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APSTAT SECTION IV PROBABILITY

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CHAPTER 14 From Justin To Kelly (or from randomness to probability)

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Randomenss and Probability Basics What is Random? Individual outcomes are unpredictable in the short run In the long run, however, outcomes are regular AND predictable (LOLN - Law of Large Numbers) Lets do a simulation Flip 10, 100, 1000, 10000 coins Find % of heads to nearest whole % sum(randint(0,1,10))

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Random Simulation Short Run Long Run 10 Flips 100 Flips 999Flips

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What is Probability Over a HUGE number of trials (probability is Long- Term), the proportion of times an outcome would occur. Typically expressed by P= and a range from 0 to 1 0 being never ever happens 1 being always happens We can only ESTIMATE real-world probabilities Can be expressed as a %, but not as cool.

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Models of Probability Two Main Thangs LIST all possible outcomes ASSIGN a probability to each outcome ie. Year in school probability @ WPS FROSHSOPHJUNIORSENIOR 60/230 60/230 70/230 40/230 P=.261.261.304.164 Should add up to 1.0, but may be a bit off due to ROUNDING ERROR

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Vocab Time Sample Space – S – Set of all possible outcomes Event – Any outcome or set of outcomes ie. Freshman ie. Juniors AND seniors

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CRAPS! – Roll Them Bones! Disclaimer – Gambling can be dangerous and addictive, plus over the LONG RUN, the casino always wins. So don’t gamble, buy a casino! Sample space when 2 die are rolled: 123456 11,11,21,31,41,51,6 22,12,22,32,42,52,6 33,13,23,33,43,53,6 44,14,24,34,44,54,6 55,15,25,35,45,55,6 66,16,26,36,46,56,6 36 potential outcomes

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CRAPS! – Roll Them Bones! Event: Rolling a 7 when pips are added “ProbSpeak”: P(Roll a 7) 123456 11,11,21,31,41,51,6 22,12,22,32,42,52,6 33,13,23,33,43,53,6 44,14,24,34,44,54,6 55,15,25,35,45,55,6 66,16,26,36,46,56,6 P(Roll 7) = 6/36 =.167

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CRAPS! – Roll Them Bones! Event: Rolling a 8 when pips are added “ProbSpeak”: P(8) 123456 11,11,21,31,41,51,6 22,12,22,32,42,52,6 33,13,23,33,43,53,6 44,14,24,34,44,54,6 55,15,25,35,45,55,6 66,16,26,36,46,56,6 P(8) = 5/36 =.139

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CRAPS! – Roll Them Bones! Event: Rolling a Hard 8 (two 4’s) “ProbSpeak”: P(Hard 8) 123456 11,11,21,31,41,51,6 22,12,22,32,42,52,6 33,13,23,33,43,53,6 44,14,24,34,44,54,6 55,15,25,35,45,55,6 66,16,26,36,46,56,6 P(8) = 1/36 =.028

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More Sample Space Same problem can have different “look” at sample space: If in craps, if all we care about are “pips” S = (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) P(2) = 1/36 =.028 P(3) = 2/36 =.056

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Key Points - Independence One outcome does not affect another outcome ie. If I roll a 6 with one die, it won’t affect my chances of rolling a 6 with the second die WITH Replacement Example: pick a card from a deck, put it back and pick another…P(2 Aces) WITHOUT Replacement Example: pick two cards from a deck without replacing the first card…P(2 Aces)

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Rules o’ Probability Probability of an event is between 0 and 1 PROBSPEAK: All possible probabilities add up to 1 PROBSPEAK: Probability of an event NOT happening is 1 minus the probability of the event happening. The probability of an event NOT happening is called the COMPLIMENT of an event and is written A c PROBSPEAK:

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Rules o’ Probability Continued If two events are disjoint (mutually exclusive), they have no outcomes in common. For example, in craps, rolling a 5 AND a 7 is disjoint, one roll can’t produce both outcomes. Therefore (for disjoint events): AND (for disjoint events)…….. S A B

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Rules o’ Probability Continued Again If two events are NOT disjoint (not mutually exclusive) but ARE independent. For example, roll 2 dice Event A: Die 1 Shows a 6 P(A)=1/6 Event B: Die 2 Shows a 6 P(B)=1/6 P(A and B)= P(A)*P(B) = 1/6 * 1/6 = 1/36 =.028ish S A B A&B

