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12.4 – Probability & Probability Distributions

12.4 – Probability & Probability Distributions

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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 - PowerPoint PPT Presentation

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Page 1: 12.4 – Probability & Probability Distributions

12.4 – Probability & Probability Distributions

Page 2: 12.4 – Probability & Probability Distributions

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

Page 3: 12.4 – Probability & Probability Distributions

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

Page 4: 12.4 – Probability & Probability Distributions

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

P(S) = s s + f

Page 5: 12.4 – Probability & Probability Distributions

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

Page 6: 12.4 – Probability & Probability Distributions

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

P(S) = s s + f

• Probability of FailureP(F) = f

s + f

Page 7: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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?

Page 8: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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.

Page 9: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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

Page 10: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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)

Page 11: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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

Page 12: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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]

Page 13: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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

Page 14: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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)

Page 15: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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

Page 16: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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]

Page 17: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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 ∙

Page 18: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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 ∙

Page 19: 12.4 – Probability & Probability Distributions

Probability with CombinationsEx. 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

Page 20: 12.4 – Probability & Probability Distributions

2. Determine the number of possibilities

Page 21: 12.4 – Probability & Probability Distributions

2. Determine the number of possibilitiess + f =

Page 22: 12.4 – Probability & Probability Distributions

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

Page 23: 12.4 – Probability & Probability Distributions

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

Page 24: 12.4 – Probability & Probability Distributions

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

Page 25: 12.4 – Probability & Probability Distributions

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

Page 26: 12.4 – Probability & Probability Distributions

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

3. Find the probability.

Page 27: 12.4 – Probability & Probability Distributions

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

3. Find the probability.P(3male,3female)

Page 28: 12.4 – Probability & Probability Distributions

2. Determine the number of possibilitiess + 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

Page 29: 12.4 – Probability & Probability Distributions

2. Determine the number of possibilitiess + 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

Page 30: 12.4 – Probability & Probability Distributions

2. Determine the number of possibilitiess + 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

Page 31: 12.4 – Probability & Probability Distributions

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

3. Find the probability.P(3male,3female) = ss + f= 123,200 376,740≈ 0.327016 ≈ 33%

Page 32: 12.4 – Probability & Probability Distributions

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

3. Find the probability.P(3male,3female) = ss + f= 123,200 376,740≈ 0.327016 ≈ 33%

NOTE: Can do all on calculator:

Page 33: 12.4 – Probability & Probability Distributions

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

3. Find the probability.P(3male,3female) = ss + f= 123,200 376,740≈ 0.327016 ≈ 33%

NOTE: Can do all on calculator:(12 nCr 3)(16 nCr 3)/(28 nCr 6)

Page 34: 12.4 – Probability & Probability Distributions

Probability with PermutationsEx. 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?

Page 35: 12.4 – Probability & Probability Distributions

Probability with PermutationsEx. 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

Page 36: 12.4 – Probability & Probability Distributions

Probability with PermutationsEx. 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

Page 37: 12.4 – Probability & Probability Distributions

Probability with PermutationsEx. 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 nPr 6)

Page 38: 12.4 – Probability & Probability Distributions

Probability with PermutationsEx. 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 nPr 6)

Page 39: 12.4 – Probability & Probability Distributions

Probability with PermutationsEx. 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 nPr 6) ≈ 0.0083

Page 40: 12.4 – Probability & Probability Distributions

Probability with PermutationsEx. 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 nPr 6) ≈ 0.0083 ≈ 0.8%

Page 41: 12.4 – Probability & Probability Distributions

Probability with BothEx. 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?

Page 42: 12.4 – Probability & Probability Distributions

Probability with BothEx. 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

Page 43: 12.4 – Probability & Probability Distributions

Probability with BothEx. 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

Page 44: 12.4 – Probability & Probability Distributions

Probability with BothEx. 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)

Page 45: 12.4 – Probability & Probability Distributions

Probability with BothEx. 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)

Page 46: 12.4 – Probability & Probability Distributions

Probability with BothEx. 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

Page 47: 12.4 – Probability & Probability Distributions

Probability with BothEx. 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%

Page 48: 12.4 – Probability & Probability Distributions

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

Page 49: 12.4 – Probability & Probability Distributions

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

Page 50: 12.4 – Probability & Probability Distributions

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 dieE(x) =

Page 51: 12.4 – Probability & Probability Distributions

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 dieE(x) = (1∙1/6)

Page 52: 12.4 – Probability & Probability Distributions

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 dieE(x) = (1∙1/6)+(2∙1/6)

Page 53: 12.4 – Probability & Probability Distributions

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 dieE(x) = (1∙1/6)+(2∙1/6)+(3∙1/6)

Page 54: 12.4 – Probability & Probability Distributions

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 dieE(x) = (1∙1/6)+(2∙1/6)+(3∙1/6)+(4∙1/6)

Page 55: 12.4 – Probability & Probability Distributions

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 dieE(x) = (1∙1/6)+(2∙1/6)+(3∙1/6)+(4∙1/6)+(5∙1/6)

Page 56: 12.4 – Probability & Probability Distributions

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 dieE(x) = (1∙1/6)+(2∙1/6)+(3∙1/6)+(4∙1/6)+(5∙1/6)+(6∙1/6)

Page 57: 12.4 – Probability & Probability Distributions

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 dieE(x) = (1∙1/6)+(2∙1/6)+(3∙1/6)+(4∙1/6)+(5∙1/6)+(6∙1/6)

= 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6

Page 58: 12.4 – Probability & Probability Distributions

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 dieE(x) = (1∙1/6)+(2∙1/6)+(3∙1/6)+(4∙1/6)+(5∙1/6)+(6∙1/6)

= 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 21/6

Page 59: 12.4 – Probability & Probability Distributions

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 dieE(x) = (1∙1/6)+(2∙1/6)+(3∙1/6)+(4∙1/6)+(5∙1/6)+(6∙1/6)

= 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 21/6 = 3.5