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Introduction to Probability Introduction to Probability and Statisticsand Statistics
Chapter 4
Probability and Probability Distributions
ProbabilityProbability• Example: If we toss a coin 10 times and get
10 heads in a row; • Question: Do you believe it is a fair coin?• Answer: No.• Reason: If the coin is fair, the chance to
have 10 heads in a row is less than 0.1% (According to probability theory).
• Tool and foundation of statistics; Evaluate reliability of statistical conclusions…
Basic ConceptsBasic Concepts
• An experimentexperiment is the process by which an observation (or measurement) is obtained.
• Experiment: Record an ageExperiment: Record an age
• Experiment: Toss a dieExperiment: Toss a die
• Experiment: Record an opinion (yes, no)Experiment: Record an opinion (yes, no)
• Experiment: Toss two coinsExperiment: Toss two coins
Basic ConceptsBasic Concepts
• A simple eventevent is the outcome that is observed on a single repetition of the experiment. – The basic element to which probability is
applied.– One and only one simple event can occur
when the experiment is performed.
• A simple event simple event is denoted by E with a subscript.
Basic ConceptsBasic Concepts
• Each simple event will be assigned a probability, measuring “how often” it occurs.
• The set of all simple events of an experiment is called the sample space, sample space, usually denoted by S.S.
ExampleExample• The die toss:The die toss:
• Simple events: Sample space:
11
22
33
44
55
66
E1
E2
E3
E4
E5
E6
S ={E1, E2, E3, E4, E5, E6}
SS•E1
•E6•E2
•E3
•E4
•E5
(or S ={1, 2, 3, 4, 5, 6})Venn Diagram
ExampleExample• Record a person’s blood type:Record a person’s blood type:
• Simple events: Sample space:
E1
E2
E3
E4
S ={E1, E2, E3, E4}AA
OO
BB
ABAB
S ={A, B, AB, O}
Basic ConceptsBasic Concepts
• An eventevent is a collection of one or more simple events. simple events.
•The die toss:The die toss:–A: an odd number–B: a number > 2
SS
A ={E1, E3, E5}
B ={E3, E4, E5, E6}
BBAA
•E1
•E6•E2
•E3
•E4
•E5
Basic ConceptsBasic Concepts
• Two events are mutually exclusivemutually exclusive if, when one event occurs, the other cannot, and vice versa.
•Experiment: Toss a dieExperiment: Toss a die–A: observe an odd number–B: observe a number greater than 2–C: observe a 6–D: observe a 3
Not Mutually Exclusive
Mutually Exclusive A and C?
A and D?B and C?
The Probability The Probability of an Eventof an Event
• The probability of an event A measures “how often” we think A will occur. We write P(A). P(A).
• Suppose that an experiment is performed n times. The relative frequency for an event A is
Number of times A occurs f
n n
n
fAP
nlim)(
n
fAP
nlim)(
• If we let n get infinitely large,
The Probability The Probability of an Eventof an Event
• P(A) must be between 0 and 1.
– If event A can never occur, P(A) = 0.
– If event A always occurs, P(A) =1.
• The sum of the probabilities for all simple events in S equals 1. P(S)=1.
The probability of an event A can be found by adding the probabilities of all the simple events in A.
The probability of an event A can be found by adding the probabilities of all the simple events in A.
–10% of the U.S. population has red hair. Select a person at random.
Finding ProbabilitiesFinding Probabilities
• Probabilities can be found using– Estimates from empirical studies– Common sense estimates based on
equally likely events.
P(Head) = 1/2
P(Red hair) = .10
•Examples: Examples: –Toss a fair coin.
ExampleExample
• Toss a fair coin twice. What is the probability of observing at least one head (event A)? Exactly one Head (event B)?
HH
1st Coin 2nd Coin Ei P(Ei)
HH
TT
TT
HH
TT
HHHH
HTHT
THTH
TTTT
1/4
1/4
1/4
1/4
P(at least 1 head)= P(A)
= P(E1) + P(E2) + P(E3)
= 1/4 + 1/4 + 1/4 = 3/4
P(at least 1 head)= P(A)
= P(E1) + P(E2) + P(E3)
= 1/4 + 1/4 + 1/4 = 3/4
Tree DiagramTree Diagram
P(exactly 1 head)=P(B)
= P(E2) + P(E3)
= 1/4 + 1/4 = 1/2
P(exactly 1 head)=P(B)
= P(E2) + P(E3)
= 1/4 + 1/4 = 1/2
ExampleExample• A bowl contains three M&Ms®, two reds, one blue.
