Download Probability Terminology What is probability?

Probability-1
A researcher claimed that there are
10% of the a large population have
disease H.
A quality control officer claimed that
10% of the products are defective.
A random sample of 5 people is taken
from this population and examined.
A random sample of 5 products is
taken from this population and
examined.
If 4 people in this random sample
have the disease, what does it mean?
How likely would this happen if the
research is right?
If 4 products in this random sample
are defective, what does it mean?
How likely would this happen if the
quality control officer is right?
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Probability Terminology
•
•
•
Random Experiment:
Experiment : an experiment
whose outcomes depend on chance.
Sample Space:
Space: collection of all
possible outcomes in random
experiment.
Event:: a collection of outcomes
Event
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Definition of Probability
What is probability?
What’s the probability
of getting a head on
the toss of a single fair
coin? Use a scale
from 0 (no way) to 1
(sure thing).
So toss a coin twice.
Do it! Did you get one
head & one tail?
What’s it all mean?
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Empirical Probability
Assignment
A rough definition:(frequentist definition)
Empirical probability assignment:
Probability of event A is the
proportion of times that the event A
would occur in a very long series of
repetitions of a random experiment.
P(A) =
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Number of times event A occurred
Number of times experiment is repeated
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Probability-2
Theorectial Probability
Assignment
Many Repetitions!*
Total Heads /
Number of Tosses
1.00
Theoretical probability assignment:
Number of equally likely outcomes in event A
0.75
P(A) =
Size of the sample space
0.50
0.25
0.00
0
25
50
75
100
Number of Tosses
125
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Counting Technique
•
•
•
•
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Tree Diagram
Tree diagram
Multiplication principle
Permutation rule
Combination rule
1
2
4
3
H
T
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Multiplication Principle
Counting Rule:
Multiplication Principle: In a sequence of k events
in which the first one has n1 possibilities and the
second event has n2 and the third has n3 , and so
forth, the total possibilities of the sequence will be
n1 × n2 × n 3 × … × n k
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Permutation Rule
Permutation Rule:
The number of possible permutations of r
objects from a collection of n distinct
objects is
n Pr = n!/(n - r)!
2x4=8
possible outcomes
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1
2
3
4
1
2
3
4
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Probability-3
Factorial
n! = 1·2·3·... ·n
Example:
Permutations
How many ways can a four-digit code be
formed by selecting 4 distinct digits from
nine digits, 1 through 9, without repeating
use of the same digit?
3! = 1·2·3 = 6
9·8·7·6 = 9!/5! = 9!/(9-4)! (formula)
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Combination Rule
Combination Rule:
The number of possible combinations of r
objects from a collection of n distinct objects is
n Cr =
n
r
= n!/[r!(n - r)!]
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Combinations
Combinations
How many ways can a combination of 4
distinct digits be selected from nine digits,
1 through 9?
9·8·7·6/4! = 9!/(4! · 5!)
= 9!/[4! · (9-4)!] (formula)
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Venn Diagram
How many ways can 6 distinct numbers
be selected from a set of 47 distinct
numbers?
A
B
47!/(6! ·41!) =10,737,573
Sample Space
A∩B
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Probability-4
Relative Frequency and
Probability
Intersection of events:
A ∩ B <=> A and B
Union of events:
A ∪ B <=> A or B
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Number of children per household
from a sample of 300 households
Class
Frequency
0
1
2
3
4
5
Total
54
117
72
42
12
3
300
Relative
Frequency
.18
.39
.24
.14
.04
.01
1.00
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Venn Diagram
If a household is random selected from
these 300 households,
A = a household has less than 3 children.
B = a household has between 2 to 4
children
P(A) = .81, P(B) = .28
P(A and B) = P(2 children) = .24
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General Addition Rule
P(A or B) = P(A) + P(B) - P(A and B)
= .81 + .28 - .24
= .85
A
B
Sample Space
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
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Mutually Exclusive
A and B are mutually exclusive (disjointed) events
A
P(A ∪ B) = P(A) + P(B)
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B
Probability-5
Special Addition Rule
In the 300 household example, if event B
is select a household that has five
children, then A and B are mutually
exclusive.
