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MATH104 Ch. 11: Probability Theory part 3

MATH104 Ch. 11: Probability Theory part 3. Probability Assignment . Assignment by intuition – based on intuition, experience, or judgment. Assignment by relative frequency – P ( A ) = Relative Frequency = Assignment for equally likely outcomes. One Die .

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MATH104 Ch. 11: Probability Theory part 3

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  1. MATH104Ch. 11: Probability Theorypart 3

  2. Probability Assignment • Assignment by intuition – based on intuition, experience, or judgment. • Assignment by relative frequency – • P(A) = Relative Frequency = • Assignment for equally likely outcomes

  3. One Die • Experimental Probability (Relative Frequency) • If the class rolled one die 300 times and it came up a “4” 50 times, we’d say P(4)= 50/300 • The Law of Large numbers would say that our experimental results would approximate our theoretical answer. • Theoretical Probability • Sample Space (outcomes): 1, 2, 3, 4, 5, 6 • P(4) = 1/6 • P(even) = 3/6

  4. Two Dice • Experimental Probability • “Team A” problem on the experiment: If we rolled a sum of “6, 7, 8, or 9” 122 times out of 218 attempts, P(6,7,8, or 9)= 122/218= 56% • Questions: What sums are possible? • Were all sums equally likely? • Which sums were most likely and why? • Use this to develop a theoretical probability • List some ways you could get a sum of 6…

  5. Outcomes • For example, to get a sum of 6, you could get: • 5, 1 4,2 3,3 …

  6. Two Dice – Theoretical Probability • Each die has 6 sides. • How many outcomes are there for 2 sides? (Example: “1, 1”) • Should we count “4,2” and “2,4” separately?

  7. Sample Space for 2 Dice 1, 1 1, 2 1, 3 1, 4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 If Team A= 6, 7, 8, 9, find P(Team A)

  8. Two Dice- Team A/B • P(Team A)= 20/36 • P(Team B) = 1 – 20/36 = 16/36 • Notice that P(Team A)+P(Team B) = 1

  9. Summary of 2 Dice (Team A/B) problem

  10. Some Probability Rules and Facts • 0<= P(A) <= 1 • Think of some examples where • P(A)=0 P(A) = 1 • The sum of all possible probabilities for an experiment is 1. Ex: P(Team A)+P(Team B) =1

  11. One Coin • Experimental • If you tossed one coin 1000 times, and 505 times came up heads, you’d say P(H)= 505/1000 • The Law of Large Numbers would say that this fraction would approach the theoretical answer as n got larger. • Theoretical • Since there are only 2 equally likely outcomes, P(H)= 1/2

  12. Two Coins • Experimental Results • P(0 heads) = • P(1 head, 1 tail)= • P(2 heads)= • Note: These all sum to 1. • Questions: • Why is “1 head” more likely than “2 heads”?

  13. Two Coins- Theoretical Answer • Outcomes: • TT, TH, HT, HH 1 2 H HH H T HT T H TH T TT

  14. 2 Coins- Theoretical answer P(0 heads) = 1/4 P(1 head, 1 tail)= 2/4 = 1/2 P(2 heads)= ¼ Note: sum of these outcomes is 1

  15. Summary of 2 coin problem

  16. Three Coins • Are “1 head” , “2 heads”, and “3 heads” all equally likely? • Which are most likely and why?

  17. Three Coins 1 2 3 H H HHH H T HHT T H HTH T HTT T H H THH T THT T H TTH 2*2*2=8 outcomes T TTT

  18. 3 coins • P(0 heads)= • P(1 head)= • P(2 heads)= • P(3 heads)=

  19. Theoretical Probabilities for 3 Coins • P(0 heads)= 1/8 • P(1 head)= 3/8 • P(2 heads)= 3/8 • P(3 heads)= 1/8 • Notice: Sum is 1.