A child selects two M&Ms at random. Probability of observing exactly two reds?
1st M&M 2nd M&M Ei P(Ei)
r1br1b
r1r2r1r2
r2br2b
r2r1r2r1
1/6
1/6
1/6
1/6
1/6
1/6
P(exactly two reds)
= P(r1r2) + P(r2r1)
= 1/6 +1/6
= 1/3
P(exactly two reds)
= P(r1r2) + P(r2r1)
= 1/6 +1/6
= 1/3
r1
b
b
r1
r1
b
br1br1
br2br2
r2
r2
r2
r2r1 b
ExampleExample• A bowl contains three M&Ms®, one red, one blue
and one green. A child takes two M&Ms randomly one at a time. What is the probability that at least one is red?
1st 2nd Ei P(Ei)
1/6
1/6
1/6
1/6
1/6
1/6
P(at least 1 red) = P(E1) +P(E2) + P(E3) + P(E6) =4/6 = 2/3
P(at least 1 red) = P(E1) +P(E2) + P(E3) + P(E6) =4/6 = 2/3
RBRB
RGRG
BRBR
BGBG
m
m
m
m
m
m
m
m
mGBGB
GRGR
ExampleExample
Simple Events Probabilities
HHHHHH
HHTHHT
HTHHTH
HTTHTT
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
P(Exactly 2 heads)= P(B)
= P(HHT) + P(HTH) + P(THH)
= 1/8 + 1/8 + 1/8 = 3/8
P(Exactly 2 heads)= P(B)
= P(HHT) + P(HTH) + P(THH)
= 1/8 + 1/8 + 1/8 = 3/8
P(at least 2 heads)=P(A)
= P(HHH)+P(HHT)+P(HTH)+P(THH)
= 1/8 + 1/8 +1/8 + 1/8 = 1/2
P(at least 2 heads)=P(A)
= P(HHH)+P(HHT)+P(HTH)+P(THH)
= 1/8 + 1/8 +1/8 + 1/8 = 1/2
THHTHH
THTTHT
TTHTTH
TTTTTT
A={HHH, HHT, HTH, THH}A={HHH, HHT, HTH, THH}
B={HHT,HTH,THH}B={HHT,HTH,THH}
• Toss a fair coin 3 times. What is the probability of observing at least two heads (event A)? Exactly two Heads (event B)?
ExampleExample
Simple Events
HHHHHH
HHTHHT
HTHHTH
HTTHTT C={HTT,THT,TTH,TTT}
C={HTT,THT,TTH,TTT}THHTHH
THTTHT
TTHTTH
TTTTTT
A={HHH,HHT,HTH,THH}
A={HHH,HHT,HTH,THH}
B={HHT,HTH,THH}B={HHT,HTH,THH}
A: at least two heads; B: exactly two heads;C: at least two tails; D: exactly one tail.Questions: A and C mutually exclusive? B and D?
D={HHT,HTH,THH}D={HHT,HTH,THH}
Mutually Exclusive
Not Mutually Exclusive
ExampleExample• Toss a fair coin twice. What is the
probability of observing at least one head (Event A)?
HH
1st Coin 2nd Coin Ei P(Ei)
HH
TT
TT
HH
TT
HHHH
HTHT
THTH
TTTT
1/4
1/4
1/4
1/4
P(at least 1 head)
= P(A)
= P(HH) + P(HT) + P(TH)
= 1/4 + 1/4 + 1/4 = 3/4
P(at least 1 head)
= P(A)
= P(HH) + P(HT) + P(TH)
= 1/4 + 1/4 + 1/4 = 3/4
A={HH, HT, TH}A={HH, HT, TH}
S
AAP
#
#)( S
AAP
#
#)(
S={HH, HT, TH, TT}S={HH, HT, TH, TT}
ExampleExample• A bowl contains three M&Ms®, two reds, one blue.
A child selects two M&Ms at random. What is the probability that exactly two reds (Event A)?
