IFYMaths2BusinessProbability2.pptx

Introduction to ProbabilityIntroduction to ProbabilityIntroduction to ProbabilityYou have calculated probabilities in GCSE work and used tree diagrams to solve some probability problems.We will now revise and extend probability work,starting with some definitions.An experiment or trial has a number of outcomes.These are the results from the experiment or trial.An event is a particular result or set of results.We usually use the word event for the outcomes we are interested in.e.g.1.If I toss a coin,the possible outcomes are a head or a tail.I could define H as the event of getting a head.e.g.2.If I roll a die,the possible outcomes are the numbers 1,2,3,4,5 or 6.An event could be getting an even number.Introduction to ProbabilityLets take the example of the die.When we roll the die,there are 6 possible outcomes,all equally likely,and these form the possibility space.You know that the probability of the event“getting an even number”isTo see how we get this result formally,we need one more definition.For equally likely outcomes,the probability of an event,E,is given byThere are 3 even numbers and 6 possible outcomes so we get the answer 3 out of 6,or .P(E)=number of ways E can occurnumber of possible outcomesIntroduction to Probabilitye.g.1.Two dice are rolled and the sum of the numbers on the uppermost faces are added.What is the probability of getting a 7?Thinking about this we realise that we can get 7 in a number of ways.For example,1 on the 1st die and 6 on the 2nd or 2 on the 1st and 5 on the 2nd.Also,what about 5 on the 1st and 2 on the 2nd?There are also other possibilities.These possibilities are equally likely so we can use the definition to find P(7)but its not easy to see in how many ways the event can arise.To solve this problem we can draw a possibility space diagram and the answer is then easy to see.Introduction to ProbabilitySolution:121110987111098761098765987654328765437654654321654321+1st die2nd dieWe now count the number of 7s.e.g.1.Two dice are rolled and the sum of the numbers on the uppermost faces are added.What is the probability of getting a 7?Introduction to ProbabilitySolution:121110987111098761098765987654328765437654654321654321+1st die2nd dieWe now count the number of 7s.and divide by the total number of possibilities.So,e.g.1.Two dice are rolled and the sum of the numbers on the uppermost faces are added.What is the probability of getting a 7?Introduction to Probability Outcomes are the results of trials or experiments.SUMMARYAn event is a particular result or set of results.A possibility space is the set of all possible outcomes.For equally likely outcomes,the probability of an event,E,is given byP(E)=number of ways E can occurnumber of possible outcomesIntroduction to ProbabilityExercise1.Two dice are rolled and the score is defined as the product of the numbers showing on the uppermost faces.Write out the possibility space and use it to find the probability of scoring 12 or more.2.Four coins are tossed.Write out the possibility space in the form of a list of all possible outcomes and use it to find the probability of 3 heads.Introduction to Probability1.Two dice are rolled and the score is defined as the product of the numbers showing on the uppermost faces.Write out the possibility space and use it to find the probability of scoring 12 or more.Solution:3630241812630252015105242016128418151296321121086426543654321654321 1st die2nd dieP(12 or more)Introduction to Probability1.Two dice are rolled and the score is defined as the product of the numbers showing on the uppermost faces.Write out the possibility space and use it to find the probability of scoring 12 or more.Solution:3630241812630252015105242016128418151296321121086426543654321654321 1st die2nd dieP(12 or more)Introduction to ProbabilitySolution:2.Four coins are tossed.Write out the possibility space in the form of a list of all possible outcomes,for example,H,H,H,H,and use it to find the probability of 3 heads.H,H,H,HT,H,H,H H,T,H,H H,H,T,H H,H,H,TT,T,H,HT,H,T,HT,H,H,TH,T,T,HH,T,H,TH,H,T,TT,T,T,HT,T,H,TT,H,T,TH,T,T,TT,T,T,TIntroduction to ProbabilitySolution:P(3 Heads)2.Four coins are tossed.Write out the possibility space in the form of a list of all possible outcomes,for example,H,H,H,H,and use it to find the probability of 3 heads.H,H,H,HT,H,H,H H,T,H,H H,H,T,H H,H,H,TT,T,H,HT,H,T,HT,H,H,TH,T,T,HH,T,H,TH,H,T,TT,T,T,HT,T,H,TT,H,T,TH,T,T,TT,T,T,TIntroduction to ProbabilityTree Diagrams1st chocolate2nd chocolateRNRNNRIntroduction to Probabilitye.g.There are 5 chocolates left in a box all looking the same.3 are raspberry creams(R),2 are nougats(N)(a)Draw a tree diagram to show the probabilities if 2 chocolates are taken at random.