February 29, 2012 Leap year day! Not another for another 4 years! AP Statistics B February 29, 2012 Leap year day! Not another for another 4 years!

AP Statistics B warm-up Wednesday, February 29, 2012 Do Problem 7, Chapter 16, p.382 What does the word “independently” mean in the last sentence of problem 7. When you’re ready, go on to the next slide.

AP Stats warm-up answers Wednesday, February 29, 2012 What 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.

Outline for today’s class Review problems 17-20, Chapter 15. Examine expected values Discrete v. continuous random variables

Comments on this type of lecture Don’t 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.

Before we begin, remember the general multiplication rule

Meaning of P(A) In problems involving choosing without replacements, it’s the first draw Examples: Odds of drawing a spade—13 spades in a deck of 52, so 13/52 = ¼ Odds of drawing an ace—4 aces out of 52, so 4/52 = 1/13. Odds of drawing anything BUT a heart—13 hearts, so 52-13=39 other cards out of 52, so probability is 39/52=3/4=0.75

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 52 It 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.

Subsequent probabilities may be affected You 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 aces Probability that you will draw a third space in a row We will now examine how these work with some of the homework problems.

Chapter 15, problem 17 (b-d) (drawing three cards) Sometimes it’s 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?

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 doesn’t count as a joker]) Probability of drawing a red card on FIRST draw is therefore 26/52, or ½.

Drawing red cards How have things changed for the second draw? How many red cards do we have left? We already drew one red card Therefore, 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 remaining Probability of drawing a red card on second draw is 25/51

Drawing red cards (still!) Third draw is the same: We are down to 24 red cards There are a total of 50 cards left in the deck (we already drew two red cards Probability 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%

17(c): drawing no spades OK, so you’ve 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.

Problem 17(c): you get no spades Similar 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 = 1-¼ = ¾ = 0.75

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 aren’t spades 50 cards left Probability of no spades on third=37/50 Probability of no spades = (3/4)(38/51)(37/50)= 0.414 = 41.4%

17(d): draw at least one ace More 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. Let’s look at the odds of drawing exactly 1 ace first.

Drawing exactly 1 acs Here’s a table that summarizes the three possibilities of drawing exactly one ace: 1st draw 2nd draw 3rd draw 1 Ace no ace 2 3

Analysis of probability for drawing exactly one ace 1st possibility: A, followed by 2 non-aces: (4/52)(48/51)(47/50)=well, we’ll see in coming slides. 2nd possibility: you draw an ace only on the 2nd draw, with non-aces on draws 1 and 3 3rd possibility: you don’t draw an ace until the 3rd draw, with non-aces in draws 1 and 2. Next, I’ll show you how to catalog these possibilities and figure out the total odds.

Summary for all 7 possibilities Pattern 1st draw 2nd draw 3rd draw A00 04/52 48/51 47/50 0A0 48/52 04/51 00A 47/51 04/50 0AA 03/50 A0A AA0 03/51 48/50 AAA 02/50

Summary of probabilities Pattern 1st draw 2nd draw 3rd draw A00 0.076923 0.941176 0.94 0A0 0.923077 0.078431 00A 0.921569 0.08 0AA 0.06 A0A AA0 0.058824 0.96 AAA 0.04

Multiply all columns and add Pattern 1st draw 2nd draw 3rd draw Product A00 0.076923 0.941176 0.94 0.068054 0A0 0.923077 0.078431 00A 0.921569 0.08 0AA 0.06 0.004344 A0A AA0 0.058824 0.96 AAA 0.04 0.000181 Total 0.217376

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: 47/51 Draw 3: 46/50. Multiply them all together and you get………………….

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.

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

Problem 19 12 batteries altogether (a) 1st two you pick are good: 5 are dead Therefore, 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 you’ll 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)

What looks hard isn’t necessarily so: read carefully 19(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.”

How to model 19(d) If you look at a representation of what happens, with B=bad and G=good, it looks like this: BBBBG Easy, right? (5/12)(4/11)(3/10)(2/9)(7/8) The book says that this comes out to 0.009.

Problem 20 Much like problems 17 and 18, except now considerably more complex We have 20 shirts (but the problem doesn’t tell you that directly) broken down as follows: 4 medium 10 large 6 XL Before going on to the next slide, calculate the initial odds of getting a medium, or a large, or an XL.

Answers to initial probabilities Medium = 4/20 = 0.20 Large = 10/20 = 0.50 XL = 6/20 = 0.30 All add up to 1, right? Note that the problem requires that you get 2 mediums, one for you and one for your sister. (It isn’t well-worded, though it’s not really ambiguous.)

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)=0.140

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

Chapter 16: Expected values Covered this in previous lecture. Now to get the fancy formula: E(X)=∑x∙P(x) A more familiar form of the equation would be to use subscripts as follows: E(X)=∑xi∙P(xi)

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

New concept: discrete v. continuous “Discrete” 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 them “Continuous” means the variable can take on an infinite number of outcomes (think about this; we’ll start here tomorrow)

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