(Chapter Probability 3)
Lecture Notes
3-1
Examples: - flip a coin - toss a die
Whats 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? Whats it all mean?
STAT 3717: Probability - 2
STAT 3717: Probability - 1
Random Experiment: an experiment whose outcomes depend on chance. Sample Point or basic Possible Outcome of an experiment. Sample Space (S): collection (or set) of all possible outcomes in random experiment. Event: a collection of outcomes.
Possible Outcomes (Sample point): 1 2 3 4 5 6 Sample Space, S : { 1, 2, 3, 4, 5, 6 }
STAT 3717: Probability - 4
STAT 3717: Probability - 3
Possible Outcomes: -HH -HT -TH -TT Sample Space, S : { HH, HT, TH, TT }
Sample Space: S = {Head, Tail} S = {Life span of a human} = {x | x0 ,xR} Event: E = {Head} E = {Life span of a human is less than 3 years}
STAT 3717: Probability - 5
STAT 3717: Probability - 6
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-2
Probability is always a value between 0 and 1. Total probability of all possible outcomes equals 1.
A rough definition:(frequentist definition) Probability of event A, denoted P(A), is the proportion of times that the event A would occur in a very long series of repetitions of a random experiment.
STAT 3717: Probability - 7
STAT 3717: Probability - 8
!
Total Heads / Number of Tosses
1.00 0.75 0.50 0.25 0.00 0 25 50 75
" #
$
100
125
STAT 3717: Probability - 10
Number of Tosses
STAT 3717: Probability - 9
$
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,
%
$
Probability Density (Mass) Distribution
0.2 0.15 0.1 0.05 0 1 2 3 4 5 6
and total probability is 1.
STAT 3717: Probability - 11
STAT 3717: Probability - 12
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-3
Possible Outcomes: -HH -HT -TH -TT
Probability (observed): P(HH) = 4/9 P(HT) = 2/9 P(TH) = 2/9 P(TT) = 1/9
!$
Event A : { observe at least one head } Event B : { observe exactly one head } P(A) = P(HH) + P(HT) + P(TH) = 4/9 + 2/9 + 2/9 = 8/9 P(B) = 2/9 + 2/9 = 4/9
STAT 3717: Probability - 13
STAT 3717: Probability - 14
&
Intersection of events: A B <=> A and B
A B
A
B
Union of events: A B <=> A or B
A
Sample Space
B
AB
STAT 3717: Probability - 15 STAT 3717: Probability - 16
Relative Frequency and Probability
Number of children per household from a sample of 300 households
Class 0 1 2 3 4 5 Total
STAT 3717: Probability - 17
'
If a household is randomly selected from these 300 households, consider events: A = a household has less than 3 children. B = a household has between 2 to 4 children P(A) = .81, P(B) = .42 P(A and B) = P(2 children) = .24
STAT 3717: Probability - 18
Frequency 54 117 72 42 12 3 300
Relative Frequency .18 .39 .24 .14 .04 .01 1.00
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-4
&
(
)
!$
A
B
P(A or B) = P(A) + P(B) - P(A and B) = .81 + .42 - .24 = .99
Sample Space
P(A B) = P(A) + P(B) - P(A B)
STAT 3717: Probability - 19 STAT 3717: Probability - 20
$$
* $
Relative Frequency and Probability
Number of children per household from a sample of 300 households
Class 0 1 2 3 4 5 Total
STAT 3717: Probability - 22
A and B are mutually exclusive (disjointed) events
Frequency 54 117 72 42 12 3 300
A
B
P(A B) = P(A) + P(B)
STAT 3717: Probability - 21
Relative Frequency .18 .39 .24 .14 .04 .01 1.00
)
!$
Ac is the complement of event A Ac A
In the 300 household example, if event A is selecting a household with 0 children, and event B is selecting a household with 5 children, then, A and B are mutually exclusive, and P(A) = .18, P(B) = 0.01 P(A B) = P(A) + P(B) = .19
STAT 3717: Probability - 23
Sample Space
P(A) + P(Ac) = 1
STAT 3717: Probability - 24
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-5
In the 300 household example, if event A is selecting a household that has 0 children and event Ac is selecting a household that has at least one child, then A and Ac are also mutually exclusive and complementary. P(A) = .18, P(Ac) = 1 P(A) = 1 .18 = .82
STAT 3717: Probability - 25
How to determine probability? Empirical Probability Theoretical Probability (Subjective approach)
STAT 3717: Probability - 26
)
Empirical study: Outcome Head Tail Total Frequency 512 488 1000
)
Empirical probability assignment for event E:
P(E) = Number of times event E occurs Number of times experiment is repeated m = n
Probability of Head: P(Head) = 512/1000 = .512 = 51.2%
STAT 3717: Probability - 27
STAT 3717: Probability - 28
Relative Frequency and Probability Distribution
Number of children per household from a sample of 300 households
Class 0 1 2 3 4 5 Total
STAT 3717: Probability - 29
Discrete Distribution
Relative Frequency Distribution
0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5
Frequency 54 117 72 42 12 3 300
Relative Frequency .18 .39 .24 .14 .04 .01 1.00
STAT 3717: Probability - 30
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-6
)
If a household is randomly selected from the 300 households, what is the probability that it has 3 children? P(3 children) = (42)/300 = .14 or 14%
Make a reasonable assumption:
We have a balanced coin!