P(A) = .81, P(B) = 0.1
P(A or B) = P(A) + P(B) = .82
Complement of Event
A’ is the complement of event A
A’
A
Sample Space
P(A) = 1 - P(A’)
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Conditional Probability
The conditional probability of A to occur
given B has occurred is denoted as P(A|B)
and is, if P(B) is not zero,
P(A|B) = P(A and B) / P(B)
or
P(A|B) = n(A and B) / n(B)
n(E) = # of equally likely outcomes in E.
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General Multiplication
Rule
General multiplication rule:
P(A and B) = P(A|B) P(B) = P(B|A) P(A)
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Conditional Probability
Cancer
C
Smoke
S
20
(.2)
Not Smoke
S’
5
P(C) = .25
(.05)
No cancer
C’
30
(.3)
45
(.45)
P(S) =.5
P(S’) = .5
P(C’) = .75
P(C|S) = .2/.5 = .4
P(C|S’) = .05/.5 = .1
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Independent Events
Events A and B are independent if
P(A|B) = P(A)
or
P(A and B) = P(A) ·P(B)
{=P(A|B) · P(B)}
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Probability-6
Independent Events
Independent Events
If events A1 , A2, …, Ak are independent,
then
Hi : the ith trial in tossing a coin turns up Head
P(H1 and H2) = P(H1 ) P( H 2) (= .5 ·.5 = .25)
P(A1 and A2 and … and Ak)
= P(A 1) ·P(A2 ) · … ·P(A
·P(Ak)
P(H1 and H2 and H3 ) = P(H1 ) P( H2) P( H3)
What is probability of seeing 2 heads in
tossing 3 coins experiment?
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Probability Distribution
If a balanced coin is tossed, Head and Tail
are equally likely to occur,
P(Head) = .5 = 1/2 and P(Tail) = .5 = 1/2
P(all possible outcomes) = P(Head or Tail)
= P(Head) + P(Tail)
= 1/2 + 1/2
= 1.0
Total probability is 1.
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Probability Distribution
If outcomes are equally likely to
occur, the distribution is
P(1) = 1/6,
P(3) = 1/6,
P(5) = 1/6,
P(2) = 1/6,
P(4) = 1/6,
P(6) = 1/6,
and total probability is 1.
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Uniform Distribution
Properties of Probability
• Probability is always a value
between 0 and 1.
Probability Density (Mass) Distribution
0.2
0.15
• Total probability equals 1.
0.1
0.05
0
0
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1
2
3
4
5
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Probability-7
Relative Frequency and
Probability
Discrete Distribution
Number of children per household
from a sample of 300 households
Class
Frequency
0
1
2
3
4
5
Total
54
117
72
42
12
3
300
Relative
Frequency
.18
.39
.24
.14
.04
.01
1.00
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0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
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Discrete Distribution
If a household is randomly selected
from the 300 household, what is the
probability that it has more than 3
children?
P(more than 3 children)
Relative Frequency Distribution
= .04 + .01
= .05
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Discrete Random Variable
Random Variable
A variable that assumes a numerical
description for the outcome of a
random experiment (by chance).
Usually is denoted by a capital letter.
X, Y, Z, ...
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Discrete Random Variable
Example: (Toss a balanced coin)
A random variable
assumes discrete values.
X = 1, if Head occurs, and X = 0 if Tail
occurs.
P(Head) = P(X=1) = P(1) = .5
P(Tail) = P(X=0) = P(0) = .5
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Probability-8
Measure of Center for a
Distribution
Discrete Random Variable
The mean value (expected value) of a
discrete random variable (distribution)
X, denoted by µX or just µ (or E[X]) is
defined as
Example: What is probability of getting a
number less than 3 when roll a balanced
die?
P( X < 3 ) = P( X ≤ 2) = ?
µX = Σx · p(x)
Example : Toss a balanced coin and interested in
number of heads turn up. { x=1 implies “head” and
x=2 implies “not head”, and p(1) = .5, p(0) = .5.}
Answer: 2/6 = 1/3
So, µX = 1 · p(1) + 0 ·p(0) = .5 + 0 = .5
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Measure of Spread for a
Distribution
Why Random Variable?
The variance of a discrete random
variable (distribution) X, denoted by σX
2 or just σ2 (or V[X]) is defined as
•
σX 2 = Σ (x - µ)2 ·p(x)
•
Example : Toss a balanced coin and interested in
number of heads turn up. { x=1 implies “head” and
x=0 implies “not head”, and p(1) = .5, p(0) = .5.}
A simple mathematical notation to
describe an event. e.g.: X < 3, X = 0, ...