1st M&M 2nd M&M Ei P(Ei)
r1br1b
r1r2r1r2
r2br2b
r2r1r2r1
1/6
1/6
1/6
1/6
1/6
1/6
P(A)
= P(r1r2) + P(r2r1)
= 1/6 +1/6 = 2/6=1/3
P(A)
= P(r1r2) + P(r2r1)
= 1/6 +1/6 = 2/6=1/3
r1
b
b
r1
r1
b
br1br1
br2br2
r2
r2
r2
r2r1 b
A={r1r2, r2r1}A={r1r2, r2r1}
S
AAP
#
#)(
S
AAP
#
#)(
Counting RulesCounting Rules
• If the simple events in an experiment are equally likely, we can calculate
events simple ofnumber total
Ain events simple ofnumber
#
#)(
S
AAP
events simple ofnumber total
Ain events simple ofnumber
#
#)(
S
AAP
• We can use counting rules to find #A and #S.
CountingCounting• How many ways from A to C?
3 2 = 63 2 = 6
3 2 2 = 123 2 2 = 12
• How many ways from A to D?
The The mn mn RuleRule• For a two-stage experiment,
m ways to accomplish the first stagen ways to accomplish the second stagethen there are mn ways to accomplish the whole experiment.
• For a k-stage experiment, number of ways equal to
n1 n2 n3 … nk
Example: Example: Toss two coins. The total number of simple events is:
2 2 = 42 2 = 4
ExamplesExamplesExample: Example: Toss three coins. The total number of simple events is:
2 2 2 = 82 2 2 = 8
Example: Example: Two M&Ms are drawn in order from a dish containing four candies. The total number of simple events is:
6 6 = 366 6 = 36Example: Example: Toss two dice. The total number of simple events is:
m
m
4 3 = 124 3 = 12
PermutationsPermutations
• n distinct objects, take r objects at a time and arrange them in order. The number of different ways you can take and arrange is
60)3)(4(5 60)3)(4(5 The order of the choice is important!
.1!0 and )1)(2)...(2)(1(! where
)!(
!
nnnn
rn
nPn
r
Example: Example: How many 3-digit lock passwords can we make by using 3 different numbers among 1, 2, 3, 4 and 5?
60)1(2
)1)(2)(3)(4(5
)!35(
!553
P 60
)1(2
)1)(2)(3)(4(5
)!35(
!553
P
ExampleExample
Example: Example: A lock consists of five parts and can be assembled in any order. A quality control engineer wants to test each order for efficiency of assembly. How many orders are there?
120)1)(2)(3)(4(5!0
!555 P 120)1)(2)(3)(4(5
!0
!555 P
The order of the choice is important!
ExampleExample• How many ways to select a student
committee of 3 members: chair, vice chair, and secretary out of 8 students?
336)6)(7(8
)1)(2)(3)(4(5
)1)(2)(3)(4)(5)(6)(7)(8(
)!38(
!883
P
336)6)(7(8
)1)(2)(3)(4(5
)1)(2)(3)(4)(5)(6)(7)(8(
)!38(
!883
P
The order of the choice is important! ---- Permutation
CombinationsCombinations• n distinct objects, select r objects at a time without
regard to the order. The number of different ways you can select is
Example: Example: Three members of a 5-person committee must be chosen to form a subcommittee. How many different subcommittees could be formed?
)!(!
!
rnr
nC n
r
101)2(
)4(5
1)2)(1)(2(3
1)2)(3)(4(5
)!35(!3
!553
C 10
1)2(
)4(5
1)2)(1)(2(3
1)2)(3)(4(5
)!35(!3
!553
CThe order of
the choice is not important!
ExampleExample• How many ways to select a student
committee of 3 members out of 8 students?• (Don’t assign chair, vice chair and
secretary).
56)1)(2(3
)6)(7(8
)]1)(2)(3)(4(5)][1)(2(3[
)1)(2)(3)(4)(5)(6)(7(8
)!38(!3
!883
C
56)1)(2(3
)6)(7(8
)]1)(2)(3)(4(5)][1)(2(3[
)1)(2)(3)(4)(5)(6)(7(8
)!38(!3
!883
C
The order of the choice is NOT important!
Combination
QuestionQuestion• A box contains 7 M&Ms®, 4 reds and 3 blues.
A child selects three M&Ms at random. • What is the probability that exactly one is red
(Event A) ?
r1 r4r3r2 b2b1 b3
• Simple Events and sample space S:
{r1r2r3, r1r2b1, r2b1b2…... }• Simple events in event A:
{r1b1b2, r1b2b3, r2b1b2……}
SolutionSolution• Choose 3 MMs out of 7. (Total number of
ways, i.e. size of sample space S)
The order of the choice is not important!
