1 / 72

AP Statistics – Probability

AP Statistics – Probability. Mrs. Lerner Charlotte Catholic High School With some updates by Dr. Davidson Mallard Creek High School. Random Phenomena. Are uncertain in the short run Exhibit a consistent pattern in the long run Note the dual aspect

Download Presentation

AP Statistics – Probability

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. AP Statistics – Probability Mrs. Lerner Charlotte Catholic High School With some updates by Dr. Davidson Mallard Creek High School

  2. Random Phenomena Are uncertain in the short run Exhibit a consistent pattern in the long run Note the dual aspect Note: This is the Condition for the Probability we study

  3. Definitions: An event is an outcome or a set of outcomes of random phenomenon (RP). Outcome(s) of interest are “Success(es)” [even if they are bad news!!!!]

  4. Probability of an Event : (note dual aspect) Probability of an event = proportion of times a success occurs in the long run (measure by experiment or simulation) [number of ways for a success to occur]/[all possible events (specified by a model of the RP)

  5. All Possible Outcomes • If there are n ways to do a first event • & m ways to do a second event • Then the number ofall possible outcomes=nm • This is called the multiplication rule

  6. Probability – Example # 1 Consider the sum of two dice that are rolled Let’s say we are interested in the Probability the sum is a 4, so SUCCESS = ? Find # of ways to get a sum = 4 [a success]is 3: 1+3, 2+2, 3+1 [3 ways]

  7. Probability The probability of a specific outcome is [# of ways to get a success]/ [all possible outcomes e.g. Probability of getting a sum =4 is: 3/36 = 1/12 ~ .0833 = 8.33 % This is the chance the next outcome is a success This is the proportion of times that successes occur in a large repetitive # of identical trials

  8. Probability Rules - 1 • The probability P(A) of any event A satisfies0≤P(A)≤1 • Thus the probability of any event is between 0 & 1

  9. Probability Rules - 2 • The Sample Space is the set of all possible outcomes • If S is the sample space in a probability model, then P(S)=1 The probabilities of all possible outcomes must add up to 1

  10. Probability Rules - 3 • The complement of any event A is • the event that A does not occur • written as Ac • The complement rule states thatP(Ac)=1-P(A) The probability that an event does not occur is 1 minus the probability that it does occur

  11. Probability Rules - 4 • Two events A and B are disjoint • if they have no outcomes in common • cannot occur at the same time. • If A and B are disjoint, then P(A or B) = P(A) + P(B). If two events have no outcomes in common, then the probability of either one occurring is the sum of their individual probabilities

  12. Probability Rules - 5 • For any two events A and B: P(A or B) = P(A) + P(B) – P(A and B) The probability of either A or B occurring is the sum of their individual probabilities minus the probability that they occur at the same time

  13. Probability Rules - 6 • Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. If A and B are independent, then P(A and B) = P(A)P(B). If two events are independent, then the probability that they both occur is the product of their individual probabilities Note – Disjoint events are always NOT INDEPENDENT, for if one occurs, then we know the other can not occur!

  14. Remember… • And = Joint = Intersection • Or = Both (1, or other, or 2) = Union

  15. Probability Rules - 7 • When P(A)>0, the conditional probability of B given A is: P(B|A)=P(A and B)/P(A) The probability of B occurring, given that A occurs, is the probability of A & B jointly occurring divided by the probability of the stated condition (A) occurring

  16. Probability Rules - 8 • For any two events A and B, P(A and B) = P(B|A) P(A) P(A and B) = P(A|B) P(B) The probability of A and B occurring jointly is the probability that B occurs given that A has already occurred multiplied by the probability that A occurs

  17. Example - 2: Suppose that for a certain Caribbean island in any 3-year period: • the probability of a major hurricane is .25 • the probability of water damage is .44 • and the probability of both a hurricane and water damage is .22. What is the probability of water damage given that there is a hurricane?

