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Probability & Statistics for P-8 Teachers

Probability & Statistics for P-8 Teachers. Chapter 4 Probability. Probability. Probability can be defined as the chance of an event occurring. It can be used to quantify what the “odds” are that a specific event will occur.

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Probability & Statistics for P-8 Teachers

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  1. Probability & Statistics for P-8 Teachers Chapter 4 Probability

  2. Probability • Probability can be defined as the chance of an event occurring. • It can be used to quantify what the “odds” are that a specific event will occur. • Some examples of how probability is used everyday would be weather forecasting, “75% chance of snow” or for setting insurance rates.

  3. Probability • Probabilities must be between 0 and 1, inclusive. • A probability of 0 indicates impossibility. • A probability of 1 indicates certainty.

  4. Sample Spaces and Probability • A probability experiment is a chance process that leads to well-defined results called outcomes. • An outcome is the result of a single trial of a probability experiment. • A sample space is the set of all possible outcomes of a probability experiment. • An event consists of outcomes.

  5. Sample Spaces Experiment Sample Space Toss a coin Head, Tail Roll a die 1, 2, 3, 4, 5, 6 Answer a true/false True, False question Toss two coins HH, HT, TH, TT

  6. Sample Spaces Find the sample space for rolling two dice. 36 different possible outcomes

  7. Sample Spaces Find the sample space for the outcomes of tossing 3 coins HHH HHT HTH HTT THH THT TTH TTT 8 different possible outcomes

  8. Tree Diagram Use a tree diagram to find the sample space for tossing three coins. HHH HHT HTH HTT THH THTTTHTTT H H T H H T T H H T T H T T

  9. P- denotes a probability. A, B, andC- denote specific events. P(A)- denotes the probability of event A occurring. Notation

  10. number of ways E can occur P (E) = number of possible outcomes Sample Spaces and Probability Probability of an event is the ratio of the outcome of the event to the sample space

  11. Probability Determine the probability of getting exactly two heads when tossing a coin three times Sample Space HHH THH HHT THT HTH THT HTT TTT 3 ways two heads occur 8 possible outcomes The probability of exactly two heads is 3/8.

  12. Probability If two dice are rolled one time, find the probability of getting a sum of 7. P(sum of 7) = 6/36

  13. Relationships Among Events P(A or B) The event “either A or B or both occur” P(A & B) The event “both A and B occur” P(not E) The event “E does not occur” Union Intersection Complement

  14. Complement Rule The complement of event Eis the set of outcomes in the sample space that are not included in the outcomes of event P (not E) = 1 – P(E)

  15. Finding Complements Find the complement of each event. Event Complement of the Event Rolling a die and getting a 4 Getting a 1, 2, 3, 5, or 6 Selecting a letter of the alphabet Getting a consonant (assume y is a and getting a vowel consonant) Selecting a month and getting a Getting February, March, April, May, month that begins with a J August, September, October, November, or December Selecting a day of the week and Getting Saturday or Sunday getting a weekday

  16. Complement Use the complement rule to find the probability of rolling a dice and getting a number larger than 1. P(1) = 1/6 P(>1) = 5/6 Complement Rule P(>1) = 1 – P(1) = 1 – 1/6 = 5/6

  17. Addition Rule • The rule of additionapplies to the following situation: • We have two events from the same sample space, and we want to know the probability that either event occurs. P (A or B) = P(A) + P(B) – P(A and B)

  18. Addition Rule A card is drawn randomly from a deck of ordinary playing cards. Find the probability of picking a heart or an ace. We know the following: • There are 52 cards in the deck. • There are 13 hearts, so P(H) = 13/52. • There are 4 aces, so P(A) = 4/52. • There is 1 ace that is also a heart, so P(H and A) = 1/52. • Therefore, based on the rule of addition: P(H or A) = P(H) + P(A) - P(H and A) P(H or A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13

  19. Mutually Exclusive Events • Two events are mutually exclusive events if they cannot occur at the same time • no outcomes in common • P(A and B) = 0

  20. Mutually Exclusive Determine which events are mutually exclusive and which are not, when a single die is rolled. a. Getting an odd number and getting an even number Getting an odd number: 1, 3, or 5 Getting an even number: 2, 4, or 6 Mutually Exclusive

  21. Mutually Exclusive Determine which events are mutually exclusive and which are not, when a single die is rolled. b. Getting an odd number and getting a number less than 4 Getting an odd number: 1, 3, or 5 Getting a number less than 4: 1, 2, or 3 Not Mutually Exclusive

  22. Special Addition Rule If two events are mutually exclusivethen the addition rule can be simplified since no intersection occurs: 0 P (A or B) = P(A) + P(B) – P(A and B) P (A or B) = P(A) + P(B)

  23. Blood Types In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities. a. A person has type O blood.

  24. Blood Types In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities. b. A person has type A or type B blood.

  25. Blood Types In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities. c. A person does not have type AB blood.

  26. Conditional Probability • Additional information or other events occurring may have an impact on the probability of an event. • Conditional probability is the probability that the second event B occurs given that the first event A has occurred.

