Probability - NKD Group · 2013. 1. 29. · Probability of an event is always between 0 & 1. 2. An impossible event has a probability of 0. 3. ... Each joint event is also mutually

Probability

The Science Of Uncertainty

BASIC PROPERTIES OF PROBABILITIES

1. Probability of an event is always between 0 & 1.

2. An impossible event has a probability of 0.

3. A guaranteed (certain) event has a probability of 1.

Chapter 4: Probability Concepts Page -2- Class Notes to accompany: Introductory Statistics, 9

th Ed, By Neil A. Weiss

Prepared by: Nina Kajiji

SOME DEFINITIONS

1. Classical Probability

All possible outcomes of the experiment are equally likely.

That is, suppose an experiment has N possible outcomes, all

equally likely then the probability that a specified event

occurs is equal to the number of ways, f, that the event can

occur, divided by the total number of possible outcomes.

Notation: f / N

2. Outcome

The results of a trial.

3. Sample Space, or Set (S)

A collection of all possible outcomes for an experiment.

4. Event

A collection of outcomes for the experiment. That is, any

subset of the sample space. Events are generally denoted

as: A, B, C, etc.

5. Probability is a generalization of the concept of percentages.

For example if we compute the probability of getting a

double when two balanced dice are rolled as 0.167 then we

can say there is a 16.7% chance of getting a double when

two balanced dice are rolled.




Example:

An experiment of selecting a card from a deck of 52 cards was

performed. The following are the some sample events and their

outcomes.

Sample Events Outcomes

A. A card selected is a king of diamonds 1

B. A card selected is a king 4

C. A card selected is a diamond 13

D. A card selected is not a face card 40

(52-12)




OPERATIONS WITH SETS

1. Subset --- A B

A set whose members belong to another larger set.

2. Union --- (A B) or (A or B)

The addition of two or more sets.

3. Intersection --- (A B) or (A & B)

The common elements of two or more sets.

4. Null

An empty set.

5. Complement --- (A

_

) or (not A)

All elements that do not belong to a set.

6. Mutually Exclusive (Disjoint)

When two sets have no common elements they are

considered to be disjoint.




Example For Set Operation

Consider an experiment to determine if people wear seatbelts.

Two possible outcomes: Y or N.

If three people are selected at random what is the sample space?

S = {YYY, YYN, YNY, NYY, YNN, NYN, NNY, NNN}

#of outcomes = 23 = 2 x 2 x 2 = 8

Define the following events:

A = At least two wear seatbelts

= {YYY, YYN, YNY, NYY}

B = Exactly two wear seatbelts

= {YYN, YNY, NYY}

C = Exactly one wears a seatbelt

= {YNN, NYN, NNY}

AB = {YYY, YYN, YNY, NYY}

A&B = {YYN, YNY, NYY}

A&C = Empty Set

not A = {YNN, NYN, NNY, NNN}

= at most one wears seatbelt




A

B

A Not B A B

B Not A

Not (A or B)




NOTATION FOR PROBABILITY

IF: E is an event

P is the probability

P(E) is the probability of the event

RELATIONSHIPS AMONG EVENTS

(not E): The event “E does not occur”

(A&B): The event “both A and B occur”

(AorB): The event “either A or B or both occur”




RULES OF PROBABILITY

1. Addition Rule

if A & B are disjoint then

P(A B) = P(A) + P(B)

if A & B are not disjoint then

P(A B) = P(A) + P(B) - P(A&B)

2. Complementation Rule

( ) 1 ( )P A P A

3. P(S) = 1

4. P(Null) = 0

5. P(A or (not A)) = 1

6. A & B are independent iff (if and only if)

P(A&B) = P(A)*P(B) (Multiplication)




Example: Seatbelt Experiment

Given: P(Y) = 0.2 & P(N) = 0.8

Then the sample space of the responses of three people is:

S = {YYY, YYN, YNY, NYY, YNN, NYN, NNY, NNN}

The probability of each outcome using rule 6 is:

{0.008, 0.032, 0.032, 0.032, 0.128, 0.128, 0.128, 0.512}

Some sample probabilities of events

A = At least two wear seatbelts

P(A) = 0.008 + (3*0.032) = 0.104

B = Exactly two wear seatbelts

P(B) = (3*0.032) = 0.096

C = Exactly one wears a seatbelt

P(C) = (3*0.128) = 0.384

P(AUC) = P(A) + P(C)

= 0.104 + 0.384 = 0.488

P(A&C) = Empty Set = 0

P(A&B) = (3*0.032) = 0.096

P(AUB) = P(A) + P(B) - P(A&B)