  20. Summary of 3 coin problem

  21. Cards • 4 suits, 13 denominations; 4*13=52 cards • picture = J, Q, K

  22. When picking one card, find… • P(heart)= • P(king)= • P(picture card)= • P(king or queen)= • P(king or heart)=

  23. Theoretical Probabilities- Cards • P(heart)= 13/52 = ¼ = 0.25 • P(king)= 4/52= 1/13 • P(picture card)= 12/52 = 3/13 • P(king or queen)= 4/52 + 4 /52 = 8/52 • P(king or heart)= 4/52 + 13/52 – 1/52 = 16/52

  24. 11.6   Not, Mutually Exclusive, Odds P(E)= 1-P(E ‘ ) where E’ = not E=complement of E  1.      If there is a 20% chance of snow tomorrow, what is the chance it will not snow tomorrow?  2.      When choosing one card from a deck, find the probability of selecting:  a.       A heart  b.      A card that is not a heart  c.       A king  d.      A card that is not a king

  25. P(A or B) 1.      When selecting one card, find the probability of: a.       king or queen  b.      king or a heart  c.       king or a 5  d.      5 or a diamond  e.       Picture card or a 7  f.       Picture card or a red card

  26. P(A or B) • Mutually exclusive events—cannot occur together • If A and B are mutually exclusive,  P(A or B) = P(A) + P(B) • If A and B are not mutually exclusive, P(A or B) = P(A) + P(B) – P(A and B)

  27. Odds Basic idea: If, when drawing one card from a deck, the probability of getting a heart is ¼, then The odds in favor of drawing a heart are 1:3 and the odds against a heart are 3:1. Another example: If, when drawing one card from a deck, the probability of getting a king is 1/13, then The odds in favor of drawing a king are 1:12, and the odds against a king are 12:1. Odds to Probability   if odds in favor of E are a:b, then P(E)=

  28. Given probabilities, find odds   1.      Recall probabilities a.       P(heart) b.      P(not a heart) c.       P(king) d.      P(picture card) e.   P(red card) 1.      Find the odds in favor of: a.       A heart b.      A card that is not a heart c.       A king d.      A picture card e.   A red card

  29. …odds 2.      If the odds in favor of winning the lottery are 1:1,000,000, find the probability of winning the lottery 3. If the odds in favor of getting a certain job are 3:4, find the probability of getting the job.

  30. 11.7:  And, Conditional Independent events- two events are independent events if the occurrence of either of them has no effect on the probability of the other If A and B are independent events, then P(A and B) = P(A)*P(B)

  31. 2 kids 1. Assuming it’s equally likely that boys and girls are born, in a family with 2 kids, find the probability of getting: a.       2 girls b.      2 boys

  32. 2. In a family with 3 kids, find the probability of getting: Assuming P(B)=P(G) a.       3 girls b.      3 boys c.       At least 1 boy

  33. 3.      In a family of 4 kids, find the probability of getting: a.       4 girls b.      4 boys c.       At least 1 boy

  34. 4. Two cards  4.      If you pick two cards out of a deck of cards and replace them in between picks, find: a.   P( 2 red cards)  b.  P(2 hearts)   c. P(2 kings)

  35. Dependent events— the occurrence of one of them has an effect on the occurrence of the other If A and B are dependent, P(A and B) = P(A)*P(B, given A)

  36. Without replacement:  1.      If you pick two cards out a deck without replacement, find the probability of getting: a.       2 red cards b.      2 kings

  37. 2. pick 3 cards without replacement find the probability of getting: a.       3 red cards b.      3 kings  c.       A king, then a queen, then a jack (in that order)

  38. Conditional Probability Find: P(driver died)= P(driver died/given no seat belt)= P(no seat belt)= P(no seat belt/given driver died)=

  39. P(driver died)= 2111/577,006 = .00366 • P(driver died/given no seat belt)= 1601/164,128 = .0097 • P(no seat belt)= 164,128/577,006= .028 • P(no seat belt/given driver died)= 1602/2111= .76

  40. Birthday problem What is the probability that two people in this class would have the same birth date?

  41. Hint Let E=at least two people have the same bday What is E’ (not E) Find P(E’)=

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