35!4!3
!773 C
3!1!2
!332 C
4!3!1
!441 C
4 3 = 12 ways to choose 1 red and 2 greens ( mn Rule)
• Event A: one red, two blues
Choose one red
Choose Two Blues
35
12#
#)(
S
AAP
S
Event Relations - UnionEvent Relations - Union The unionunion of two events, A and B, is the
event that either A or B or bothor both occur when the experiment is performed. We write
A B
A BA B
S
A B
Event Relations-IntersectionEvent Relations-Intersection The intersection of two events, A and B, is
the event that both A and B occur.
We write A B.
A B
• If A and B are mutually exclusive, then P(A B) = 0.
SS
Event Relations - ComplementEvent Relations - Complement The complement of an event A consists of
all outcomes of the experiment that do not result in event A. We write AC ( The event that event A doesn’t occur).
A
AC
ExampleExample
Select a student from a college – A: student is colorblind– B: student is female– C: student is male
•What is the relationship between events B and C?•AC: •BC: •BC:
Mutually exclusive and B = CC
Student is not colorblind
Student is both male and female =
Student is either male or female = all students = S
ExampleExample
Toss a coin twice – A: At least one head {HH, HT, TH};– B: Exact one head {HT, TH};– C: At least one tail {HT, TH, TT}.
•AC:•AB:•AC:
{TT} No head
{HT, TH} Exact one head
{HH, HT, TH, TT}=S -- Sample space
Probabilities for UnionsProbabilities for Unions
The Additive Rule for Unions:The Additive Rule for Unions:
• For any two events, A and B, the probability of their union, P(A B), is
)()()()( BAPBPAPBAP )()()()( BAPBPAPBAP
A B
Example: Additive RuleExample: Additive RuleExample: Suppose that there were 1000 students in a college, and that they could be classified as follows:
Male (B) Female
Colorblind (A) 40 2
Not Colorblind 470 488
A: Colorblind P(A) = 42/1000=.042B: Male P(B) = 510/1000=.51
P(AB) = P(A) + P(B) – P(AB)= 42/1000 + 510/1000 - 40/1000 = 512/1000 = .512
P(AB) = P(A) + P(B) – P(AB)= 42/1000 + 510/1000 - 40/1000 = 512/1000 = .512 Check: P(AB)
= (40 + 2 + 470)/1000=.512
Check: P(AB)= (40 + 2 + 470)/1000=.512
A Special CaseA Special CaseWhen two events A and B are mutually exclusive, mutually exclusive, P(AB) = 0 and P(AB) = P(A) + P(B).
A: male and colorblindP(A) = 40/1000
B: female and colorblindP(B) = 2/1000
P(AB) = P(A) + P(B)= 40/1000 + 2/1000= 42/1000=.042
P(AB) = P(A) + P(B)= 40/1000 + 2/1000= 42/1000=.042
A and B are mutually exclusive, so that
Male Female
Colorblind 40 2
Not Colorblind 470 488
Probabilities for ComplementsProbabilities for Complements
• We know that for any event A:A:
P(A AC) = 0
• Since either A or AC must occur,
P(A AC) =1
• so that P(A AC) = P(A)+ P(AC) = 1
P(AC) = 1 – P(A)P(AC) = 1 – P(A)
A
AC
ExampleExample
A: male P(A) = 510/1000=.51B: female
P(B) = 1- P(A)= 1- .51=.49
P(B) = 1- P(A)= 1- .51=.49
A and B are complementary, so that
Select a student at random from the college. Define:
Male Female
Colorblind 40 2
Not Colorblind 470 488
ExampleExample• Toss a fair coin twice. Define
– A: head on second toss– B: head on first toss– If B occurred, what is probability that A
occurred?– If B didn’t occur, what is probability that A
occurred?
HTHT
THTH
TTTT
1/4
1/4
1/4
1/4
P(A given B occurred) = ½
P(A given B did not occur) = ½
P(A given B occurred) = ½
P(A given B did not occur) = ½ HHHH
P(A) does not change, whether B happens or not…
A and B are independent!
