TIMES 3Virtual Laboratory in Probability and StatisticsSlide 3CountingExamplesSlide 6Slide 7Calculator PracticeThe Binomial DistributionThe Binomial Random VariableBinomial ProbabilityCentral Tendency and VariabilitySlide 13Slide 14Slide 15Chuck – a - LuckVirtual Lab: Pepys’ ProblemSlide 18Slide 19Slide 20TIMES 3TIMES 3Technological Integrations in Mathematical Environments and StudiesJacksonville State University 2010Virtual Laboratory in Virtual Laboratory in Probability and Probability and StatisticsStatisticsGoogle Virtual Lab UAHGo to the AppletsLet’s play Roulette!1.Find the mean for each bet.2.Find the standard deviation for each bet.3.What information is given by these statistics?4.If you had a limited amount of money, but you wanted to play roulette for a long time, what bet would accomplish your goal?5.Would your choice of bet make any difference on your net winnings in the long run?6.Choose any bet and run the simulation 500 times. Based on your mean did you leave a winner?7CountingCountingPermutation: An ordered arrangement of all or part of a set of objects.These can almost always be solved using the multiplication rule.The number of arrangements of n things is n! (By Definition 0! = 1)nPr, the number of arrangements of n things taken r at a time is: )!(!rnnExamplesExamplesSix photos can be arranged on a shelf _____ ways.The top two finishers in order in an 8 dog race can come in _____ ways. (Exacta)A Trifecta is the top 3 in order, a Superfecta is the top 4 in order. Count these.Three different flavored scoops of ice cream can be arranged on a cone _____ ways.There are ____ different possible 3 scoop cones using 31 different flavors if order mattersCountingCountingCombination: An unordered selection of all or part of a set of objects.There will always be fewer combinations than permutations given the same n and r.nCr, the number of ways to select r objects from n is given by:!)!(!rrnnExamplesExamplesThere are _____ possible foursomes out of 7 golfers.How many ways are there to fit 6 kids into a van if you don’t care where they sit? _____The top two finishers in an 8 dog race without regard to order can come in _____ ways. (Quinella)There are _____ different possible 3 scoop cones using 31 different flavors is order doesn’t matter.Compare the years required to taste every 3 scoop cone using 31 flavors when order matters and when it doesn’t.Calculator PracticeCalculator Practice7! 14!(Factorials are the fastest growing functions.)7P0 7P7 12P5 10P67C0 7C7 5C1 20C8The Binomial DistributionThe Binomial DistributionA binomial experiment has:1. A fixed number of trials, n2. Two outcomes: success and failure3. A constant probability of success, p(If there are only two outcomes and the probability of success is p, then the probability of failure is 1 – p.)4. Independent trialsThe Binomial Random The Binomial Random VariableVariableX = the number of successes out of n trialsThe value of X must be an integer between 0 and n inclusive.Binomial ProbabilityBinomial ProbabilityGiven a binomial random variable, x, in a distribution with parameters n and p:xnxxnppCXP )1()(Recall the example of tossing 3 coins and guessing the outcome.Central Tendency and Central Tendency and VariabilityVariabilityThe mean of the binomial distribution is given by:μ = npThe standard deviation of the binomial distribution is given by:)1( ppn At a carnival game you are offered the opportunity to bet one dollar on a number from one to six, on a single roll of two dice. If your number comes up on one die, you win $2. If it comes up on both dice, you win $5. Only if the number does not appear on either die do you lose your dollar.If we think this one out as a binomial distribution, there are n = 2 trials in the die rolls. Let’s say a success is that we roll a one. Then the probability of success is p =1/6 and the probability of failure is 5/6. The random variable, x, counts the number of successes.x P(x)0 2C0 (1/6)^0 (5/6)^2 = .6941 2C1 (1/6)^1 (5/6)^1 = .2782 2C2 (1/6)^2 (5/6)^0 = .028We would come to the same probabilities by counting the 25 ways to lose and the 11 ways to win out of the 36 outcomes.Now that we know the probabilities for rolling a one either once or twice or not at all, we can define a random variable, Y, to see if the game is fair.$Y P(Y)-1 .6942 .2785 .028E(x) = 0 SD = 1.58This game is fair.Note that X is binomial and Y is not. Why?Chuck – a - LuckChuck – a - LuckIt costs $1 to play this game.Pick a number from 1 to 6.The house rolls 3 fair dice.If your number shows up once, you win $1.If your number shows up twice, you win $2.If your number shows up three times, you win $3.So, do you want to play?Virtual Lab: Pepys’ Virtual Lab: Pepys’ ProblemProblemIn 1693, Samuel Pepys (a member of the British Navy and later Parliament) asked Isaac Newton whether it is more likely to get at least one ace in 6 rolls of a die or at least two aces in 12 rolls of a die. This problems is known a Pepys' problem; naturally, Pepys had fair dice in mind.From the Virtual Lab home page, go to the list of Applets.Under the Dice & Spinners heading, select Dice Experiment.With n=6 dice, run the experiment 2000 times and find the binomial probability of rolling Z = 1 ace on 6 dice. “Distribution” gives the classical probability. “Data” gives the empirical probability based on your simulation. Remember to change the random variable from the default Y = sum to Z = number of aces.7P(one ace on 6 dice) = ________________ 7Now with n=12, run the simulation 2000 times and find the binomial probability of rolling Z = 2 aces on 12 dice. Compare the results.7P(two aces on 12 dice) = _______________So, what’s the answer to Pepys’ question? Calculate the following probabilities. (Use the formula or the applet Distribution values.)7P(3 aces with 18 dice) = __________ P(4 aces with 24 dice) = __________ P(5 aces with 30 dice) = __________7Make a conjecture that extends Pepys’ problem to more than 30 dice.Questions?Japanese, Korean, and Chinese dice represent the ace as a large red