  18. Suppose that for a certain Caribbean island in any 3-year period the probability of a major hurricane is .25, the probability of water damage is .44 and the probability of both a hurricane and water damage is .22. What is the probability of water damage given that there is a hurricane?

  19. If three people, Joe, Betsy, and Sue, play a game in which • Joe has a 25% chance of winning • Betsy has a 40% chance of winning • What is the probability that Sue will win?

  20. If three people, Joe, Betsy, and Sue, play a game in which Joe has a 25% chance of winning and Betsy has a 40% chance of winning, what is the probability that Sue will win?

  21. A summer resort rents rowboats to customers but does not allow more than four people to a boat. Each boat is designed to hold no more than 800 pounds. • Suppose the distribution of the weight of adult males who rent boats, including their clothes and gear, is normal with a mean of 190 pounds and standard deviation of 10 pounds. • If the weights of individual passengers are independent, what is the probability that a group of four adult male passengers will exceed the acceptable weight limit of 800 pounds?

  22. The following table shows the frequencies of political affiliations in the age ranges listed from a random sample of adult citizens in a particular city. Dem. Repub. Indep. 18-30 25 18 12 31-40 32 21 10 41-50 17 25 17 Over 50 14 32 15  _____________________________________________________________ 5. What proportion of the Republicans are over 50? • 61/238 • 32/96 • 96/238 • 32/61 • Cannot be determined.

  23. The following table shows the frequencies of political affiliations in the age ranges listed from a random sample of adult citizens in a particular city. Dem. Repub. Indep. 18-30 25 18 12 31-40 32 21 10 41-50 17 25 17 Over 50 14 32 15  _____________________________________________________________ 5. What proportion of the Republicans are over 50? • 32/96

  24. The following table shows the frequencies of political affiliations in the age ranges listed from a random sample of adult citizens in a particular city. Dem. Repub. Indep. 18-30 25 18 12 31-40 32 21 10 41-50 17 25 17 Over 50 14 32 15  _____________________________________________________________ 6. If one adult citizen is chosen at random, what is the probability that this person is a Democrat between the ages of 41 and 50? • 17/238 • 17/88 • 61/238 • 17/61 • 88/238

  25. The following table shows the frequencies of political affiliations in the age ranges listed from a random sample of adult citizens in a particular city. Dem. Repub. Indep. 18-30 25 18 12 31-40 32 21 10 41-50 17 25 17 Over 50 14 32 15  _____________________________________________________________ 6. If one adult citizen is chosen at random, what is the probability that this person is a Democrat between the ages of 41 and 50? • 17/238

  26. The following table shows the frequencies of political affiliations in the age ranges listed from a random sample of adult citizens in a particular city. Dem. Repub. Indep. 18-30 25 18 12 31-40 32 21 10 41-50 17 25 17 Over 50 14 32 15  _____________________________________________________________ 7. Given that a person chosen at random is between 31 and 40, what is the probability that this person is an Independent? • 10/238 • 10/63 • 10/54 • 54/238 • 63/238

  27. The following table shows the frequencies of political affiliations in the age ranges listed from a random sample of adult citizens in a particular city. Dem. Repub. Indep. 18-30 25 18 12 31-40 32 21 10 41-50 17 25 17 Over 50 14 32 15  _____________________________________________________________ 7. Given that a person chosen at random is between 31 and 40, what is the probability that this person is an Independent? • 10/63

  28. The following table shows the frequencies of political affiliations in the age ranges listed from a random sample of adult citizens in a particular city. Dem. Repub. Indep. 18-30 25 18 12 31-40 32 21 10 41-50 17 25 17 Over 50 14 32 15  _____________________________________________________________ 8. What proportion of the citizens sampled are over 50 or Independent? • 54/238 • 61/238 • 100/238 • 115/238 • Cannot be determined

  29. The following table shows the frequencies of political affiliations in the age ranges listed from a random sample of adult citizens in a particular city. Dem. Repub. Indep. 18-30 25 18 12 31-40 32 21 10 41-50 17 25 17 Over 50 14 32 15  _____________________________________________________________ 8. What proportion of the citizens sampled are over 50 or Independent? • 100/238

  30. Example - • Assume an multiple-choice examination consists of questions, each having five possible answers. • Linda estimates that she has probability 0.75 of knowing the answer to any question that may be asked. • If she does not know the answer, she will guess, with conditional probability 1/5 of being correct. What is the probability that Linda gives the correct answer to a question? (Draw a tree diagram to guide the calculations.)