Hence, all possible outcomes are equally likely.
STAT 3717: Probability - 31
STAT 3717: Probability - 32
)
Theoretical probability assignment for event E:
P(E) = Number of equally likely outcomes in event E Size of the sample space n( E ) = n( S )
)
$
In a randomly observed sample of 100 individuals, 25 have cancer (event A), and 50 are smokers (event B). Also, 20 are smokers and have cancer.
STAT 3717: Probability - 34
Probability of Head: P(Head) = 1/2 = .50 = 50%
STAT 3717: Probability - 33
&
25
&
50 A
A
B
B
Sample Space n(S) = 100
STAT 3717: Probability - 35
n(A) = 25
Sample Space n(S) = 100
STAT 3717: Probability - 36
n(A) = 25 n(B) = 50
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-7
+$
25 ?
&
50 20 ? ?
+$
25 5
&
50 20 30 45
A
B
A
B
Sample Space
Sample Space
n(A B) = 20
STAT 3717: Probability - 37 STAT 3717: Probability - 38
AB
+$
&
45 5 A 20 30 B .05 A
&
.45 .30 .20 B
Sample Space n(S) = 100
STAT 3717: Probability - 39
Sample Space
AB
STAT 3717: Probability - 40
P(A B) = .20
,
Smoke B Cancer A No cancer c A 50 100 20 Not Smoke c B 25 Cancer A No cancer c A Smoke B 20 (.2) 30 50 Not Smoke c B 5 45 50 25 75 100
n(A B) = 20 (Smoke and have cancer) n(A) = 25, n(B) = 50, n(S) = 100
STAT 3717: Probability - 41
n(A B) = 20 (Smoke and have cancer) P(A B) = n(A B) / n(S) = 20/100 = .2
STAT 3717: Probability - 42
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-8
,
Smoke B Cancer A No cancer c A 20 (.2) 30 (.3) 50 P(B) =.5 Not Smoke c B 5 (.05) 45 (.45) 50 c P(B ) = .5 25 P(A) = .25 75 P(A ) = .75
c
&
A
B
= 1 or 100%
100
n(A B) = 20 (Smoke and have cancer) P(A B) = n(A B) / n(S) = 20/100 = .2
STAT 3717: Probability - 43
Sample Space
STAT 3717: Probability - 44
%
%
Smoke B Cancer A 20 (.2) 30 (.3) 50 P(B) =.5 Not Smoke c B 5 (.05) 45 (.45) 50 c P(B ) = .5 25 P(A) = .25 75 c P(A ) = .75
= 1 or 100%
A
B
No cancer c A
100
Sample Space
P(A B) = P(A) + P(B) - P(A B)
STAT 3717: Probability - 45
n(A B) = 55 (Smoke or have cancer) P(A B) = n(A B) / n(S) = 55/100 = .55
STAT 3717: Probability - 46
(
)
!$
The conditional probability of A to occur given B has occurred is denoted as P(A|B) and is,
P(A | B) =
P(A) = .25
P(A B) = P(A) + P(B) - P(A B) = .25 + .50 - .20 = .55
Smoke B Cancer A No cancer c A P(B) =.5
STAT 3717: Probability - 47
Not Smoke c B
(.2)
20
25
if P(B) is not zero,
50
= 1 or 100%
n(A B) n(B) P(A B) = P(B)
100
n(E) = number of equally likely outcomes in event E.