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Disjoint Events Are NOT Independent Hurting your brain? Just think…If I roll two die and add up the pips, what are the chances that I get a 5 and a 7. That’s why (in disjoint events) P(A and B)=0 S Roll 5 Roll 7

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CHAPTER 15 Probability Goes Crazy with the Cheez Whiz

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Tree Diagrams ie. Flip 3 Coins, Count # of Heads H T H T H T H T H T H T H T FLIP 1 FLIP 3 FLIP 2 OUTCOMES 3H, 0T 2H, 1T 1H, 2T 2H, 1T 1H, 2T 0H, 3T

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3 Flip outcomes 3 Heads2 Heads1 Head0 Heads 1/8 3/8 3/8 1/8 P=.125.375.375.125

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More (and evilererer) Probability Rules Addition Rule – 3 or more disjoint events S A B C

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Addition Rule – Non-Disjoint Events Find P(A or B) If I do P(A)+P(B) the area A&B gets counted twice To make it work, I do P(A) + P(B) and then subtract one P(A&B) S A B A&B

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Example – P(Male Or Senior Citizen) P(Male)=0.5 P(65+)=0.2 Assume independence for Maleness & Oldness S A B A&B =.5 +.2 - (.5)(.2) =.7 -.1 =.6

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Joint Events Not always independent - Can’t assume Example – Survey of music tastes at WPS Probability of student liking hip-hop (A) P(A)=0.5 Probability of student liking rap (B) P(B)=0.4 Think! Isn’t there a decent chance that people who like hip hop may be more likely to like rap as well…. Proportion of students who like BOTH rap and hip hop P(A and B)=0.3

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Joint Events - Continued Probability of student liking hip-hop (A) P(A)=0.5 Probability of student liking rap (B) P(B)=0.4 Proportion of students who like BOTH rap and hip hop P(A and B)=0.3 S Hip Hop RAP BOTH 0.3 0.2 0.1 0.4

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Joint Events – Same thing – Using a table Probability of student liking hip-hop (A) P(A)=0.5 Probability of student liking rap (B) P(B)=0.4 Proportion of students who like BOTH rap and hip hop P(A and B)=0.3 LikeDislikeTotal Like0.30.5 Dislike Total0.41.0 Rap Hip-Hop 0.1 0.5 0.6 0.4 0.2 S Hip Hop RAP BOTH 0.3 0.2 0.1 0.4

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Conditional Probability Main Idea: Probability can change if we know some other event has occurred World Poker Championships Flushes are good – All same suit You get 2 cards that are secret, then 5 cards are dealt for the community You make the best 5-card hand you can

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World Poker Championships My Hand Community Cards ♠ ♠ ♠ ♠ ♦ ♥ Wow, I’m close to a flush! What is the probability that the last card (the river) is a ♠? Overall, the chance of a ♠ is 13/52 or.25, but we already know what 6 cards are and that 4 of them are ♠s… Find Probability(♠ given that 4 of 6 visible cards are ♠s) ProbSpeak: P(Spade 4 of 6 visible spades) Think: ?

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General Rule for Any Two Events P (A and B) = P(A)P(BІA) Example: Probability of getting 2 aces in two successive draws (no replacement) P(Ace on 1 st and Ace on 2 nd ) =P(Ace on 1 st )P(Ace on 2 nd І Ace on 1 st ) =4/52*3/51=0.0045 Notice if replacement (independence), the formula still works since P(Ace on 2 nd І Ace on 1 st )= P(Ace on 1 st ) Therefore 4/52*4/52=0.0059

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Definition for Conditional Probability P (A and B) = P(A)P(BІA) Take this old Formula and solve for P(BІA)

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Using Decision Trees The following information gives information on DVD players sold by a certain electronics store Let B1 = Event that Brand 1 is purchased Let B2 = Event that Brand 2 is purchased Let E = Event that Warranty is purchased Therefore: P(B1)=.7 P(B2)=.3 AND!!!!!! P(E І B1)=.2 P(E І B2)=.4 Percent of Customers Purchasing Of those who purchase, Percentage who purchase extended warranty Brand 17020 Brand 23040

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Using Decision Trees - Continued P(B1)=.7 P(B2)=.3 P(E І B1)=.2 P(E І B2)=.4 B1 B2 E Not E E.7.2.3.4.8.6 (.7)(.2)=.14 (.7)(.8)=.56 (.3)(.4)=.12 (.3)(.6)=.18