Conditional ProbabilitiesConditional Probabilities
The probability that A occurs, given that event B has occurred is called the conditional probabilityconditional probability of A given B and is defined as
0)( if )(
)()|(
BP
BP
BAPBAP 0)( if
)(
)()|(
BP
BP
BAPBAP
“given”
Probabilities for IntersectionsProbabilities for Intersections
In the previous example, we found P(A B) directly from the table. Sometimes this is impractical or impossible. The rule for calculating P(A B) depends on the idea of independent and dependent events.
Two events, A and B, are said to be independent if and only if the probability that event A occurs is not changed by occurrence of event B, or vice versa.
Two events, A and B, are said to be independent if and only if the probability that event A occurs is not changed by occurrence of event B, or vice versa.
Example 1Example 1• Toss a fair coin twice. Define
– A: head on second toss– B: head on first toss– A B: head on both first and second
HTHT
THTH
TTTT
1/4
1/4
1/4
1/4
P(A|B) = P(AB)/P(B)=(1/4)/(1/2)=1/2
P(A|not B) = 1/2
P(A|B) = P(AB)/P(B)=(1/4)/(1/2)=1/2
P(A|not B) = 1/2HHHH
P(A) does not change, whether B happens or not…
A and B are independent!
Example 2Example 2• A bowl contains five M&Ms®, two red and three blue.
Randomly select two candies, and define– A: second candy is red.– B: first candy is blue.
m
m
m
m
m
P(A|B) =P(2nd red|1st blue)= 2/4 = 1/2
P(A|not B) = P(2nd red|1st red) = 1/4
P(A|B) =P(2nd red|1st blue)= 2/4 = 1/2
P(A|not B) = P(2nd red|1st red) = 1/4
P(A) does change, depending on whether B happens or not…
A and B are dependent!
Defining IndependenceDefining Independence• We can redefine independence in terms
of conditional probabilities:
Two events A and B are independent if and only if
P(AB) = P(A) or P(B|A) = P(B)
Otherwise, they are dependent.
Two events A and B are independent if and only if
P(AB) = P(A) or P(B|A) = P(B)
Otherwise, they are dependent.
• Once you’ve decided whether or not two events are independent, you can use the following rule to calculate their intersection.
Multiplicative Rule for IntersectionsMultiplicative Rule for Intersections
• For any two events, A and B, the probability that both A and B occur is
P(A B) = P(A) P(B given that A occurred) = P(A)P(B|A)P(A B) = P(A) P(B given that A occurred) = P(A)P(B|A)
• If the events A and B are independent, then the probability that both A and B occur is
P(A B) = P(A) P(B) P(A B) = P(A) P(B)
Example 1Example 1In a certain population, 10% of the people can be classified as being high risk for a heart attack. Three people are randomly selected from this population. What is the probability that exactly one of the three is high risk?
Define H: high risk N: not high risk
P(exactly one high risk) = P(HNN) + P(NHN) + P(NNH)
= P(H)P(N)P(N) + P(N)P(H)P(N) + P(N)P(N)P(H)
= (.1)(.9)(.9) + (.9)(.1)(.9) + (.9)(.9)(.1)= 3(.1)(.9)2 = .243
P(exactly one high risk) = P(HNN) + P(NHN) + P(NNH)
= P(H)P(N)P(N) + P(N)P(H)P(N) + P(N)P(N)P(H)
= (.1)(.9)(.9) + (.9)(.1)(.9) + (.9)(.9)(.1)= 3(.1)(.9)2 = .243
Example 2Example 2Suppose we have additional information in the previous example. We know that only 49% of the population are female. Also, of the female patients, 8% are high risk. A single person is selected at random. What is the probability that it is a high risk female?
Define H: high risk F: female
From the example, P(F) = .49 and P(H|F) = .08. Use the Multiplicative Rule:
P(high risk female) = P(HF)
= P(F)P(H|F) =.49(.08) = .0392
From the example, P(F) = .49 and P(H|F) = .08. Use the Multiplicative Rule:
P(high risk female) = P(HF)
= P(F)P(H|F) =.49(.08) = .0392
Example 3Example 3
2 green and 4 red M&Ms are in a box; Two of them are selected at random.
A: First is green;
B: Second is red.
