• Have any questions?
  • info.zbook.org@gmail.com

OCR Statistics 1 Probability Section 1: Introducing .

2m ago
5 Views
0 Downloads
379.07 KB
11 Pages
Last View : 6d ago
Last Download : n/a
Upload by : Halle Mcleod
Share:
Transcription

OCR Statistics 1 ProbabilitySection 1: Introducing probabilityNotes and ExamplesThese notes contain subsections on Notation Sample space diagrams The complement of an event Mutually exclusive events Probability tree diagrams Conditional probability Independent events Problems involving “at least one”NotationMany students lose a lot of marks on probability questions, as their solutionsare difficult to follow. P(A) is a quick way of writing down ‘the probability of event Aoccurring.’ P(H) can be interpreted as a quick way of writing down ‘the probabilityof a head occurring.’ P(HH) can be interpreted as a quick way of writing down ‘theprobability of a head followed by a head occurring.’Try to make your notes clear. Not only will your work become easier to mark,but it will also make it easier for you to check your solution and find anyerrors. Good presentation during this chapter will help you produce goodsolutions in examinations.Sample space diagramsAnother way of displaying the outcomes of 2 events is on a sample spacediagram. You may have met this at GCSE. It is still a clever way of displayingoutcomes on more complicated problems at AS level.Example 1A regular tetrahedron has one of the numbers 1, 2, 3, 4 on each face.This is rolled with an ordinary die.The score is the sum of the numbers showing.Find the probability of each score1 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and ExamplesSolutionDIEscoreTETRAHEDRONscore12 3423 4534 5645 6756 7867 8978 9 10123456Careful listing could solve thisproblem, but with 24 outcomesit is easy to miss out cases orduplicate others.The sample space diagramclearly and efficiently shows allof the possible outcomes.There are 24 possible outcomes.You can now work out the probabilities of each score.For example there are three 8s in the table, so the probability of getting an 8 is3 1 .24 8Score2Probability1243224 1214324 815424 166424 167424 168324 819224 12110124It is simple to calculatethese probabilities usingthe sample space diagram.It is usually best not tosimplify the fractions, as ifyou need to do any furthercalculations it will be easierto work with a commondenominator.In this case all the separate scores on the die or tetrahedron were equallylikely, so we could easily calculate their probabilities. The calculations whenthe original probabilities are not the same are much more complicated.The complement of an eventYou will have met examples at GCSE using the complement, such as findingthe probability of not getting a 6 on the throw of a die, finding the probability ofnot passing a driving test (given in the question the probability of passing thetest) or finding the probability a train is on time (given in the question theprobability of the train being late) or 2 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and ExamplesIn this chapter and in later chapters you will meet many examples based onthe words ‘at least’. In most of these questions it will be easier to work out theproblem using the probability of the complementary event.If A is the event of getting at least one six when you throw 5 dice, A' is theevent of getting no sixes when you throw 5 dice.To calculate P(A) it will much be simpler to calculate P(A') first and then useP A 1 P A .Similarly, if B is the event of getting at least one head when you throw 4 coins,B' is the event of getting no heads when you throw 4 coins. To calculate P(B)it will be much simpler to calculate P(B') first and then use P B 1 P B .Mutually exclusive eventsIt is important to realise that in Probability work the word or has a specialmeaning.P(A or B) means that A or B or both A and B can occur.If two events are mutually exclusive, then they cannot both occur. This meansthat P(A or B) is simply equal to P(A) P(B).You can work out whether two events A and B are mutually exclusive if youknow P(A), P(B) and P(A or B).Example 2(i)Given that P(A) 0.3, P(B) 0.15 and P A or B 0.45, are events A and Bmutually exclusive?(ii)Given that P(C) 0.5, P(D) 0.32 and P C or D 0.7, are events C and Dmutually exclusive?SolutionP A P B 0.3 0.15(i) 0.45 P( A or B)So A and B are mutually exclusive.(ii)P C P D 0.5 0.32 0.82 P(C or D)So C and D are not mutually exclusive.3 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and ExamplesProbability Tree DiagramsA tree diagram allows you to highlight all the possible outcomes for two ormore events and systematically work out the corresponding probabilities.Example 3An ‘A’ level teaching group has 7 boys and 3 girls.2 students are selected at random from this group.Find the probability that the students selected are(i)two boys(ii)two girls(iii) one boy and one girlSolutionIf the first student selected is a boy, there are 6 boys and 3 girls left from 9 students.6So the probability that the second student is a boy is and the probability that the93second student is a girl is .9If the first student selected is a girl, there are 7 boys and 2 girls left from 9 students.7So the probability that the second student is a boy is and the probability that the92second student is a girl is9These probabilities can now be shown on a tree diagram.69P( B1 ) 7103P(G1 ) 10BoyBoy39Girl79Boy29GirlGirl7642217 10 99045 1573217P(Boy, Girl) 10990 3037217P(Girl, Boy) 10990 30P(Boy, Boy) 4 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and ExamplesP(Girl, Girl) 32 109 61 90 15427or901561(ii) The probability 2 girls will be selected isor1590(i) The probability 2 boys will be selected is(iii) The probability that 1 boy and 1 girl will be selected is77 14 7 30 30 30 15Conditional probabilityIn Example 3 above the idea of conditional probability was used. Theprobabilities for the second selection depend on the probabilities of the firstselection.The probabilities for the second selection can be expressed formally like this:Let B1 be the event that the first selection is a boy.Let G1 be the event that the first selection is a girl.Let B2 be the event that the second selection is a boy.Let G2 be the event that the second selection is a girl.6i.e.P(B2 B1) 93P(G2 B1) 97i.e.P(B2 G1) 92P(G2 G1) 9In Example 3, the conditional probabilities can be simply deduced from theinformation given in the question.However, sometimes a little more work is needed.Example 4In a small sixth form of 50 students, Maths and English are the two most popularsubjects.30 students are studying Maths.25 students are studying English.10 students are studying Maths and English.Find the probability that a student studies Maths given that he/she studies English.Solution 1Let M be the event that the student