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Chapter 6 Rules of Probability

Chapter 6 Rules of Probability. 6.1 Sample spaces and events 6.2 Postulates of probabilities 6.4 Additive rules 6.5 Conditional probability 6.6 Multiplication rules 6.7 Bayes theorem. 6.1 Sample Spaces and Events. In a probability experiment the possible outcomes form the samples space, S

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Chapter 6 Rules of Probability

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  1. Chapter 6 Rules of Probability • 6.1 Sample spaces and events • 6.2 Postulates of probabilities • 6.4 Additive rules • 6.5 Conditional probability • 6.6 Multiplication rules • 6.7 Bayes theorem

  2. 6.1 Sample Spaces and Events • In a probability experiment the possible outcomes form the samples space, S 1. Pick a card and note the suit S={spade, club, heart, diamond} 2. Pick a card and note the color S={red, black} 3. Pick a card and note if it is an ace S={ace, not ace}

  3. More examples 4. Roll a die and note the dots S={1, 2, 3, 4, 5, 6} 5. Roll 2 dice and note the sum S={2, 3, 4, … …, 11, 12}

  4. Event • A subset of a sample space is called an event For example, roll one die. Even={2, 4, 6} Odd= {1, 3, 5} Greater than 6=empty set=

  5. Operations • Union A∪B=all outcomes in A or B • Intersection A∩B=all outcomes in both A and B • Complement A’=all outcomes not in A (like S-A)

  6. Operations of Events • If two events have no outcomes in common they are mutually exclusive (or disjoint) i.e., A∩ B= S={1, 2, 3, 4, 5, 6} Odd ={1, 3, 5}=A Even={2, 4, 6}=B more than 2={3, 4, 5, 6}=C ‘Odd’ and ‘Even’ are mutually exclusive; ‘Even’ and ‘C’ are not mutually exclusive.

  7. Example 6.1 • Roll a red die and a white die S= {(i,j): i, j=1,2,…, 6} A={(i,j): i+j=7}={sum=7} B={(3,j): j=1,..,6}={red=3} Get A∪B A∩B A’

  8. Another example • S={52 cards in deck} Club ∪Spade = Black Club ∩Spade=  Red Ace= {Ace of hearts, ace of diamond} Red’= Black

  9. Venn Diagram A Venn Diagram shows events as potentially intersecting circles. e.g., R=Republican F=Female S F R

  10. F F F R R R • R U F • R ∩ F S S

  11. 6.2 Postulates of probabilities • Roll a die S={1, 2, 3, 4, 5, 6} • P(1)=P(2)=…=P(6)=1/6 For any event A: 0  P(A)  1 P(S)=1 P(A’)=1-P(A)

  12. Example 6.2 5, 6 • A={1, 2}, B={ 3, 4} A and B are disjoint. (or mutually exclusive) P(A∪B)=2/6+2/6=2/3 P(A∪B)=P(A)+P(B) if A and B are mutually exclusive 3, 4 B 1, 2 A

  13. Example 6.3 • S={52 cards} P(ace ∪ king)=P(ace or king) =4/52+4/52=2/13 • P(rain)=0.7 then P(not rain)=1-0.7=0.3

  14. Exercise • Which of the following pairs of events are mutually exclusive? Explain. • A driver getting a ticket for speeding and a ticket for going through a red light. • Being foreign-born and being President of the United States. • A person wearing black shoes and green socks. • Having rain and sunshine on the 4th of July, 2005.

  15. Exercise • If A and B are the events that Consumer Union will rate a car stereo good or poor, P(A)=0.24 and P(B)=0.35, determine the following probabilities: • P(A’) • P(AUB) • P(A ∩B)

  16. Exercise • Let E, T, and N be the events that a car brought to a garage needs an engine overhaul, transmission repairs, or new tires. Draw a Venn diagram for the following • E U T • E U T U N • E ∩T ∩N • E ∩T ∩N’

  17. 6.3. Probability and odds-skip

  18. 6.4 Addition Rules • For mutually exclusive events P(A1∪A2 ∪•••∪Ak)=P(A1)+P(A2)+•••+P(Ak) P(ace or king or queen) =P(ace)+P(king)+P(queen) =4/52+4/52+4/52=3/13

  19. When individual outcomes are mutually exclusive, The probability of an event A is the sum of the probabilities for all outcomes in A For a dice, S={1, 2, 3, 4, 5, 6} P(even)=P(2)+P(4)+P(6)=1/6+1/6+1/6=1/2

  20. General Addition Rule • For any two events A and B P(A∪B)=P(A)+P(B)–P(A∩B) A ∩ B B A

  21. Example 6.4 • Draw a card from a deck. What is the probability of getting an ace or a spade? P(ace or spade) =P(ace)+P(spade) – P(ace and spade) =4/52+13/52 –1/52 =16/52 U

  22. 6.5 Conditional Probability • Roll a die. The probability of getting any number in the set of {1,2,3,4,5,6} is 1/6 ! Try to predict the outcome when I roll the die. We predict that any number will have the same probability of showing up. • How about if you know some prior information?

  23. If you know the number is even • Re-evaluate the probabilities: 0 probabilities for all 3 odd numbers equal probabilities for 3 even numbers (each even number has a probability of 1/3) This is called the conditional probability.

  24. Notation: P(A|B) • The conditional probability of event A, given event B (occurs) P(1|even)=P(3|even)=P(5|even)=0 P(2|even)=P(4|even)=P(6|even)=1/3

  25. L E 1, 3 4, 6, 2 5 Conditional Probability • S={1,2,3,4,5,6} • E=even • L=less than or equal to 3 • P(E|L) is the probability of “E given L”. • P(even given that point total≤3) =P(E|L)=1/3.

