Transcript 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 Failure
P(F) = f
s+f
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?
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.
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
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)
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
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]
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
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)
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
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]
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
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
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
2. Determine the number of possibilities
2. Determine the number of possibilities
s+f=
2. Determine the number of possibilities
s + f = C(16+12,6)
2. Determine the number of possibilities
s + f = C(16+12,6) = C(28,6)
2. Determine the number of possibilities
s + f = C(16+12,6) = C(28,6) = 376,740
2. Determine the number of possibilities
s + f = C(16+12,6) = C(28,6) = 376,740
[On calc: 28, MATH, PRB, nCr, 6]
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.
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)
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
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
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
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%
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:
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)
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?
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
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
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 nPr 6)
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 nPr 6)
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 nPr 6) ≈ 0.0083
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 nPr 6) ≈ 0.0083 ≈ 0.8%
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?
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
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
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)
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)
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
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%
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 die
E(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 die
E(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 die
E(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 die
E(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 die
E(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 die
E(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 die
E(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 die
E(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 die
E(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 die
E(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