STAT 3717: Probability - 48
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-9
Smoke B Cancer A No cancer c A 20 (.2) 30 (.3) 50 P(B) =.5
Not Smoke c B 5 (.05) 45 (.45) 50 c P(B ) = .5 25 P(A) = .25 75 c P(A ) = .75
= 1 or 100%
Smoke B Cancer A No cancer c A 20 (.2) 30 (.3) 50 P(B) =.5
Not Smoke c B 5 (.05) 25 P(A) = .25
100
75 45 c (.45) P(A ) = .75 50 100 c P(B ) = .5 = 1 or 100%
P(A|B) = ? P(A|Bc) = ?
STAT 3717: Probability - 49
P(A|B) = P(A B)/P(B) = .2/.5 = .4 P(A|Bc) = ?
STAT 3717: Probability - 50
,
Smoke B Cancer A No cancer c A 20 (.2) 30 (.3) 50 P(B) =.5 Not Smoke c B 5 (.05) 25 P(A) = .25
75 45 c (.45) P(A ) = .75 50 100 c P(B ) = .5 = 1 or 100% 4 times
If two events A and B are independent then P(A|B) = P(A) P(B|A) = P(B) and P(A B) = P(A) P(B) . Reason: Since P(A|B) = P(A B) / P(B), P(A B) = P(A|B) P(B) = P(A) P(B)
STAT 3717: Probability - 52
P(A|B) = P(A B)/P(B) = .2/.5 = .4 P(A|Bc) = P(A B)/P(Bc) = .05/.5 = .1
STAT 3717: Probability - 51
)
Ten percent of the people in a large population have disease H. If one person is randomly selected what is the probability that this person has disease H? P(H) = 10% ?.
.,
Ten percent of the people in a large population have disease H. If two people are randomly selected what is the probability that both of them have disease H? P(H1 H2) = ? .
H1 denotes the event that the first people chosen has disease H. H2 denotes the event that the 2nd people chosen has disease H.
STAT 3717: Probability - 53
STAT 3717: Probability - 54
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-10
,
Ten percent of the people in a large population have disease H. If two people are randomly selected what is the probability that both of them have disease H? P(H1 H2) = P(H1| H2) P(H2) = P(H1) P(H2) (H1 and H2 are almost independent.)
STAT 3717: Probability - 55
,
Ten percent of the people in a large population have disease H. If two people are randomly selected what is the probability that both of them have disease H? P(H1 H2) = P(H1) P(H2) = .10 x .10 = .01 = 1% .
STAT 3717: Probability - 56
,
Ten percent of the people in a large population have disease H. If three people are randomly selected what is the probability that both of them have disease H? P(H1 H2 H3) = P(H1) P(H2) P(H3) = .10 x .10 x .10 = .001 = 0.1% .
STAT 3717: Probability - 57
$
Tree diagram Multiplication principle Permutation rule Combination rule
+$
STAT 3717: Probability - 58
Multiplication Principle
Coin Spinner 1 4
1 2 3 4 1 2 3 4
2 3
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 number of possible outcomes of the sequence of events will be n1 n2 n3 nk
Example:
H T
STAT 3717: Probability - 59
2x4=8
possible outcomes
STAT 3717: Probability - 60
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-11
$
How many ways can a three-letter code be formed by using A, B, and C without repeating use of the same digit? n1 = 3, n2 = 2, n3 = 1
number of ways = 3 x 2 x 1 = 3! = 6
n! = 123... n Example: 3! = 123 = 6
STAT 3717: Probability - 61
STAT 3717: Probability - 62
$
How many ways can a four-digit code be formed by selecting 4 distinct digits from the nine digits, 1 through 9, without repeating the use of any digit? 9876 = 9! / 5! = 9! / (9-4)! (formula)
Permutation Rule
Permutation Rule: The number of possible permutations of selecting r objects from a collection of n distinct objects is
Prn =
n! (n r )!
Note: 1. No repetitions. 2. Order of selection is important.