Mathematical function can be used to
model the distribution through the use
of random variable. e.g.: Binomial,
Poisson, Normal, …
So, σX 2 = (1-.5) 2 ·p(1) + (0-.5)2 · p(0) = .125 + .125
= .25
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Bernoulli Trial
Bernoulli Probability
Definition: Bernoulli trial is a random
experiment whose outcomes are
classified as one of the two categories.
(S , F) or (Success, Failure) or (1, 0)
Example: In a random experiment of
tossing an unbalanced coin, the
probability of Head is 0.3, what is the
probability distribution?
P(S) = P(X=1) = π , P(F) = P(X=0) = 1 - π .
P(Head) = P(X=1) = 0.3,
P(Tail) = P(X=0) = 1 - 0.3 = 0.7.
Example: (Head, Tail), (Died, Survived)
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Probability-9
Bionomial Experiment
A random experiment involving a
sequence of independent and
identical Bernoulli trials.
Example:
•Toss a coin ten times and observing Head or
Tail turns up.
Binomial Probability Model
How to model the number
successes, x, in a sequence of n
independent and identical
Bernoulli trials?
•Roll a die 3 times and observing a 6 or not 6
turns up.
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Binomial Probability Model
In a binomial experiment involving n
independent and identical Bernoulli trials
each with probability of success π , the
probability of having x successes can be
calculated with the binomial probability
mass function, and it is, for x = 0, 1, …, n,
n− x
 n x
P ( X = x ) =   ⋅ π ⋅ (1 − π )
 x
x
n− x
n!
=
⋅ π ⋅ (1 − π )
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x!⋅( n − x )!
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Parameters of Binomial
Distribution
Parameters of the distribution:
Mean of the distribution, µ = n· π
Variance of the distribution, σ2 = n· π ·(1- π)
Standard deviation, σ , is the square root of
variance.
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Binomial Probability
Binomial Probabillity
Example: A balanced die is rolled three
times (or three balanced dice are rolled),
what is the probability to see two 6’s?
Example: If there are10% of the population in a
community have a certain disease, what is the
probability that 4 people in a random sample of 5
people from this community has the disease?
(Parameters: µ = 3·(1/6) = 1/2, σ 2 = 3·(1/6)·(5/6) = 5/12)
n = 3, π = 1/6, x = 2
Identify n = 5, x = 4, π = .10
P(X=2) = { 3! / [2!·1!] } ·(1/6)2·(5/6)3-2
= 3·(1/6)2·(5/6)1
= .069
P(X=4) = { 5! / [4!·(5-4)!] }·(.10)4·(1-.10)5-4
= 5·(.10) 4(.90)1
= .0004
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Probability-10
Binomial Probabillity
Poisson Distribution
Example: In the previous problem, what is the
probability that 4 or more people have the
disease?
The Poisson distribution is used to
model discrete events that occur
infrequently in time or space.
Identify n = 5, x = 4, π = .15
P(X≥4) = P(X=4) + P(X=5)
= .0004 + { 5! / [5!(5-5)!] }·(.10)5·(1-.10)5-5
= .0004 + .00001 = .00041
(What this number is telling us?)
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Model the number of successes in a
given time period or in a given unit
space.
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Poisson Distribution
Let X represents the number of occurrences
of some event of interest over a given
interval from a Poisson process, and the λ is
the mean and also the variance of the
distribution, the probability of X assumes the
value x is, for x = 0, 1, 2, …,
P( X = x) =
−λ
e λ
x!
x
Poisson Process
•
•
•
The probability that a single event occurs
within an interval is proportional to the length
of the interval.
Within a single interval, an infinite number of
occurrences is possible.
The events occurs independently both within
the same interval and between consecutive
non-overlapping intervals.
Can be used to approximate Binomial prob, with large n.
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Poisson Proability
If on average there are 4 cars stop by the gas
station A in a given 15-minute around noon,
what is the probability of observing 2 cars at this
station in a given 15-minute period around
noon? (Assume the arrivals of cars follow a
Poisson Process.)
A (Simple) Random Sample of size n
consists of n individuals from the
population chosen in such a way that
every set of n individuals has an equal
chance to be the sample actually
selected.
λ = 4, x = 2
P(X=2) =(e -4·42)/2! = .1465
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