(b)Find the probabilities of(i)both being nougats(ii)the 1st being a raspberry cream and the 2nd being a nougat(iii)one of each typeWe use the word event in statistics to refer to a possible result from a trial or experiment.e.gs.A seed growing into a red flower.A six showing when we throw a die.A tree diagram shows the probabilities of 2 or more events.Introduction to Probability1st chocolate2nd chocolate5 chocolates:3Rs,2NsSolution:Tip:To get to this branch,weve come up here.2 of the 5 chocolates are NsThe 1st chocolate can be R or N so we need 2 branches3 of the 5 chocolates are RsRNRIntroduction to Probability1st chocolate2nd chocolateSolution:Tip:To get to this branch,weve come up here.RNSo an R has gone.R5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNRN5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNBoth the Ns are leftRN5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNTo reach here,an N has gone so all 3 Rs are left.RNR5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNand only 1 N is leftRNNR5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNRNNWe need to multiply to find the probability of reaching the end of each branch.R5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNRNNR5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNRNNR5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNRNNR5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNRNNR5 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNRNNRTip:The four probabilities added together equal 15 chocolates:3Rs,2NsIntroduction to Probability1st chocolate2nd chocolateSolution:RNRNNR5 chocolates:3Rs,2NsIntroduction to Probability(b)Find the probabilities of(i)both being Ns(ii)the 1st being R and the 2nd N:Solution:(i)1st chocolate2nd chocolateRNRNNRIntroduction to Probability(b)Find the probabilities of(i)both being Ns(ii)the 1st being R and the 2nd N:Solution:(i)(ii)1st chocolate2nd chocolateRNRNNRIntroduction to Probability(b)Find the probabilities of(i)both being Ns(ii)the 1st being R and the 2nd N(iii)one of each typeSolution:(i)(ii)1st chocolate2nd chocolateRNRNNRIntroduction to Probability(b)(iii)probability of one of each typeSolution:There are 2 ways to get one of each type:1st chocolate2nd chocolateRNRNNRIntroduction to Probability(b)(iii)probability of one of each typeSolution:There are 2 ways to get one of each type:1st chocolate2nd chocolateRNRNNRIntroduction to Probability(b)(iii)probability of one of each typeSolution:There are 2 ways to get one of each type:or1st chocolate2nd chocolateRNRNNRWe need to add to find the probability of moving along one branch or another.+Introduction to ProbabilitySUMMARYA tree diagram shows the probabilities of more than one event.To find probabilities for a branch always assume the event on the branch leading to it has occurred.To find the probability of 2 events,multiply the individual probabilities.continuede.g.Assumes A has occurredABABe.g.Introduction to ProbabilitySUMMARYTo find the probability of 1 event or another,add the individual probabilities.e.g.ABCDP(A and C or B and D)Introduction to ProbabilityExercise2.Components for a machine are supplied by 3 factories A,B and C.45%come from A,25%from B and the rest from C.Data shows that 4%of those from A,5%of those from B and 2%of those from C are faulty.Draw a tree diagram showing all the probabilities and use it to find the probability that a randomly selected component is faulty.(a)The 1st is replaced before the 2nd is taken,and1.There are 4 red and 6 blue biros in a drawer.Two are taken out at random.Draw a tree diagram showing the probabilities of the different events if(b)The 1st is not replaced before the 2nd is taken.In each case find the probability of one red and one blue.Introduction to Probability1.There are 4 red and 6 blue pens in a drawer.Two are taken out at random.Draw a tree diagram showing the probabilities of the different events if(a)The 1st is replaced before the 2nd is taken.1st 2ndRBRBBRP(one red and one blue)=Introduction to Probability1st 2ndRBRBBRP(one red and one blue)=(b)The 1st is not replaced before the 2nd is taken.1.There are 4 red and 6 blue pens in a drawer.Two are taken out at random.Draw a tree diagram showing the probabilities of the different events ifIntroduction to ProbabilityACFBNFNFN2.Components for a machine are supplied by 3 factories A,B and C.45%come from A,25%from B and the rest from C.Data shows that 4%of those from A,5%of those from B and 2%of those from C are faulty.Introduction to ProbabilityACFBNFNFN