• Find P(AB).
m
m
mmm m
Method 1Method 1• Choose 2 MMs out of 6. Order is recorded. (Total
number of ways, i.e. size of sample space S)
The order of the choice is important! Permutation 30)5(6
!4
!6
)!26(
!662
P
441 C
221 C
2 4 = 8 ways to choose first green and second red
( mn Rule)
• Event AB: First green, second red
First green
Second Red
30
8#
#
)(
S
BA
BAP
m
m
mmm m
Method 2Method 2A: First is green;
B: Second is red;AB: First green, second red
m
m
mmm m
P(A)P(B|A)
P(A B) = P(A)P(B|A)P(A B) = P(A)P(B|A)
P(A B) = 2/6(4/5)=8/30P(A B) = 2/6(4/5)=8/30
2/6
P(Second red | First green)=4/5
Example 4Example 4
A: Male B: Colorblind
Find P(A), P(A|B)Are A and B independent?
P(A) = 510/1000=.51P(A) = 510/1000=.51
Select a student at random from the college. Define:
Male (A) Female
Colorblind (B) 40 2
Not Colorblind 470 488
P(A|B) = P(AB)/P(B)=.040/.042=.95P(A|B) = P(AB)/P(B)=.040/.042=.95
P(B) = 42/1000=.042P(B) = 42/1000=.042 P(AB) = 40/1000=.040P(AB) = 40/1000=.040
P(A|B) and P(A) are not equal. A, B are dependent
Probability Rules & Relations of EventsProbability Rules & Relations of Events
Complement Event
Additive Rule
Multiplicative Rule
Conditional probability
Mutually Exclusive Events
Independent Events
)(1)( APAP c
)(
)()|(
BP
BAPBAP
)()()()( BAPBPAPBAP
)|()()( ABPAPBAP
0)( BAP)()()( BPAPBAP
)()()( BPAPBAP )()|( APBAP
)()()()( CPBPAPCBAP
Random VariablesRandom Variables• A numerically valued variable x is a random random
variable variable if the value that it assumes, corresponding to the outcome of an experiment, is a chance or random event.
• Random variables can be discrete discrete or continuous.ontinuous.
• Examples: Examples: x = SAT score for a randomly selected studentx = number of people in a room at a randomly
selected time of dayx = weight of a fish drawn at random
Probability Distributions for Probability Distributions for Discrete Random VariablesDiscrete Random Variables
Probability distribution robability distribution of a discrete of a discrete random variable random variable xx , is a graph, table or formula that gives
• possible values of x • probability p(x) associated with each
value x.
1)( and 1)(0
havemust We
xpxp 1)( and 1)(0
havemust We
xpxp
Example 1Example 1• Toss a fair coin once,• Define x = number of heads.• Find distribution of x
1/2
1/2
P(x = 0) = 1/2P(x = 1) = 1/2
P(x = 0) = 1/2P(x = 1) = 1/2
HH
TT
x
1
0 x p(x)
0 1/2
1 1/2
Example 2Example 2• Toss a fair coin three times and
define x = number of heads.
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
P(x = 0) = 1/8P(x = 1) = 3/8P(x = 2) = 3/8P(x = 3) = 1/8
P(x = 0) = 1/8P(x = 1) = 3/8P(x = 2) = 3/8P(x = 3) = 1/8
HHHHHH
HHTHHT
HTHHTH
THHTHH
HTTHTT
THTTHT
TTHTTH
TTTTTT
x
3
2
2
2
1
1
1
0
x p(x)
0 1/8
1 3/8
2 3/8
3 1/8
Probability Histogram for x
Probability Histogram for x
ExampleExample• In a casino game, it has probability 0.5 of
winning $2 and probability 0.5 of winning $3.• x denotes the money won in a game. Find its
probability distribution.• How much should be paid for a game?
x p(x)
$2 0.5
$3 0.5
Expected Value: 2(.5)+3(.5)=$2.5
Expected Value of Expected Value of Random VariableRandom Variable
• Let x be a discrete random variable with probability distribution p(x). Then the expected value, denoted by E(x), is defined by
)()(
)Mean on,(Expectati Value Expected
xxpxE
)()(
)Mean on,(Expectati Value Expected
xxpxE
ExampleExample• Toss a fair coin 3 times and
record x the number of heads.x p(x) xp(x)
0 1/8 0(1/8)=0
1 3/8 1(3/8)=0.375
2 3/8 2(3/8)=0.75
3 1/8 3(1/8)=0.375
Total 1.5
1.5
)()(
xxpxE1.5
)()(
xxpxE
ExampleExampleIn a lottery, 8,000 tickets are sold at $5 each. The prize is a $12,000 automobile and only one ticket will be the winner. If you purchased two tickets, your expected gain?
