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12.4 – Probability & Probability Distributions

12.4 – Probability & Probability Distributions. If an event can succeed in s ways and fail in f ways, then:. If an event can succeed in s ways and fail in f ways, then: Probability of Success. If an event can succeed in s ways and fail in f ways, then: Probability of Success

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12.4 – Probability & Probability Distributions

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  1. 12.4 – Probability & Probability Distributions

  2. If an event can succeed in s ways and fail in f ways, then:

  3. If an event can succeed in s ways and fail in f ways, then: • Probability of Success

  4. If an event can succeed in s ways and fail in f ways, then: • Probability of Success P(S) = s s + f

  5. If an event can succeed in s ways and fail in f ways, then: • Probability of Success P(S) = s s + f • Probability of Failure

  6. If an event can succeed in s ways and fail in f ways, then: • Probability of Success P(S) = s s + f • Probability of Failure P(F) = f s + f

  7. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female?

  8. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s.

  9. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male

  10. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3)

  11. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3) = 220

  12. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3) = 220 [On calc: 12, MATH, PRB, nCr, 3]

  13. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3) = 220 [On calc: 12, MATH, PRB, nCr, 3] Female

  14. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3) = 220 [On calc: 12, MATH, PRB, nCr, 3] Female = C(16,3)

  15. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3) = 220 [On calc: 12, MATH, PRB, nCr, 3] Female = C(16,3) = 560

  16. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3) = 220 [On calc: 12, MATH, PRB, nCr, 3] Female = C(16,3) = 560 [On calc: 16, MATH, PRB, nCr, 3]

  17. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3) = 220 [On calc: 12, MATH, PRB, nCr, 3] Female = C(16,3) = 560 [On calc: 16, MATH, PRB, nCr, 3] So 220 ∙ 560

  18. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3) = 220 [On calc: 12, MATH, PRB, nCr, 3] Female = C(16,3) = 560 [On calc: 16, MATH, PRB, nCr, 3] So 220 ∙ 560 = 123,200

  19. Probability with Combinations Ex. 1 12 male and 16 female students have been selected as equal qualifiers for 6 college scholarships. If the qualifiers interviewed on the first day are to be chosen at random, what is the probability that 3 will be male and 3 female? 1. Determine the number of successes, s. Male = C(12,3) = 220 [On calc: 12, MATH, PRB, nCr, 3] Female = C(16,3) = 560 [On calc: 16, MATH, PRB, nCr, 3] So 220 ∙ 560 = 123,200 = s

  20. 2. Determine the number of possibilities

  21. 2. Determine the number of possibilities s + f =

  22. 2. Determine the number of possibilities s + f = C(16+12,6)

  23. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6)

  24. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740

  25. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740 [On calc: 28, MATH, PRB, nCr, 6]

  26. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740 [On calc: 28, MATH, PRB, nCr, 6] 3. Find the probability.

  27. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740 [On calc: 28, MATH, PRB, nCr, 6] 3. Find the probability. P(3male,3female)

  28. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740 [On calc: 28, MATH, PRB, nCr, 6] 3. Find the probability. P(3male,3female) = s s + f

  29. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740 [On calc: 28, MATH, PRB, nCr, 6] 3. Find the probability. P(3male,3female) = s s + f = 123,200 376,740

  30. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740 [On calc: 28, MATH, PRB, nCr, 6] 3. Find the probability. P(3male,3female) = s s + f = 123,200 376,740 ≈ 0.327016

  31. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740 [On calc: 28, MATH, PRB, nCr, 6] 3. Find the probability. P(3male,3female) = s s + f = 123,200 376,740 ≈ 0.327016 ≈ 33%

  32. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740 [On calc: 28, MATH, PRB, nCr, 6] 3. Find the probability. P(3male,3female) = s s + f = 123,200 376,740 ≈ 0.327016 ≈ 33% NOTE: Can do all on calculator:

  33. 2. Determine the number of possibilities s + f = C(16+12,6) = C(28,6) = 376,740 [On calc: 28, MATH, PRB, nCr, 6] 3. Find the probability. P(3male,3female) = s s + f = 123,200 376,740 ≈ 0.327016 ≈ 33% NOTE: Can do all on calculator: (12 nCr 3)(16 nCr 3)/(28 nCr 6)

  34. Probability with Permutations Ex. 2 Courtney has a playlist of 6 songs on her MP3 player. What is the probability that the player will randomly play her favorite song first, then her second favorite song, and the three least favorite songs last?