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Using Decision Trees – Continued 2 NOW ANSWER QUESTIONS!!!! B1 B2 E Not E E.7.2.3.4.8.6 (.7)(.2)=.14 (.7)(.8)=.56 (.3)(.4)=.12 (.3)(.6)=.18 What proportion of DVD purchasers also purchased the warranty? P(B1 and E) + P(B2 and E) P(E)=.14+.12=.26

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Using Decision Trees – Bayes’s Rule What is probability of B1 given E B1 B2 E Not E E.7.2.3.4.8.6 (.7)(.2)=.14 (.7)(.8)=.56 (.3)(.4)=.12 (.3)(.6)=.18 What proportion of DVD purchasers also purchased the warranty? P(B1 І E) =.14/.26 P(B1 and E) =.14 P(E)=.14+.12=.26 P(B1 І E) = 0.539

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CHAPTER 16 Random Variables

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Discrete Random Variables Discrete???? Just means that there are a reasonable (countable) number of options. What do we do with them? List outcomes and then probabilities Answer questions Easy as pie…

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Example – Rolling 2 dice Outcomes (X) 23456789101112 Probability.027.055.083.121.139.167.139.121.083.055.027 Find P(X>9) = P(10)+P(11)+P(12)=.083+.055+.027 =.165 Find P(X 9) = P(9)+P(10)+P(11)+P(12) =.121+.083+.055+.027 =.286 Find P(5

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Continuous Random Variables Continuous? Not countable Example, think of all possible decimals between 0 and 1…Boy that’s a lot! If we threw down a histogram of a gazillion random numbers between 0 and 1, we’d get this: 0.0 1.0 Uniform Distribution Density Curve Area underneath is 1.0

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Find Probabilities (Just area of rectangle): P(X>0.8) = 0.0 1.0 0.8 0.2

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Find Probabilities (Just area of rectangle): P(.2

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Check this out! What is the probability P(X=0.8)? 0.0 1.0 0.8 0! ZERO! ZIP! NADA! Why? Area of a straight line is Zero, Yeah?

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SO…… P(X>0.8) is the same as P(X 0.8) With Continuous Variables, It does not matter which one you use… Cool, Huh??? 0.0 1.0 0.8

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What’s the sassiest density curve? NORMAL DISTRIBUTION. YEAH!!!

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Male Height N(68,2) Let X=Ht in Inches Find P(X>71) Normalcdf(71,100000000,68,2) P(X>71)=_____

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Means and Variances of Random Variables Example: 2005 AP Stat Scores Outcome12345 Probabiity.13.23.25.19.20 Remember x-bar is a sample mean, but we are talking about the entire population of AP Stat test takers, so we must use (population mean) x = 1(.13)+2(.23)+3(.25)+4(.19)+5(.20) = 3.1

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Variances of Random Variables Recall: Variance is (Standard Deviation) 2 Here is the formula, It looks icky, but it’s pretty easy to use… Variance of X Sum Outcome Probability Mean of outcomes Outcome Value

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Means and Variances of Random Variables Outcome12345 Probabiity.13.23.25.19.20 x = 3.1 Standard deviation would be the square root of this. 1.31ish 1.7903

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Law of Large Numbers How can we find the actual of men’s heights? Not really realistic to measure every man in the world Use x-bar as a reasonable estimate Gets more reasonable as the sample size increases – That’s the LAW OF LARGE NUMBERS. The larger the sample size, the more likely x-bar will approach the .

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Rules for Means If I taught in Canada, they would not dig the average height of males in inches, they like centimeters. Plus, all men there measure their heights while wearing 8cm high pumps. Very stylish! How does that change the mean???

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Rules for Means Here is the rule: Here is what we do with those Canadian heights (1in 2.54cm)

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Rules for Means 2 In volleyball there are two main blocking statistics, solo blocks (by self) and assisted blocks (with a buddy). If Chrissa Trudelle averaged.5 solo blocks and 1.3 assisted blocks per match, how many total blocks did she average? Rule: DO IT!

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Rules For Variances OK, the last rule was ridiculously easy, but this next stuff is a bit rough. Think about the men’s height and the changes in Canada with the centimeters and 8cm pumps. How would these change the variance?

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Rules For Variances If we add the same value (8cm) to every height, how does variance (and standard deviation) change? Right, variance does not change if I add the same value to each height…

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Rules For Variances If we multiply each observation by the same amount what will that do? Right, multiplying the variance by a factor will change the variance. Greater if >1 or if <-1. Less if between 1 and -1

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Rules For Variances – Linear Transform If given N(68,2) for average male height, and we transform it again with 8+2.5X, what happens to the variance? Rule DO IT! Standard deviation would be the square root of this. 5

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Rules For Variances – Add/Subtract Here are the formulas: p (rho) is like r, it shows the correlation between X and Y and is between -1 and 1. Should be stated unless X and Y are independent Don’t these look familiar???

