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AP Statistics BFebruary 29, 2012Leap year day! Not another for another 4 years!1AP Statistics B warm-upWednesday, February 29, 2012Do Problem 7, Chapter 16, p.382What does the word independently mean in the last sentence of problem 7.

When youre ready, go on to the next slide. 2AP Stats warm-up answersWednesday, February 29, 2012What are the probabilities of getting the contracts? 30% for the first, 60% for the second. So the expected value is the product of the probabilities times each contract value (.3)($50,000)+(.6)($20,000) =$15,000+$12,000=$27,000.Independently here means that the probability of getting the second contract is unrelated to the probability of getting the first. Alternatively stated, winning the first contract neither increases nor decreases the likelihood that the company will win the second, smaller contract.3Outline for todays classReview problems 17-20, Chapter 15.Examine expected values Discrete v. continuous random variables

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Comments on this type of lectureDont worry about writing everything down. You can go to my home page on garfieldhs.org, look under Statistics(all classes) and download the presentation for today (colleges are doing this a lot).Follow along the general path of the ideas. You can sweat the details later. Give me feedback on what would make the presentation better for you.5

Before we begin, remember the general multiplication rule

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Meaning of P(A)In problems involving choosing without replacements, its the first drawExamples:Odds of drawing a spade13 spades in a deck of 52, so 13/52 = Odds of drawing an ace4 aces out of 52, so 4/52 = 1/13.Odds of drawing anything BUT a heart13 hearts, so 52-13=39 other cards out of 52, so probability is 39/52=3/4=0.757

Understanding P(B|A) (B|A) means that something has happened to change the probability of B from happening.In these cases, that something is that a card has been drawn. It can have two effects:It ALWAYS decreases the deck by 1 card, so your denominator is now 51, not 52It MAY affect the remaining odds. For example, if the first odds are drawing an ace, the aces in the deck are reduced, and the total cards are reduced by one, so the odds drawing the next ace on the second hand is 3/51.8

Subsequent probabilities may be affectedYou could continue to draw cards, and the probabilities will change and may be affected by P(A) and subsequent odds.Examples:Probability that the third card is NOT an ace after drawing 2 acesProbability that you will draw a third space in a rowWe will now examine how these work with some of the homework problems.9

Chapter 15, problem 17 (b-d)(drawing three cards)Sometimes its better to verbalize rather than look for formulas.With (b), all 3 cards you draw are red.# of red cards in the deck: 26 (13 hearts, 13 diamonds)So for the first card you draw, what are the odds that it will be red?10

Problem 17(b)(drawing red cards only)Probability = # of red cards available divided by # of ALL cards available# of all cards= 52 on first draw (13 of each suit; no jokers [Jason Romero doesnt count as a joker])Probability of drawing a red card on FIRST draw is therefore 26/52, or .11

Drawing red cardsHow have things changed for the second draw?How many red cards do we have left?We already drew one red cardTherefore, we have 26-1=25 red cards still available to be drawn.How many cards altogether do we have left?Again, we drew one card, so we have 52-1=51 cards remainingProbability of drawing a red card on second draw is 25/5112

Drawing red cards (still!)Third draw is the same:We are down to 24 red cardsThere are a total of 50 cards left in the deck (we already drew two red cardsProbability of red card on third draw is therefore 24/50=12/25, if you want to simplify.Total probability is PRODUCT of the three: (1/2)(25/51)(12/50)=.118 = 11.8%13

17(c): drawing no spadesOK, so youve now seen how to approach this.Work in groups on 17(c): You are dealt a hand of three cards, one at a time. Find the probability that.(c) you get no spades.After 8-10 minutes, report out, and advance to the next slide for my explanation.14

Problem 17(c): you get no spadesSimilar in approach to the last problem.First draw: the probability of getting no spaces is 39/52. (if you like, you can look at this as the complement of getting spades.you have a 13/52 chance of getting spades, and the complement is therefore 1-13/52 = 1- = = 0.75 15

Problem 17(c): you get no spades (slide 2)For second draw, there are still 13 spades, and 38 non-spades. 51 cards total (13+38), so probability = 38/51.Third draw: similar to second draw:37 cards arent spades50 cards leftProbability of no spades on third=37/50Probability of no spades = (3/4)(38/51)(37/50)= 0.414 = 41.4%

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17(d): draw at least one aceMore complicated, because there are many possible outcomes. You could draw 1 ace, 2 aces, or three aces. The total probability will be the sums of all the individual probabilities.Lets look at the odds of drawing exactly 1 ace first.17

Drawing exactly 1 aceHeres a table that summarizes the three possibilities of drawing exactly one ace:1st draw2nd draw3rd draw1Aceno aceno ace2no aceAceno ace3no aceno aceAce18