  35. Probability with Permutations Ex. 2 Courtney has a playlist of 6 songs on her MP3 player. What is the probability that the player will randomly play her favorite song first, then her second favorite song, and the three least favorite songs last? P(desired order) = s s + f

  36. Probability with Permutations Ex. 2 Courtney has a playlist of 6 songs on her MP3 player. What is the probability that the player will randomly play her favorite song first, then her second favorite song, and the three least favorite songs last? P(desired order) = s s + f = (2 favorite 1st in order)(3 least last, any order) total possible order

  37. Probability with Permutations Ex. 2 Courtney has a playlist of 6 songs on her MP3 player. What is the probability that the player will randomly play her favorite song first, then her second favorite song, and the three least favorite songs last? P(desired order) = s s + f = (2 favorite 1st in order)(3 least last, any order) total possible order = (1 nPr 1)(3 nPr 3)/(6 nPr6)

  38. Probability with Permutations Ex. 2 Courtney has a playlist of 6 songs on her MP3 player. What is the probability that the player will randomly play her favorite song first, then her second favorite song, and the three least favorite songs last? P(desired order) = s s + f = (2 favorite 1st in order)(3 least last, any order) total possible order = (1 nPr 1)(3 nPr 3)/(6 nPr6)

  39. Probability with Permutations Ex. 2 Courtney has a playlist of 6 songs on her MP3 player. What is the probability that the player will randomly play her favorite song first, then her second favorite song, and the three least favorite songs last? P(desired order) = s s + f = (2 favorite 1st in order)(3 least last, any order) total possible order = (1 nPr 1)(3 nPr 3)/(6 nPr6) ≈ 0.0083

  40. Probability with Permutations Ex. 2 Courtney has a playlist of 6 songs on her MP3 player. What is the probability that the player will randomly play her favorite song first, then her second favorite song, and the three least favorite songs last? P(desired order) = s s + f = (2 favorite 1st in order)(3 least last, any order) total possible order = (1 nPr 1)(3 nPr 3)/(6 nPr6) ≈ 0.0083 ≈ 0.8%

  41. Probability with Both Ex. 3 Suppose Hernanda pulls 5 marbles without replacement from a bag of 28 marbles in which there are 7 red, 7 black, 7 blue, & 7 white. What is the probability that 2 are of one color and 3 are of another color?

  42. Probability with Both Ex. 3 Suppose Hernanda pulls 5 marbles without replacement from a bag of 28 marbles in which there are 7 red, 7 black, 7 blue, & 7 white. What is the probability that 2 are of one color and 3 are of another color? P(3 one color, 2 another) = s s + f

  43. Probability with Both Ex. 3 Suppose Hernanda pulls 5 marbles without replacement from a bag of 28 marbles in which there are 7 red, 7 black, 7 blue, & 7 white. What is the probability that 2 are of one color and 3 are of another color? P(3 one color, 2 another) = s s + f = (2 colors from 4)(2 same from 7)(3 same from 7) total possible

  44. Probability with Both Ex. 3 Suppose Hernanda pulls 5 marbles without replacement from a bag of 28 marbles in which there are 7 red, 7 black, 7 blue, & 7 white. What is the probability that 2 are of one color and 3 are of another color? P(3 one color, 2 another) = s s + f = (2 colors from 4)(2 same from 7)(3 same from 7) total possible = (4 nPr 2)(7 nCr 2)(7 nCr 3)/(28 nCr 5)

  45. Probability with Both Ex. 3 Suppose Hernanda pulls 5 marbles without replacement from a bag of 28 marbles in which there are 7 red, 7 black, 7 blue, & 7 white. What is the probability that 2 are of one color and 3 are of another color? P(3 one color, 2 another) = s s + f = (2 colors from 4)(2 same from 7)(3 same from 7) total possible = (4 nPr 2)(7 nCr 2)(7 nCr 3)/(28 nCr 5)

  46. Probability with Both Ex. 3 Suppose Hernanda pulls 5 marbles without replacement from a bag of 28 marbles in which there are 7 red, 7 black, 7 blue, & 7 white. What is the probability that 2 are of one color and 3 are of another color? P(3 one color, 2 another) = s s + f = (2 colors from 4)(2 same from 7)(3 same from 7) total possible = (4 nPr 2)(7 nCr 2)(7 nCr 3)/(28 nCr 5) ≈ 0.0897

  47. Probability with Both Ex. 3 Suppose Hernanda pulls 5 marbles without replacement from a bag of 28 marbles in which there are 7 red, 7 black, 7 blue, & 7 white. What is the probability that 2 are of one color and 3 are of another color? P(3 one color, 2 another) = s s + f = (2 colors from 4)(2 same from 7)(3 same from 7) total possible = (4 nPr 2)(7 nCr 2)(7 nCr 3)/(28 nCr 5) ≈ 0.0897 ≈ 9%

  48. Expected Value • Expected Value – The weighted average of the values in a probability distribution if the weight applied to each value is its theoretical probability – tells what will happen in long run

  49. Expected Value • Expected Value – The weighted average of the values in a probability distribution if the weight applied to each value is its theoretical probability – tells what will happen in long run Ex. 4 Find the expected value of one roll of a die

  50. Expected Value • Expected Value – The weighted average of the values in a probability distribution if the weight applied to each value is its theoretical probability – tells what will happen in long run Ex. 4 Find the expected value of one roll of a die E(x) =

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