  31. An examination consists of multiple-choice questions, each having five possible answers. Linda estimates that she has probability 0.75 of knowing the answer to any question that may be asked. If she does not know the answer, she will guess, with conditional probability 1/5 of being correct. What is the probability that Linda gives the correct answer to a question? (Draw a tree diagram to guide the calculations.) P(correct) = .75 +.25*.2 = .8

  32. Random Variables • A random variable assumes any of several different values as a result of some random phenomenon

  33. Discrete Random Variables [RV] • A Discrete RV – has a countable number of possible values • The probabilities must satisfy two requirements • Every probability is between 0 and 1 • The sum of all probabilities is 1 • We can use a probability histogram to look at the probability distribution.

  34. Discrete Random Variables • Mean of a Discrete R. V. – (also called expected value) –

  35. Discrete Random Variables • Variance of a Discrete R. V. –

  36. Continuous Random Variables • Continuous R. V. – takes all values in an interval of numbers • We look at its distribution using a density curve • The probability of any event is the area under the density curve in that interval.

  37. Rules for Means • If X is an R. V. and a & b are fixed numbers, then themeanμa+bX = a +bμX • If X and Y are R. V.‘s, then μX±Y = μX±μY

  38. Rules for Variances • If X is an R. V. and a and b are fixed numbers, then σ2a+bX = b2σX2 • Note that multiplying by a constant changes the variance but adding a constant does not. • If X and Y are independent R. V.’s, thenσ2X±Y =σX2+σY2“Pythagorean Theorem of Statistics” For STANDARD DEVIATION: square ‘em, add ‘em, take the square root

  39. 11. Suppose X and Y are random variables with μX = 10, σX = 3, μY = 15, and σY = 4. Given that X and Y are independent, what are the mean and standard deviation of the random variable X+Y?

  40. 11. Suppose X and Y are random variables with μX = 10, σX = 3, μY = 15, and σY = 4. Given that X and Y are independent, what are the mean and standard deviation of the random variable X+Y? μX+Y = μX + μY σX+Y = √σ2X + σ2Y =10 + 15 = √9+16 = 25 = √25 = 5

  41. You roll a die. If it comes up a 6, you win $100. If not, you get to roll again. If you get a 6 the second time, you win $50. If not, you lose. a. Create a probability model for the amount you will win at this game.

  42. You roll a die. If it comes up a 6, you win $100. If not, you get to roll again. If you get a 6 the second time, you win $50. If not, you lose. a. Create a probability model for the amount you will win at this game.

  43. You roll a die. If it comes up a 6, you win $100. If not, you get to roll again. If you get a 6 the second time, you win $50. If not, you lose. b. Find the expected amount you’ll win.

  44. You roll a die. If it comes up a 6, you win $100. If not, you get to roll again. If you get a 6 the second time, you win $50. If not, you lose. b. Find the expected amount you’ll win.

  45. Law of Large Numbers • Law of Large Numbers – The long run relative frequency of repeated independent trials gets closer and closer to the true relative frequency as the number of trials increases.

  46. Binomial Distributions • Binomial Distribution – the distribution of the count X successes in the binomial setting. • B(n,p) where n is the number of observations and p is the probability of success

  47. Binomial Distributions • Use binompdf(n,p,X) to find the probability of a single value of X, such as P(X = 3). • Use binomcdf(n,p,X)to find the probability of at most X successes • for example P(X ≤ 3).

More Related