  27. Conditional Probability • Additional information may have an impact on the probability of an event. • Example: • Consider rolling a dice and getting a 6 • Sample space is {1, 2, 3, 4, 5, 6} • P(rolling a 6) = 1/6 • Suppose we know an even number was rolled • New sample space is {2, 4, 6} • P(rolling a 6) = 1/3

  28. Notation: P(B|A) represents the probability of event B occurring after it is assumed that event A has already occurred. (read P(B|A) as “B given A”) Conditional Probability

  29. Consider a standard 52 card deck Let A = drawing a card less than five (2, 3, or 4) B = drawing a spade Find P(B|A) Conditional Probability Given a card less than five New sample space (12 cards) Spade includes 3 cards P(B|A) = 3/12

  30. Consider a standard 52 card deck Let A = drawing a card less than five (2, 3, or 4) B = drawing a spade Find P(B|A) using the formula Conditional Probability P(A) = 12/52 P(A and B) = 3/52 3/52 = = 3/12 12/52

  31. Parking Tickets The probability that Sam parks in a no-parking zone and gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.20. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a parking ticket. N= parking in a no-parking zone, T= getting a ticket

  32. Women in the Military A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results of the survey are shown.

  33. Women in the Military Find the probability that the respondent answered yes (Y), given that the respondent was a female (F).

  34. Multiplication Rule • We can create a multiplication rule for the intersection of two events A and B by rewriting the conditional probability formula • Solving for the intersection yields: P(A and B) P(B|A) = P(A) P(A and B) = P(A) P(B|A) or P(A and B) = P(B) P(A|B)

  35. Multiplication Rule • From the bag of marbles, find the probability of picking two red marbles from a bag with 4 red and 3 blue marbles. • Pick a red marble from the first draw is 4/7 • Since a red has already been drawn, then 3 red marbles are left from the remaining 6 marbles P(red and red) = P(red) P(red|red) = (4/7) (3/6) = 6/21

  36. Multiplication Rule At a university in western Pennsylvania, there were 5 burglaries reported in 2003, 16 in 2004, and 32 in 2005. If a researcher wishes to select at random two burglaries to further investigate, find the probability that both will have occurred in 2004.

  37. Independence • Two events A and B are independent events if the fact that A occurs does not affect the probability of B occurring. • Examples: • Tossing a coin and rolling a die • Choosing a 3 from a deck of cards, replacing it, and then choosing an ace as the second card

  38. Independence • If two events are independent, then the events do not affect the probability outcomes • One way to represent this is by: • since event A has no affect on event B regardless of whether A was given to us or not P(B|A) = P(B)

  39. Special Multiplication Rule If two events are independentthen the multiplication rule can be simplified since the conditional probability has no affect on the initial probability P(B) P (A and B) = P(A) P(B|A) P (A and B) = P(A) P(B)

  40. Independent events A coin is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 on the die. This problem could be solved using sample space. H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6

  41. Independent events A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week.

  42. Independent Events Determine if the following events, based on a standard deck of cards, are independent of each other A = Drawing an ace B = Drawing a heart Show P(B|A) = P(B) P(B) = 13/52 = 1/4 P(B|A) = 1/4 Since P(B|A) = P(B) = 1/4 then the events are independent

  43. Basic Counting Principles • Counting problems are of the following kind: • “How many different 8-letter passwords are there?” • “How many possible ways are there to pick 11 soccer players out of a 20-player team?” • “What is the probability of winning the lottery?”

  44. Fundamental Counting Rule • Stan is about to order dinner at a restaurant. He has a choice of: • two appetizers (soup or salad) • three main courses (pasta, steak, or fish) • two desserts (cake or pie) • How many different meal combinations can Stan choose?

  45. Using a Tree Diagram to Determine the Number of Combinations Appetizers Main Courses Desserts cake pie pasta steak fish cake pie soup cake pie 12 Combinations cake pie pasta steak fish cake pie salad cake pie

  46. Counting Rules • The fundamental counting rule is also called the multiplication of choices. • If a task is made up of many stages, the total number of possibilities for the task is given by • m × n× p× . . . • where m is the number of choices for the first stage, and n is the number of choices for the second stage, p is the number of choices for the third stage, and so on.

  47. Fundamental Counting Rule A paint manufacturer wishes to manufacture several different paints. The categories include Color: red, blue, white, black, green, brown, yellow Type: latex, oil Texture: flat, semigloss, high gloss Use: outdoor, indoor How many different kinds of paint can be made if you can select one color, one type, one texture, and one use? # of Colors # of Types # of Textures # of Uses × × × 7 × 2 × 3 × 2 = 84 84 different kinds of paint

  48. Fundamental Counting Rule Colleen has six blouses, four skirts and four sweaters. How many different outfits can she choose from, assuming she wears three different items at once? Blouses Skirts Sweaters = 96 outfits 4 × 6 4 × The final score in a soccer game is 5 to 4 for Team A. How many different half-time scores are possible? Team B (0 - 4) Team A (0 - 5) = 30 different possible half-time scores. 6 × 5

  49. Fundamental Counting Rule How many four-letter arrangements are possible? use any of the 26 letters: A - Z 26 1st 26 2nd 26 3rd 26 4th = 456,976 × × × How many four-letter arrangements are possible if a letter cannot be repeated? use any of the 26 letters: A - Z 26 1st 25 2nd 24 3rd 23 4th = 358,800 × × × only 25 letters left only 24 letters left only 23 letters left

  50. Permutations Apermutation is an arrangement of a set of objects for which the order of the objects is important. Show the number of ways of arranging the letters of the word CAT: CAT CTA ACT ATC TCA TAC There are six arrangements or permutations of the word CAT. By changing the order of the letters, you have a different permutation. Using the Fundamental Counting Principle, we would have 3 x 2 x 1 number of distinct arrangements or permutations.

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