μ = E(x) = Σ xp(x)
= (-10) (7998/8000)+(11,990)(2/8000)= -$7
Define x = your gain. x = -10 or 11,990
x p(x)
-$10 7998/8000
$11,990 2/8000
Mean & Standard DeviationMean & Standard Deviation• Let x be a discrete random variable with
probability distribution p(x). Then the mean, variance and standard deviation of x are given as
2
22
:deviation Standard
)()( :Variance
)( :Mean
xpx
xxp
2
22
:deviation Standard
)()( :Variance
)( :Mean
xpx
xxp
222 )( :Variance xpx222 )( :Variance xpx
ExampleExample• Toss a fair coin 3
times and record x the number of heads.
• Find variance by the definition formula.
x p(x) (x-2p(x)
0 1/8 (0-1.5)2(1/8)=.28125
1 3/8 (1-1.5)2(3/8)=.09375
2 3/8 (2-1.5)2(3/8)=.09375
3 1/8 (3-1.5)2(1/8)=.28125
Total .75
5.1)()( xxpxE 5.1)()( xxpxE
0.75
)()( 22
xpx
0.75
)()( 22
xpx
ExampleExample
• Toss a fair coin 3 times and record x the number of heads.
• Find the variance by the computational formula.
x p(x) x2p(x)
0 1/8 02(1/8)=0
1 3/8 12(3/8)=0.375
2 3/8 22(3/8)=1.5
3 1/8 32(1/8)=1.125
Total 3
5.1)()( xxpxE 5.1)()( xxpxE
75.0
5.13
)(2
222
xpx
75.0
5.13
)(2
222
xpx
ExampleExample• For a casino game, it has
probability .2 of winning $5 and probability .8 of nothing.
• x is money won in a game.• Calculate the variance of x.
x p(x) (x-2p(x)
0 0.8 (0-1)2(0.8)=0.8
5 0.2 (5-1)2(0.2)=3.2
Total 4
4
)()( 22
xpx
4
)()( 22
xpx
x p(x) x2p(x)
0 0.8 02(0.8)=0
5 .2 52(0.2)=5
Total 5
4
15
)(2
222
xpx
4
15
)(2
222
xpx
1)2.0(5)8.0(0
:Value Expected
1)2.0(5)8.0(0
:Value Expected
Key ConceptsKey ConceptsI. Experiments and the Sample SpaceI. Experiments and the Sample Space
1. Experiments, events, mutually exclusive events, simple events
2. The sample space
3. Venn diagrams, tree diagrams, probability tables
II. ProbabilitiesII. Probabilities
1. Relative frequency definition of probability
2. Properties of probabilities
a. Each probability lies between 0 and 1.
b. Sum of all simple-event probabilities equals 1.
3. P(A), the sum of the probabilities for all simple events in A
Key ConceptsKey ConceptsIII. Counting RulesIII. Counting Rules
1. mn Rule, extended mn Rule2. Permutations:
3. Combinations:IV. Event RelationsIV. Event Relations
1. Unions and intersections2. Events
a. Disjoint or mutually exclusive:
b. Complementary:)(1)( APAP c
)!(!
!
)!(
!
rnr
nC
rn
nP
nr
nr
0)( BAP
)()()( BPAPBAP
Key ConceptsKey Concepts3. Conditional probability:
4. Independent events
5. Additive Rule of Probability:
6. Multiplicative Rule of Probability:
)()()( BPAPBAP
)(
)()|(
BP
BAPBAP
)()()()( BAPBPAPBAP
)|()()( ABPAPBAP
)()|( APBAP
)()()()( CPBPAPCBAP
Key ConceptsKey ConceptsV. Discrete Random Variables and Probability V. Discrete Random Variables and Probability
DistributionsDistributions
1. Random variables, discrete and continuous
2. Properties of probability distributions
3. Mean or expected value of a discrete random variable:
4. Variance and standard deviation of a discrete random variable:
1)( and 1)(0 xpxp 1)( and 1)(0 xpxp
2
2222
:deviation Standard
)()()( :Variance
xpxxpx2
2222
:deviation Standard
)()()( :Variance
xpxxpx
)( :Mean xxp )( :Mean xxp