STAT 3717: Probability - 63 STAT 3717: Probability - 64
Combination Rule
How many ways can a combination of 3 distinct letters be selected from the three letters A, B, C and D? Permutations = 4! / (4 3)! = 24
Order is not important, the possible permutations of 3 letters = 3! = 6 Combination Rule: The number of possible combinations of r objects from a collection of n distinct objects is
Crn =
n Prn n! = = r r! r!(n r )!
Combinations = 24 / 6 = 4.
STAT 3717: Probability - 65
Note: 1. No repetition. 2. Order is not important.
STAT 3717: Probability - 66
STAT 3717 Yates / Chang
Probability 3)
Lecture (Chapter Notes
3-12
How many ways can a combination of 4 distinct digits be selected from nine digits, 1 through 9? 9876 / 4! = 9! / (4! 5!) = 9! / [4! (9-4)!] = 126
STAT 3717: Probability - 67
How many ways can 6 distinct numbers be selected from a set of 47 distinct numbers? 47! / (6! 41!) = 10,737,573
(formula)
STAT 3717: Probability - 68
!
If n elements are selected from a population so that every set of n elements selected from the population has a equal probability of being selected, the n elements are called a (simple) random sample.
A random sample is likely to be representative of the population.
STAT 3717: Probability - 69
!
Selecting a random sample: - picking numbers from a hat - using a table of random numbers (random number generator)
STAT 3717: Probability - 70
!
-
&
/ .
!
&
What is Random Variable?
A random variable assumes a numerical value associated with the outcome of a random experiment (by chance). Usually is denoted by a capital letter. X, Y, Z, ... Very useful for mathematically modeling the distribution of variables.
STAT 3717: Probability - 71
STAT 3717: Probability - 72
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-13
$ $ !
&
!
&
Continuous Random Variable: A random variable that assumes continuous values in one or more intervals.
Discrete Random Variable: A random variable that assumes discrete values by chance.
STAT 3717: Probability - 73
STAT 3717: Probability - 74
!
X = 1, if Head occurs, and X = 0, if Tail occurs.
&
$
If a balanced coin is tossed, Head and Tail are equally likely to occur, P(X=1) = .5 = 1/2 and P(X=0) = .5 = 1/2
1/2 0 1
Example: (Toss a balanced coin)
P(Head) = P(X=1) = P(1) = .5 P(Tail) = P(X=0) = P(0) = .5
Probability distribution function: p(X) = .5 , if X = 0, 1, and p(X) = 0 , otherwise.
STAT 3717: Probability - 75
P(all possible outcomes) = P(X=1 or 0) = P(X=1) + P(X=0) = 1/2 + 1/2 = 1.0 Total probability is 1.
STAT 3717: Probability - 76
& $ $ *
Table
0 $
Count p(X)
Experiment: Toss 2 Coins. Count # Tails.
Probability Distribution Values, X Probabilities, P(X) 0 1 2
1984-1994 T/Maker Co.
# Tails
# times
0 1 2
1 2 1
.25 .50 .25
1/4 = .25 2/4 = .50 1/4 = .25
Graph p(X) .50 .25 .00
STAT 3717: Probability - 78
0
1
2
X
STAT 3717: Probability - 77
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-14
$
1. List of All possible [x, p(x)] pairs
x = Value of Random Variable (Outcome) p(x) = Probability Associated with Value
!
Example: Toss of a single die.
&
1/6
1 2 3 4 5 6
Probability distribution function: p(X) = 1/6, if X = 1, 2, 3, 4, 5, or 6, p(X) = 0 otherwise.
2. 0 p(x) 1, for all possible values of x. 3. p(x) = 1, sum over all possible values of x.
Example: What is probability of getting a number less than 3 when roll a balanced die? p( X < 3 ) = p( X 2) = ? Answer: 2/6 = 1/3
STAT 3717: Probability - 79
STAT 3717: Probability - 80
!
&
$
The probability distribution of a discrete random variable X gives the probability associated with each possible X value. Each probability is the limiting relative frequency of occurrence of the corresponding X value when the experiment is repeatedly performed. The probability distribution may be given by means of a table, formula, or graph.
STAT 3717: Probability - 82
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,
STAT 3717: Probability - 81
!
Histogram for categorical variable x.