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Check this out Rearrange the formulas a bit….. Perfect square trinomials???

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Speaking of Rho That little p only affects things if there is some correlation between the variables If the problem lists a rho, you gotta use it If it doesn’t list a rho, but it should have, do the problem without it, but talk about how there could be some correlation which would affect the variance (or standard deviation) If no correlation p = 0. Therefore:

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Let’s use it now! Coach Boff sweeps the gym floor in N(10,2) minutes and mops the floor in N(15,3) minutes. Assume that the time sweeping and mopping are independent. Find the mean and standard deviation of the combined time. Mean is easy. 10+15= 25 minutes Now let’s find the standard deviation…

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Let’s use it now! REMEMBER! You can not add standard deviations, you must square them to get variances, add the variances and then square root the sum!

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But if X and Y had p =.5 Why more? If rho is positive, X more likely to be higher if Y is also higher. Variation moving in the same way will increase the variance.

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Now you try! Mr. Riebhoff and Mr. Marsheck are the nation’s 1030 th best partner biathlon team. Mr. Riebhoff will do the running leg which is a 10k road race where he has historically had a time of N(48,5) in minutes. Marsheck will do the bike ride of 50k where he has historically had a time of N(106, 10) in minutes. What are the mean and standard deviation of their combined finish times?

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Remember… Show formula(ae) first Talk about any assumptions you are making Don’t forget that standard deviation is the square root of variance Have fun!

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CHAPTER 17 Probability Models

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Binomial Distribution 4 Requirements for a Binomial Distribution 2 outcomes – Success/Failure I.e. – Heads or Tails, Boy or Girl Baby, Make or Miss a Shot Independent observations Probability does not change when you learn the result of a previous event Probability for success (p) is constant for all observations FIXED NUMBER OF OBSERVATIONS!!!!!!!!! 5 Free throws, 17 exam questions, 20 Students THE KEY

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Important parts n = # of Observations Fixed # for a binomial distribution p = Probability of success Defined by you or the question x = # of successes Can be from 0 to n

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Do you remember… Normal Distribution: N( , ) ex. N(68,2) NOW! Binomial Distribution: B(n,p) Example: A 70% free throw shooter shoots 10 Free throws B(10,0.7)

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Which of these would be Binomial? Flip a fair coin and count number of flips until a head appears. 350 students at WPS. 10% are 6 th graders. Choose 10 names at random with no replacement and count # of 6 th graders. Shaq is a 52% free throw shooter. Observe next 10 free throws and count # of makes.

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Binomial PDF Remember Normal CDF? NOW – Binomial PDF Cumulative Distribution Function Probability Distribution Function

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Binomial PDF – 10 FT @ 70% B(10,0.7) X = 0 to 10 0 1 2 3 4 5 6 7 8 9 10 Binompdf(10,0.7,0) Binompdf(10,0.7,1).3.2.1

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Cumulative Distribution Function Cumulative It accumulates, adds up… EXAMPLE – 70% FT Shooter, 10 FTs 012345678910 X =.000.001.009.037.103.200.267.233.121.029 PDF.000.001.010.047.150.350.617.850.9711.00 CDF

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Graph the CDF 0 1 2 3 4 5 6 7 8 9 10.25 1.0.75.50

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Formulae for Binomial Distribution Mean For Binomial Distribution = np Makes sense yeah? Example, I flip a coin 16 times, how many heads? = 16(.5) = 8

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Formulae for Binomial Distribution Standard Deviation For Binomial Distribution = Why? Sausage. Just deal and know where it is on the Formulae Sheet. Ex. Find SD of 10 FT Problem = = 1.449

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LET’S DO IT! Find Mean and Standard Deviation on 20 Free Throws if… p=0.7 p=0.8 p=0.9 p=0.99 What happens to as p approaches 1.0?

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Math Attack Remember Factorials? -- n! Examples: 5! = 5*4*3*2*1 = 120 3! = 3*2*1 = 6 Now the crazy stuff: 0! = 1 Kinda Like a 0 = 1, yah? We’ll need these in a minute, you’ll see why.