studies Maths.5 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and ExamplesLet E be the event that the student studies English.We want to find P(M E).Using the formula for conditional probability:P( M and E )P( M E ) P( E )There are 10 students studying both Maths and English, so P( M E ) There are 25 students studying English, so P( E ) Therefore P( M E ) 115210 1 .50 525 1 50 212 2 55Solution 2The given information can be presented in a table like the one below.MathsNot MathsTotalEnglish101525Not English20525Total302050To find the probability that a student takes Maths given that he/she takes English, youare considering only the 25 students in the column headed “English”.Out of these 25 students 10 take Maths.102So the probability of a student studying Maths given they study English is 25 5From a table like this you can work out many conditional probabilities.You can find the probability of a student studying English given that they studyMaths by looking at the “Maths” row.10 1 P(E M) 30 3You can find the probability of a student studying English given they do notstudy Maths by looking at the “Not Maths” row.15 3 P(E M') 20 4From Example 4 above, you should be able to see that you can write theprobability of a student studying Maths given that they study English, P(M E),as:P( M and E )P( M E ) .P( E )This can be generalised to any events A and B:6 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and ExamplesP( A and B)P( B)and can also be written in the formP( A and B) P( A B) P( B) .P( A B) This is called the multiplication law.In some problems, you need to use this law to work out conditional probability,if it is not obvious from the information in the question.Example 5A Head of Mathematics has analysed the results of students taking the modulesStatistics 1 (S1) and Statistics 2 (S2) over the last 10 years at her centre, collectinginformation on students who have achieved a grade A or B.She has estimated the following probabilities.The probability that a student gets an A or B grade on S1 is 0.6.If the student has achieved a grade A or B at S1, the probability that a student gets anA or B grade on S2 is 0.7.If the student has not achieved a grade A or B at S1, the probability that a student getsan A or B grade on S2 is 0.35.Calculate the probability that an A or B grade is achieved on S1, given that an A or Bgrade has been achieved on S2.Excluding re-sits, the large majority of students takeS1 before S2; the exception is a few students whotake the modules at the same time. However, thequestion asks us to analyse the information in thereverse order. Using the conditional probabilityformula means that this is not a problem.SolutionDisplay the probabilities on a tree diagram.0.6A or Bon S10.70.30.350.4A or B on S2Not A or B on S2A or B on S2Not A orB on S10.65Not A or B on S2Let X be the event that the result on S1 was an A or B.Let Y be the event that the result on S2 was an A or B.7 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and ExamplesWe require the probability that an A or B grade is achieved on S1, given that an A orB grade has been achieved on S2. This is P(X Y).P( X and Y )P(X Y) P(Y )P( X and Y ) P(A or B grades on both modules) 0.6 0.7 0.42P(Y) 0.6 0.7 0.4 0.35 0.42 0.14 0.56P(X Y) P( X and Y ) 0.42 42 3 P(Y )0.56 56 4Independent eventsIf events A and B are independent then the outcome of A has no influence onthe outcome B. For example if you get a Head on the first throw of a coin, thisdoes not affect the probability that you get a Head on the second throw of acoin.If events A and B are independent:P(A and B) P(A) x P(B)You can see that this is consistent with the multiplication law involvingconditional probability, since if A and B are independent then P(A B) is thesame as P(A).Example 6In a game, two dice are thrown.Let A be the event the first die is a 6.Let B be the event the second die is a 6.Find the probability that you get 2 sixes.Solution1 1 1P 2 sixes P A and B P A P B 6 6 36This multiplication rule can be extended to more than 2 independent events.If there are three independent events A, B and C you multiply the threeprobabilities.Many students lose a lot of marks on probability questions because theirsolutions are difficult to follow. Show clearly how you are adding or multiplyingprobabilities, even when you work them out on your calculator.We can use conditional probability to find out whether two events areindependent.8 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and ExamplesExample 7The number of students selecting Maths and French is shown in the table.MathsNot MathsTotalFrench9615Not French211435Total302050Find the probability that:(i)a student chosen at random studies Maths(ii)a student studies Maths given that he/she studies French(iii) a student studies Maths given that he/she does not study French.What can you deduce from these results?SolutionLet M be the event that the student studies Maths.Let F be the event that the student studies French.30 3 50 5(i)P( M ) (ii)P( M F ) Look at the “French” column. Out ofthese 15 students, 9 are studying Maths.9 3 15 5Look at the “Not French” column. Out ofthese 35 students, 21 are studying Maths.(iii)P( M F ) 21 3 35 5In this case, P(M F ) P(M F ) P(M )This means that the probability that a student studies Maths is not affected by thechoice of whether they study French or not.Therefore the events M and F are independent.Tree diagrams can still be useful in problems involving independent events.Example 8A form tutor wants to find the probability that a student, Myles, will not be late oneither of Monday or Tuesday, given that he will be on time for at least one of thedays. From previous records he finds that:1P(Myles is late) 10Assume that Myles’ patterns of latenessare independent.9 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and ExamplesSolutionLet A be the event that Myles was on time for at least one of the days.Let B be the event that Myles was not late on either day.Monday110910Tuesday110Late910On TimeLate110LateOn Time910On Time111 1010 100199P(Late, On Time) 1010 100919P(On Time, Late) 1010 1009981P(On Time, On Time) 1010 100P(Late, Late) 1.10081The probability Myles will be on time on 2 consecutive days is.10099189 The probability that Myles will be late on exactly 1 occasion is100 100 100 50So the probability Myles will be late on 2 consecutive days isP( B and A)P( A)P(A) P(on time for at least one of thedays)We require P(B A) 1 – P(late on both days)199 1 100 100Has to satisfy both conditions. In thefirst Myles is on time on both days, inthe second Myles is on time for at leastone day. So Myles must be on time onboth days.P( B and A ) P(Myles was not late on either day and Myles was on time for at leastone of the days) P(Myles was on time on both days)10 of 1104/03/13 MEI