  26. Conditional Probability • Verify: P(1|even)=0 P(2|even)=1/3

  27. Example 6.5 • Hair Eye Probability brown blue 0.3 brown brown 0.4 blond blue 0.2 blond brown 0.1 -------------------------------------------- P(blue eyes|blond hair) =P(blue eyes and blond hair)/P(blond hair) =0.2/(0.2+0.1)=2/3=0.67 P(blond hair|blue eyes) = P(blond hair and blue eyes)/P(blue hair) =0.2/(0.3+0.2)=2/5=0.40

  28. A simpler way: tabulate the probabilities P(blue eyes|blond hair) =0.2/0.3

  29. Exercise • The probability that Henry will like a new movie is 0.70 and the probability that Jane, his girlfriend, will like it is 0.60. If the probability is 0.28 that he will like it and she will dislike it, what is the probability that he will like it given that she is not going to like it?

  30. 6.6 Multiplication Rules • Conditional probability formula • This is equivalent to P(A∩B)=P(A|B)P(B) Similarly P(A∩B)=P(B|A)P(A)

  31. Example 6.6 Pick up two cards from a deck without replacement What is the probability of getting two aces?

  32. Method 1: like in Example 5.13

  33. Method 2 • A=first card is an ace • B=second card is an ace P(A)=4/52 P(B|A)=3/51

  34. Toss 2 dice • A= 6 first die • B= 6 second die

  35. For more than 2 events multiply conditional probabilities in a similar way • P(A∩B∩C)=P(A)P(B|A)P(C|A∩B )

  36. Example 6.7 • Pick up 3 cards without replacement. Find the probability of getting 3 aces. A=1st card is an ace; B=2nd card is an ace; C=3rd card is an ace. P(A)=4/52 P(B|A)=3/51 P(C|A∩B)=2/50 P(A∩B∩C)=(4/52)(3/51)(2/50)

  37. Example 6.8 • Probability of getting 4 aces in 4 cards

  38. Independent events vs Dependent events • Roll a red die and white die. A=red is a 6; B= white is a 6 Then P(A)=P(B)=1/6; P(A∩B)=1/36; P(B|A)=P(A∩B )/P(A)=1/6 =P(B) P(white is 6|red is 6) = P(white is 6) The probability that white is a 6 is independent of whether red is a 6. “red is a 6” and “white is 6” are independent events.

  39. Independence If the conditional probability of B given A is equal to the unconditional probability of B P(B|A)=P(B) then, A and B are said to be independent This is equivalent to P(A|B)=P(A) Or P(A∩B )=P(B)P(A|B)=P(B)*P(A) O.W., A and B are independent

  40. For independent events A and B P(A∩B )=P(A)*P(B) P(A|B)=P(A) P(B|A)=P(B) If any of these 3 conditions holds, then A & B are independent. If any of these 3 conditions does not hold, then A & B are dependent.

  41. Example 6.9 • Flip two fair coins. Probability of getting two heads? Method 1: Sample space: {HH, HT, TH, TT} P(HH)=1/4 Method 2: A=head on 1st flip={HH, HT} B=head on 2nd flip={HH, TH} Show P(B|A)=P(B)  P(HH)=P(A∩B)=P(A)P(B)=(1/2)(1/2)=1/4

  42. Example 6.10 • Pick a card, set H=heart A=ace K=king P(H)=13/52=1/4 P(H|K)=1/4  H and K are independent P(A)=4/52 P(A|K)=0  A and K are dependent

  43. Independent and mutually exclusive • They are totally different concepts! Do not mix these up. • Difference?

  44. How to get the probability? • The solutions of most probability problems use the following facts. • For equally likely outcomes #of outcomes in A P(A)=———————————— total # of possible outcomes • P(not A)=1–P(A)

  45. More • For mutually exclusive events P(A1 or A2 or … or Ak)= P(A1)+P(A2)+… P(Ak) • For any two events P(A or B)=P(A∪B)=P(A)+P(B)–P(A∩B) • P(A|B)=P(A∩ B)/P(B)

  46. More • P(A and B)=P(A∩B)=P(A)P(B|A) or P(B)P(A|B) • For independent events P(A|B)=P(A); P(B|A)=P(B); P(A∩B)=P(A)P(B) • For independent events A1, A2,…, Ak P(A1∩ A2 ∩ …∩ Ak)=P(A1)…P(Ak)

  47. More • For any events P(A1∩ A2 ∩ …∩ Ak) =P(A1)P(A2|A1)…P(Ak|A1∩…∩Ak-1)

  48. Example 6.11 Toss a coin 3 times. Find P(2 heads)=? HHT (½)(½)(½)=1/8 HTH (½)(½)(½)=1/8 THH (½)(½)(½)=1/8 P(2 heads)=P(HHT∪HTH∪THH)=P(HHT)+P(HTH)+P(THH)=3/8 P(at least one head)=? =1–P(no head)=1–P(TTT) =1-1/2*1/2*1/2 =1-1/8 =7/8

  49. Example 6.12 Pick 5 cards P(2 aces|1 king)= P(2 aces and 1 king)/P(1 king) #of ways to pick up 2 aces and 1 king #of ways to pick up 1 king

  50. Example 6.13 • Deal 5 cards. What is the probability that you get 5 face cards? (J,Q, or K) Method 1: Method 2: multiply conditional probability

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