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

P(S) = s s + f

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

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

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?

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.

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

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)

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

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]

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

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)

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

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]

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 ∙

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 ∙

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

2. Determine the number of possibilities

2. Determine the number of possibilitiess + f =

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

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

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

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

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.

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)

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

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

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

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%

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:

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)

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?

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

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

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)

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)

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

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%

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?

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

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

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)

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)

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

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%

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

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

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) =

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)

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)

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)

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)

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)

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)

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

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

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