OCR S1 Probability 1 Notes and Examples81100P( B and A)P(B A) P( A) 819981 100 100 99Problems involving “at least one”In the last section the use of the complementary event in ‘at least one’problems was discussed. You can now solve these problems.Example 9(i)Find the probability of getting at least one six when you throw 5 dice.(ii)Find the probability of getting at least one head when you throw 4 coins.Solution(i)Let A be the event that you get at least one six when you throw 5 dice.So A' is the event that you get no sixes when you throw 5 dice.The probability of not getting a 6 is 56 .Using the multiplication rule:5 5 5 5 5 5 P(A') 6 6 6 6 6 6 55 5 So P(A) 1 – 0.598 (3 s.f.) 6 (ii)Round decimalsto 3 significantfigures and stateclearly how youhave rounded.Let A is the event of getting at least one head when you throw 4 coins,So A' is the event of getting no heads when you throw 4 coins.The probability of not getting a Head is 12 .Using the multiplication rule:41 1 1 1 1 1P(A') 2 2 2 2 2 16 1 15So P(A) 1 – 16 16For some practice in the work in this section, try the Probability matchingactivity and the Probability puzzle. Note that in these activities, the notationP( A B) means “the probability of A, or B, or both, occurring), and P( A B)means “the probability of both A and B occurring).11 of 1104/03/13 MEI

Maths by looking at the “Maths” row. P(E M) 10 1 30 3 You can find the probability of a student studying English given they do not study Maths by looking at the “Not Maths” row. P(E M') 15 3 20 4 From Example