∪B=PA+PB=PA∩B=34+13-14=56Many students made little headway into solving this problem due to their lack of understanding of independent events. The additional rule was then applied using an incorrect value for P(A), resulting in final answers well outside the interval [0,1]. Students must note that a probability must lie with [0,1] and is never a negative answer.Source: ? VCAA 2015 Mathematical Methods (CAS): Written examination 2 reportSample question 16Question 9aPW=PA×PW|A+PB×PW|BPW=p×35+(1-p)×15PW=2p+15Many students made good use of a tree diagram in their formulation of a solution. Some students left their answer unsimplified as a sum of two products. A significant number of students offered a final expression not in terms of p.Question 9biPB|W=P?(B∩W)P?(W)=1-p52p+15=1-p2p+1While most students recognised that this question involved conditional probability, many could not apply it within the context of the specific question. Algebraic fractions were not handled well.Question 9biiPB|W=0.3=1-p2p+10.6p+0.3=1-pp=716Many students missed the specific connection of this part with the previous part.Source: ? VCAA 2015 Mathematical Methods: Written examination 1 reportSample question 17Question 9aP(at least 1 morning walk) = 1 – P(no walk)P(at least 1 morning walk) = 1-14×23=56, ORw = walkP(w’w+ww’+ww)=34×13+34×23+14×13=56A tree diagram or listing the sample space were the best options to solve this problem.Question 9biPW=58×34+38×13=1932Many students made a good attempt at this question. Most students correctly identified the required sum of two products; however, made errors in the evaluation of the final fraction.Question 9biiPP | W=P?(W∩P)P?(W)=58×341932=1519Many students were able to identify the conditional probability and use their answer to part bi. in the denominator, however, used an incorrect numeratorSource: ? VCAA 2014 Mathematical Methods: written examination 1 reportSample question 18Question 7aPX=x=0.2+0.6p2+0.1+1-p+0.1=10.6p2-p+0.4=0 ?p=1±1-0.961.2=1±0.21.2=1 or 0.81.2=23 OR6p2-10p+4=0 ?3p2-5p+2=03p-2p-1=0 ∴p=23or p=1This question presented a range of fundamental problems for students: mixed operations with fractions and decimals, factorising and solving quadratics, using substitution to ‘show’ and not eliminate or determine other possible solutions, and poor notation.Question 7biEX=xPX=x=1×0.6×49+0.2+31-23+0.4=35×49+15+1+25=11315EX=11315=1.86Many students had difficulty adding decimals and fractions to give an answerQuestion 7biiPX≥EX=PX=2+PX=3+PX=4=110+13+110=815Many students gave 0.5 as the answerSource: ? VCAA 2013 Mathematical Methods (CAS): Written examination 1 reportSample question 19Question 10ai.Probabilities add to 10.2 + a + b + 0.1 = 1a + b = 0.7(1)Calculation of the mean0.2(0) + a (1) + b(2) + 0.1(3) = 1.3a + 2b = 1(2)Question 10aii.From first equationA = 0.7 – bSubstituting into the second equation(0.7 – b) + 2b = 1b = 0.3a = 0.4Question 10bi.X0123P(X=x)0.20.30.40.1N40200 x 0.3 = 60200 x 0.4 = 80200 x 0.1 = 20Question 10bii.μ=0×0.2+1×0.3+2×0.4+3×0.1=1.4E(X2) = 02 × 0.2 + 12 × 0.3 + 22 × 0.4 + 32 × 0.1 = 2.8Var(X) = E(X2) - μ2 = 2.8 – 1.42= 0.84σ=0.84=0.9165Source: ? WA SCSA 2019 Mathematics Methods marking keySample question 20Question 4an0123Pr(N=n)210410310110Question 4bE(X) = 0 × 210+1×410+2×310+3×110=1310=1.3Source: ? WA SCSA 2018 Mathematics Methods marking keySample question 21Question 13aAmount won20100Probability63610362036Question 13bLet the random variable X be the amount of money won by a player:E(X) = 20 × 2036+10×1036 = 22036 = $6.11Question 13cExpected payout = 6.11 – 5 = 1.11Lui Yang is better off in the long term. In the long term Liu Yang will likely win $1.11 per game.Question 13dLet amount to be paid be $PE(X) = -0.2P-0.2P = 20 × 636+ 10 × 1036 – P0.8P = 6.11P=$7.64Source: ? WA SCSA 2017 Mathematics Methods marking keySample question 22Question 15aRoll twoSum of two rolls1234Roll One11 + 1 = 2345234563456745678Question 15biX2345678Pr(X=x)116216316416316216116Question 15biiPS≤5= 416+316+216+116=1016=58=0.625Question 15biiiμ=2×116+3×216+4×316+5×416+6×316+7×216+8×116=5E(X2) = 22×116+32×216+42×316+52×416+62×316+72×216+82×116=552Var(X) = E(X2) - μ2=552-25=2.5σ=2.5=1.58Source: ? WA SCSA 2016 Mathematics Methods marking key ................
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