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Section 5.3

Section 5.3. Day 2. Page 311, E15.

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Section 5.3

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  1. Section 5.3 Day 2

  2. Page 311, E15 About 10% of high school girls report that they rarely or never wear a seat belt while riding in motor vehicles. Suppose you randomly sample four high school girls. Estimate the probability that no more than one of the girls says that she rarely or never wears a seat belt. • Assumptions? • Model?

  3. Assumptions • probability of a randomly selected high school girl reporting that she rarely or never wears a seat belt is 10%

  4. Assumptions probability of a randomly selected high school girl reporting that she rarely or never wears a seat belt is 10% the girls are selected independently of each other.

  5. Simulation About 10% of high school girls report that they rarely or never wear a seat belt while riding in motor vehicles. Suppose you randomly sample four high school girls. Estimate the probability that no more than one of the girls says that she rarely or never wears a seat belt. (2) Model?

  6. Model Outcomes: Reports she rarely or never wears a seat belt Does not report she rarely or never wears a seat belt

  7. Model Digits assigned to outcomes: Reports she rarely or never wears a seat belt 0 Does not report she rarely or never wears a seat belt 1 – 9 How many random digits constitute one run of the simulation?

  8. Simulation About 10% of high school girls report that they rarely or never wear a seat belt while riding in motor vehicles. Suppose you randomly sample four high school girls. Estimate the probability that no more than one of the girls says that she rarely or never wears a seat belt. How many random digits constitute one run of the simulation?

  9. Simulation About 10% of high school girls report that they rarely or never wear a seat belt while riding in motor vehicles. Suppose you randomly sample four high school girls. Estimate the probability that no more than one of the girls says that she rarely or never wears a seat belt. How many random digits constitute one run of the simulation?

  10. Model Digits assigned to outcomes: Reports she rarely or never wears a seat belt 0 Does not report she rarely or never wears a seat belt 1 – 9 Select four digits, allowing repeats, and record the number of digits that are 0

  11. A tetrahedral die has the numbers 1, 2, 3, and 4 on its faces. Suppose you roll a pair of tetrahedral dice. How many possible outcomes are there?

  12. A tetrahedral die has the numbers 1, 2, 3, and 4 on its faces. Suppose you roll a pair of tetrahedral dice. How many possible outcomes are there? 4 ● 4 = 16

  13. a. Make a two-way table of all 16 possible outcomes

  14. b.Find P(sum of 6 or sum of 7) =

  15. b. P(sum of 6 or sum of 7) = P(sum of 6) + P(sum of 7) =

  16. c. Find P(doubles or sum of 7)

  17. c. P(doubles or sum of 7) = P(doubles) + P(sum of 7) =

  18. d. Can you use the Addition Rule of Disjoint Events to find P(doubles or a sum of 6)?

  19. Can you use the Addition Rule of Disjoint Events to find P(doubles or a sum of 6)? No, these events are not disjoint. The outcome (3, 3) is common to both.

  20. The Addition Rule How do you determine P(A or B) if A and B are not disjoint events?

  21. The Addition Rule For any two events A and B, P(A or B) = P(A) + P(B) – P(A and B)

  22. P(doubles or a sum of 6)= P(doubles) + P(sum of 6) - P(doubles and a sum of 6) = ?

  23. P(doubles or a sum of 6)= P(doubles) + P(sum of 6) - P(doubles and a sum of 6) = 4/16 + 3/16 – 1/16 = 6/16

  24. Fill in the missing cells and marginal totals.

  25. Fill in the missing cells and marginal totals.

  26. Using the table, calculate the probability that a student randomly selected doesn’t have a dog or doesn’t own a laptop.

  27. Using the table, calculate the probability that a student randomly selected doesn’t have a dog or doesn’t own a laptop. No double counting this way.

  28. Use Addition Rule to calculate the probability that a student randomly selected doesn’t have a dog or doesn’t own a laptop.

  29. Use Addition Rule to calculate the probability that a student randomly selected doesn’t have a dog or doesn’t own a laptop.

  30. The Addition Rule If you roll a pair of dice, find the probability you get doubles or a sum of 8.

  31. The Addition Rule If you roll a pair of dice, find the probability you get doubles or a sum of 8. These are not disjoint as they have the common outcome (4, 4).

  32. The Addition Rule If you roll a pair of dice, find the probability you get doubles or a sum of 8. P(doubles or sum of 8) = P(doubles) + P(sum of 8) – P(doubles and sum of 8) = ?

  33. The Addition Rule If you roll a pair of dice, find the probability you get doubles or a sum of 8. P(doubles or sum of 8) = P(doubles) + P(sum of 8) – P(doubles and sum of 8) =

  34. The Addition Rule If you roll a pair of dice, find the probability you get doubles or a sum of 8. P(doubles or sum of 8) = P(doubles) + P(sum of 8) – P(doubles and sum of 8) =

  35. Computing P(A and B) In a local school, 80% of the students carry a backpack (B) or a wallet (W). Forty percent carry a backpack, and 50% carry a wallet. If a student is selected at random, find the probability that the student carries both a backpack and a wallet.

  36. Computing P(A and B) Not disjoint, so P(B or W) = P(B) + P(W) – P(B and W) In a local school, 80% of the students carry a backpack (B) or a wallet (W). Forty percent carry a backpack, and 50% carry a wallet. If a student is selected at random, find the probability that the student carries both a backpack and a wallet.

  37. Computing P(A and B) Not disjoint, so P(B or W) = P(B) + P(W) – P(B and W) Substitute known values: 0.80 = 0.40 + 0.50 – P(B and W) P(B and W) = ?

  38. Computing P(A and B) Not disjoint, so P(B or W) = P(B) + P(W) – P(B and W) Substitute known values: 0.80 = 0.40 + 0.50 – P(B and W) P(B and W) =0.10 Thus, the probability that a student carries both a backpack and a wallet is 10%.

  39. Use the Addition Rule to compute the probability that if you flip two fair coins, you get heads on the first coin or you get heads on the second coin.

  40. P(heads on first or heads on second ) = P(heads on first) + P(heads on second ) – P(heads on both) = ?

  41. P(heads on first or heads on second ) = P(heads on first) + P(heads on second ) – P(heads on both) =

  42. Page 324, E33

  43. Page 324, E33 Whistle? Yes No Total Yes Swim? No Total

  44. Page 324, E33

  45. Page 324, E33

  46. Page 324, E33

  47. Page 324, E33

  48. Questions? Quiz 5.1 – 5.3 Wednesday • One side of 4 x 6 note card

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