(n = 1000) 500 250 0 1 2
x
&
p(x)
frequency
x 0 1 2
p(x) .25 .50 .25
.50 .25 0 1 2
x
Mean = xi / n = (250 x 0 + 500 x 1 + 250 x 2) / 1000 = 1
STAT 3717: Probability - 83
(250x0 + 500x1 + 250x2)/1000 = 1 Mean of the (.25x0 + .5x1 + .25x2) = 1
distribution
STAT 3717: Probability - 84
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-15
$ $
The mean value (expected value) of a discrete random variable (distribution) X, denoted by X or just , or E[X], is defined as = x p(x)
Example: Toss a balanced coin and interested in number of heads that turn up. {x = 1 implies Head and x = 0 implies not Head, and p(1) = .5, p(0) = .5.}
*
$
.50 .25 0 1 2
So, = 1 p(1) + 0 p(0) = 1 .5 + 0 .5 = .5
STAT 3717: Probability - 85
If the distribution is p(0) = .25, p(1) = .5, p(2) = .25, then the mean of the distribution is = (.25x0 + .5x1 + .25x2) = 1
STAT 3717: Probability - 86
*
x -4 -1 11 p(x) .25 .50 .25
p(x)
$
Probability distribution (or line chart) x -100 -1 11 -4 -1 11
x
*
p(x) .10 .25 .65
p(x) .75
& $
.50 .25
.50 .25 -100 -1 11
x
The mean of the distribution is = (.25x(-4) + .5x(-1) + .25x11) = 1.25
STAT 3717: Probability - 87
The mean of the distribution is = (-100) x.1+ (-1)x.25+ 11x.65 = -3.1
STAT 3717: Probability - 88
$ $
The variance of a discrete random variable (distribution) X, denoted by X 2 or just 2 (or V[X]) is defined as
$ $
Example: Toss a balanced coin and interested in number of heads that turn up. { x = 1 implies head and x = 0 implies not head, and p(1) = .5, p(0) = .5.} The mean value, = 1p(1) + 0 p(0) = .5
2 = (x - )2 p(x)
The standard deviation of a discrete random variable (distribution) X, denoted by X or just is defined as 2
2 = (1-.5)2 p(1) + (0-.5)2 p(0) = .125 + .125
= .25
STAT 3717: Probability - 90
The variance,
=
STAT 3717: Probability - 89
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-16
x (x-)2 0 1 1 0 2 1
p(x) .25 .50 .25
p(x) .50 .25 0 1 2 x
$ ! & .
Mean = = 0x0.25 + 1x.5 + 2x.25 = 1 2 = (0-1)2x.25+ (1-1)2x.5+ (2-1)2x.25 = .5
STAT 3717: Probability - 91 STAT 3717: Probability - 92
Bernoulli Trial
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) P(S) = P(X=1) = p, P(F) = P(X=0) = 1 p. Example: (Head, Tail), (Died, Survived)
1
$
Example: In a random experiment of tossing an unbalanced coin, the probability of Head is 0.3, what is the probability distribution? P(Head) P(Tail) = P(X=1) = 0.3, = P(X=0) = 1 0.3 = 0.7.
(A discrete distribution.)
STAT 3717: Probability - 94
STAT 3717: Probability - 93
1
*
1
How to model the number successes, x, in a sequence of n independent and identical Bernoulli trials?
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. Roll a die 3 times and observing a 6 or not 6 turns up. A random sample of 5 people and observing 2 of them have disease H. (Approximately binomial.)
STAT 3717: Probability - 96
STAT 3717: Probability - 95
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-17
,
If two events A and B are independent then P(A|B) = P(A) P(B|A) = P(B) and P(A B) = P(A) P(B) . Reason: Since P(A|B) = P(A B) / P(B), P(A B) = P(A|B) P(B) = P(A) P(B)
STAT 3717: Probability - 97
,
The probability of getting all 6s in roll a balanced die 2 times experiment. P(S1 S2) = P(S1) P(S2 | S1) = P(S1) P(S2) = (1/6)2
STAT 3717: Probability - 98
,
If events A1, A2, , Ak are independent, then P(A1 and A2 and and Ak) = P(A1) P(A2) P(Ak)
,
The probability of getting all 6s in rolling a balanced die 5 times experiment. P(S1 S2 S3 S4 S5) = P(S1) P(S2) P(S3) P(S4) P(S5) = (1/6)5
STAT 3717: Probability - 99
STAT 3717: Probability - 100
,
The probability of getting 6s in the first two trials and non-6s from the last three trials from rolling a balanced die 5 times experiment. P(S1 S2 F3 F4 F5) = P(S1) P(S2) P(F3) P(F4) P(F5) = (1/6)2 (5/6)3
STAT 3717: Probability - 101
1
The probability of getting two 6s in rolling a balanced die 5 times experiment. P(SSFFF) = (1/6)2 x (5/6)3 P(SFSFF) = (1/6)2 x (5/6)3 P(SFFSF) = (1/6)2 x (5/6)3 5 5! How many of them? = = 10 2 2! 3!