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Binomial Coefficient # of ways I can get k successes in n tries. Example: How many ways can I get three tails in 5 flips? Old Skool Way: (easy to mess up) TTTHHTTHTHTTHHTTHTTH THTHTTHHTTHTHTTHTTTH HTTHTHHTTT

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Impress your friends at the next math party way… Pascal’s Triangle 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 5 Choose 0 5 Choose 2 5 Choose 1 5 Choose 3

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Formula Way Formula: Use it! 5 Choose 3 “n choose k”

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Binomial Probability Recall 3 coins flipped, X = # of Heads H T H T H T H T H T H T H T FLIP 1 FLIP 3 FLIP 2 3H, 0T 2H, 1T 1H, 2T 2H, 1T 1H, 2T 0H, 3T X 0 1 2 3 1 3 3 1 P(X=0) = 1*P(H C ) 3 =.125 P(X=1) = 3*P(H C ) 2 P(H) =.375 P(X=2) = 3*P(H) 2 P(H C ) =.375 P(X=3) = 1*P(H) 3 =.125

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Imagine doing P(5 heads in 9 flips) What we need is a formula… Insert binomial coefficient here…

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Let’s do 3 flips P(2 heads)

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Now You Try! In a previous chapter, we found that the probability of rolling a 7 (craps!) with two fair die is 0.167. Let X be the number of 7’s rolled in a series of 10 rolls

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Now You Try! #1 – Find the probability that 3 7’s will be rolled in the 10 attempts

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Now You Try! #2 – Use your TI-83 and find the distribution of X binompdf(trials,p,x) 0 1 2 3 4 5 6 7 8 9 10

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Now You Try! #3 – Find the and of the number of 7’s that would be rolled in 10 attempts = =

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Geometric Distribution 4 Requirements for a Geometric Distribution 2 outcomes – Success/Failure I.e. – Heads or Tails, Boy or Girl Baby, Make or Miss a Shot Independent observations Probability does not change when you learn the result of a previous event Probability for success (p) is constant for all observations Looking for # of trials needed for 1 success!!!!!!! Flip a coin, how many flips until 1 st Head? THE KEY

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Geometric vs. Binomial Binomial Shoot 10 FT’s with p(make)=0.7 find p(8 makes) Geometric With p(make) = 0.7, Shoot until 1 st make, count the number of attempts

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Identify the Geometric Distributions A – Flip a coin until you get a head B – Record the number of times a player makes both shots in a one-and-one foul-shooting situation. (In this situation, you get to attempt a second shot only if you make the first) C – Draw a card from the deck, observe it and replace it into the deck. Count the number of times you draw a card in this manner until you observe a jack.

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Identify the Geometric Distributions D – Buy a “pick 6” lottery ticket every week until you win the lottery. Count the # of weeks it takes for you to win. E – There are 10 red marbles and 5 blue marbles in a jar. You reach in, and without looking, select a marble. You want to know how many marbles you need to draw (without replacement), on average, in order to be sure that you have 3 red marbles.

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CRAPS! Roll till a 7 shows up 7 Not 7 P=1/6 P=5/6 P(X=1) = 1/6 =.167 7 Not 7 P=1/6 P=5/6 P(X=2) = (5/6)(1/6) =.139 7 Not 7 P=1/6 P=5/6 P(X=3) = (5/6) 2 (1/6) =.116 7 Not 7 P=1/6 P=5/6 P(X=4)= (5/6) 3 (1/6)=.097 FORMULA FOR GEOMETRIC PROBABILITIES

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Let’s try it! Mr. Riebhoff is U-G-L-Y (he ain’t got no alibi…). In college, he had only a 20% chance of a randomly selected woman (he used a random # table) agreeing to meet him for a soda.

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Let’s try it! #1 – Find a probability distribution from x = 1 to x = 5 that shows x = the # of females he would ask before getting a “yes” 12345

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Let’s try it! #2 – Make a CDF of the data from #1 1 2 3 4 5 1.0 0.0

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Let’s try it! #3 – What is the probability that after 5 girls asked, Riebhoff would still be dateless?

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Using the TI-83 geometpdf(p,x) In Riebhoff Date Example: geometpdf(0.2,1) = geometpdf(0.2,2) = geometpdf(0.2,3) = geometpdf(0.2,4) = Number if trials ‘till success

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Mean of Geometric Random Variable Common Sense Guess what the mean number of rolls I would need to roll a 5 on a fair die? Guess the mean number of flips I would need to get a head on a fair coin? = 1/p

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