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Ch. 6: Discrete Probability-- Questions

Ch. 6: Discrete Probability-- Questions. 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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Ch. 6: Discrete Probability-- Questions

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  1. Ch. 6: Discrete Probability--Questions

  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)= _____ • 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) = ____ • P(even) = ___

  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:

  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 List the outcomes in the sample space If Team A= 6, 7, 8, 9, find P(Team A)

  8. Two Dice- Team A/B • P(Team A)= ___ • P(Team B) = ___ • Notice that P(Team A)+P(Team B) = ___

  9. 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

  10. One Coin • Experimental • If you tossed one coin 1000 times, and 505 times came up heads, you’d say P(H)= ___ • 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)= ___

  11. 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”?

  12. Two Coins- Theoretical Answer • Outcomes:

  13. 2 Coins- Theoretical answer P(0 heads) = ___ P(1 head, 1 tail)= 2/4 = ___ P(2 heads)= ___ Note: sum of these outcomes is ___

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

  15. Three Coins 1 2 3

  16. 3 coins • P(0 heads)= • P(1 head)= • P(2 heads)= • P(3 heads)= • Note: sum is ____

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

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

  19. P(A or B) • If A and B are mutually exclusive (can’t happen together, as in the king/queen example), then P(A or B)=P(A) + P(B) • If A and B are NOT mutually exclusive (can happen together, as in the king/heart example), P(A or B)=P(A) + P(B) –P(A and B)

  20. P (A and B) • For independent events: P(A and B) • P(A and B) = P(A) * P(B) • In General: • P(A and B) = P(A) * P(B/given A)

  21. 2 cards (independent) -questions • Example: Pick two cards, WITH replacement from a deck of cards, • P(king and king)= • P(2 hearts) =

  22. P(A and B) Example-- Independent • For independent events: P(A and B) • P(A and B) = P(A) * P(B) • Example: Pick two cards, WITH replacement from a deck of cards, • P(king and king)= ___ • P(2 hearts) = ____

  23. P(A and B) – Dependent (without replacement) • In General: • P(A and B) = P(A) * P(B/given A) • Example: Pick two cards, WITHOUT replacement from a deck of cards, • P(king and king)= ____ • P(heart and heart)= ____ • P(king and queen) = ___

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

  25. P(driver died)= ___ • P(driver died/given no seat belt)= ___ • P(no seat belt)= ___ • P(no seat belt/given driver died)= ___

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