STAT 3717: Probability - 102
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-18
1
The probability of getting two 6s in rolling a balanced die 5 times experiment is
1
In a binomial experiment involving n independent and identical Bernoulli trials each with probability of success p, the probability of having x successes can be calculated with the binomial probability distribution function, and it is, for x = 0, 1, , n,
P(2 Ss) =
( 2 ) x (1/6) x (5/6)
2
5
3
P (X = x ) = =
STAT 3717: Probability - 104
n x
p x (1 p) n x
= 5!/(2!3!) x (1/6)2 x (1 1/6)3
STAT 3717: Probability - 103
n! p x (1 p ) n x x!(n x )!
1 $
Parameters of the distribution: Mean of the distribution, = np Variance of the distribution, 2 = np(1 p)
Standard deviation, , is the square root of variance.
1
Example: A balanced die is rolled three times (or three balanced dice are rolled), what is the probability to see two 6s?
[ = 3(1/6) = 1/2,
2 = 3(1/6)(5/6) = 5/12 ]
Identify n = 3, p = 1/6, x = 2 P(X=2) = { 3! / [2!1!] } (1/6)2(5/6)3-2 = 3(1/6)2(5/6)1 = .069
P(X = x) = n! p x (1 p) n x x!(n x )!
= np(1 p)
STAT 3717: Probability - 105
STAT 3717: Probability - 106
Random Sample
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.
1
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? (Approximately Binomial)
Identify n = 5, x = 4, p = .10 P(X=4) = { 5! / [4!(5-4)!] }(.10)4(1.10)5-4 = 5(.10) 4(.90)1 = .00045
P(X = x) =
STAT 3717: Probability - 108
STAT 3717: Probability - 107
n! p x (1 p) n x x!(n x )!
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-19
1
Example: In the previous problem, what is the probability that 4 or more people have the disease?
Identify n = 5, x = 4, p = .10 P(X4) = P(X=4) + P(X=5) = .00045 + { 5! / [5!(5-5)!] }(.10)5(1 .10)5-5 = .00045 + .00001 = .00046
$
The Poisson distribution is used to model discrete events that occur infrequently in time or space. Model the number of successes in a given time period or in a given unit space.
STAT 3717: Probability - 110
(What is this number telling us?)
STAT 3717: Probability - 109
$
Let X represents the number of occurrences of some event of interest over a given interval from a Poisson process, and the is the mean of the distribution, the probability of X assumes the value x is, for x = 0, 1, 2, ,
P( X = x) = e
x!
STAT 3717: Probability - 111
x
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.
It can be used to approximate Binomial prob, with large n.
STAT 3717: Probability - 112
If on average there are 4 people catch flu in a given week in a community during a certain season, what is the probability of observing 2 people catch flu in this community in a given week period during the season? (Assume the number of people catching flu in a given period of time follow a Poisson Process.) =4 x=2 P(X=2) =(e-442)/2! = .1465
STAT 3717: Probability - 113
If on average there are 4 people catch flu in a given week in a community during a certain season, what is the probability of observing less than 2 people catch flu in this community in a given week period during the season? (Assume the number of people catching flu in a given period of time follow a Poisson Process.) =4 x = 0, 1 P(X<2) = P(X=0) + P(X=1) = (e-440)/0! + (e-441)/1! = .0916
STAT 3717: Probability - 114
STAT 3717 Yates / Chang
Probability (Chapter 3)
Lecture Notes
3-20
P ( X = x) =
n! p x (1 p) n x x!(n x)!
x
2 $
e-5 =? 4! = ?
$ "" "
$
!
STAT 3717: Probability - 115
STAT 3717: Probability - 116
P ( X = x) = e
x
x!
x
"
" "
"
STAT 3717: Probability - 117
STAT 3717 Yates / Chang