Analysis of probability for drawing exactly one ace1st possibility: A, followed by 2 non-aces: (4/52)(48/51)(47/50)=well, well see in coming slides.2nd possibility: you draw an ace only on the 2nd draw, with non-aces on draws 1 and 33rd possibility: you dont draw an ace until the 3rd draw, with non-aces in draws 1 and 2.Next, Ill show you how to catalog these possibilities and figure out the total odds.19

Summary for all 7 possibilitiesPattern1st draw2nd draw3rd drawA0004/5248/5147/500A048/524/5147/5000A48/5247/5104/500AA48/524/5103/50A0A04/5248/5103/50AA004/523/5148/50AAA04/523/5102/5020

Summary of probabilitiesPattern1st draw2nd draw3rd drawA000.0769230.9411760.940A00.9230770.0784310.9400A0.9230770.9215690.080AA0.9230770.0784310.06A0A0.0769230.9411760.06AA00.0769230.0588240.96AAA0.0769230.0588240.0421

Multiply all columns and addPattern1st draw2nd draw3rd drawProductA000.0769230.9411760.940.0680540A00.9230770.0784310.940.06805400A0.9230770.9215690.080.0680540AA0.9230770.0784310.060.004344A0A0.0769230.9411760.060.004344AA00.0769230.0588240.960.004344AAA0.0769230.0588240.040.000181Total0.21737622

Problem 18 (similar to 17, often with the complements, or flip sides)For example, 18(a) has no aces. Flip side of 17(c), which was no spades.Draw 1: 48/52. Draw 2: 47/51Draw 3: 46/50.Multiply them all together and you get.23

Problem 18(b):all heart(s)13 hearts. Just keep taking them away as you reduce the deck.Work on this at your table for a few minutes and see what you come up with. My analysis is on the next slide.

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Answer to 18(b)First odds of drawing hearts: 13/52.Second draw odds: 12/51Third draw odds: 11/50.Altogether: (13/52)(12/51)(11/50) = 25

Problem 1912 batteries altogether5 are deadTherefore, 7 are good. (a) 1st two you pick are good:(7/12)(6/11)=0.318(b) At least one of the first three works. Set this up like you did the at least one ace of 17(d), except youll have different odds (answer is 0.955)(c) The first 4 all work: (7/12)(6/11)(5/10)(4/9)= something small (7.1% per the book)26

What looks hard isnt necessarily so: read carefully19(d) sounds awful: you have to pick 5 batteries to find one that works.Sounds like something worse than the at least one ace from problem 17.What it ACTUALLY means is you have to pick 4 batteries BEFORE you find one that works.27

How to model 19(d)If you look at a representation of what happens, with B=bad and G=good, it looks like this: BBBBGEasy, right?(5/12)(4/11)(3/10)(2/9)(7/8)The book says that this comes out to 0.009.28

Problem 20Much like problems 17 and 18, except now considerably more complexWe have 20 shirts (but the problem doesnt tell you that directly) broken down as follows:4 medium10 large6 XLBefore going on to the next slide, calculate the initial odds of getting a medium, or a large, or an XL.29

Answers to initial probabilitiesMedium = 4/20 = 0.20Large = 10/20 = 0.50XL = 6/20 = 0.30All add up to 1, right?

Note that the problem requires that you get 2 mediums, one for you and one for your sister. (It isnt well-worded, though its not really ambiguous.)30

Analysis of problem 20(a) says first two are wrong sizes, i.e., either L or XL. Together those are 16 (10Ls + 6XLs), so probability is (16/20)(15/19) = 0.632(b) says the first medium shirt you grab is the third. In other words , you got the wrong sizes the first 2 times, so you can use the results from (a). The odds of getting a medium on the 3rd grab is 4/18, so answer is 0.632(4/18)=0.14031

20(c) and (d)The first four shirts you grab are all extra largeThis one is easy: (6/20)(5/19)(4/18)(3/17)= 0.003(d), by comparison, is quite complicated: 17 outcomes by my calculationsIll give extra credit worth a homework assignment for anybody who completes it by next Monday32

Chapter 16: Expected valuesCovered this in previous lecture.Now to get the fancy formula: E(X)=xP(x)A more familiar form of the equation would be to use subscripts as follows: E(X)=xiP(xi)33

Expected values from the lottery of yesterdayLets use this formula with the lottery we did yesterday to see how it works.E(X)=xiP(xi) = 5(.7)+10(0.1)+20(.2)= $8.50Here, i=1 corresponds to the 14 $5-bills, i=2 to the 2 $10-bills, and i=3 to the 4 $20-bills.

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New concept: discrete v. continuousDiscrete means that you can list all the outcomes, e.g.SAT scores (all SAT scores are integers, so you can list each of them)Heights of a population (difficult, but you COULD list themContinuous means the variable can take on an infinite number of outcomes (think about this; well start here tomorrow)35

Homework tonight! Due tomorrow!Chapter 16, problems 1, 2, 4, 5, 8Grades are due next week, and all I may have for you is homework, so be sure its all in by Monday!36