= 0.104 + 0.096 - 0.096 = 0.104

not A = 1 - P(A)

= 1 - 0.014 = 0.896




Example Using Grouped Data Table

Classes Frequency CM Rel.Freq

910-929 1 919.5 0.033

930-949 1 939.5 0.033

950-969 3 959.5 0.100

970-989 9 979.5 0.300

990-1009 7 999.5 0.233

1010-1029 6 1019.5 0.200

1030-1049 2 1039.5 0.067

1050-1069 1 1059.5 0.033

TOTALS 30 1.000

Grouped Data Table

P(E) = (f / N)

Probability of x < 950

= (1 + 1) / 30 = 0.067

Probability of 950 <= x <= 1009

= (3 + 9 + 7) / 30 = 0.633




CONTINGENCY TABLES (N-WAY GROUPING)

When you wish to group more than one variable you would use contingency tables. Examples: Sex by Age, Height by Weight, Sales by Product Line, etc.

Sample Contingency Table Sex\Age Under 21 21-25 Over 25 Total

Male Freq uency Counts #of Males

Female #of Females

Total #<21 #of21-25 #>25 GrandTotal

Cross-classified data is generally presented in contingency tables.

The numbers presented in the body of the table provide the

frequencies for various contingencies. These numbers are

presented in individual CELLS.




Example

Age & gender data for freshmen in a calculus class is presented.

The contingency table is shown below (yellow highlight is not part

of the table):

R1 R2 R3

21< 21-25 >25 Total

A1 Males 8 12 2 22

A2 Females 12 13 3 28

Total 20 25 5 50

INDIVIDUAL EVENTS

A1 = Event the student selected is male

A2 = Event the student selected is female

R1 = Event the student selected is under 21

R2 = Event the student selected is between 21-25

R3 = Event the student selected is over 25

NOTE:

A1 & A2 are mutually exclusive

R1, R2, & R3 are mutually exclusive




JOINT EVENTS

(A1&R1): The student selected is a male under 21.

(A2&R3): The student selected is a female over 25.

In the above table there are: 2 x 3 = 6 joint events. That is, all

CELLS in the table are considered joint events and therefore joint

probabilities for each cell can be calculated.

Each joint event is also mutually exclusive from the other joint

event.

Probability of a joint event is calculated as:

P(A&B) = (f / N)

Where: f is the cell frequency.

N is the population size.

Example:

P(A1&R1) = 8/50 = 0.16 = 16%

P(A2&R3) = 3/50 = 0.06 = 6%




MARGINAL PROBABILITY

From the row and column totals of a contingency table it is

possible to calculate marginal probabilities.

Example:

P(A1) = (Total of A1) / N

= 22 / 50

= 0.44

Note: The sum of all joint probabilities on a row or a column

equals the marginal probability in that row or column.




CONDITIONAL PROBABILITY

The probability of an event given that another event has occurred.

Notation: P(A|B) = f/N

-or-

Generally: P(A|B) = P(A&B) / P(B)

Where: A is the event of interest.

B is the event that has occurred.

f is the conditional frequency.

N is the conditional population size.

Example:

Given that a student selected is male, what is the probability of

him being less than 21 years old?

P(R1|A1) = 8/22 = 0.36

-or-

P(R1|A1) = P(A1&R1) / P(A1)

= 0.16 / 0.44

= 0.36




INDEPENDENCE

Recall: For any non-trivial (that is, the probability is not zero)

events independence is:

P(A&B) = P(A)P(B)

Further, two events are independent iff the following two

conditions are also true.

1) P(B|A) = P(B)

2) In terms of marginal & joint probabilities:

P(A&B) = P(A)P(B|A)

Example:

Using the card drawing experiment. Note: Once a card is picked it

is replaced back in the deck.

Events P(E)

A. A card selected is a king of diamond 1/52 = 0.019

B. A card selected is a king 4/52 = 0.077

C. A card selected is a diamond 13/52 = 0.25

D. A card selected is not a face card 40/52 = 0.769

Q: What is the probability of selecting a card that is the king of

diamond?

A: Since there is only one king of diamond in a deck of cards the

joint probability is calculated as:

P(B&C) = 1 / 52 = 0.019




Q: Are events B & C independent?

Standard formula if independent is assumed:

1) P(B&C) = P(B)P(C)

P(BC) = (0.077)(0.25) – calculated using (f/ N)

= 0.019

Additional Condition using marginal and joint probabilities

2a) Recall: P(C|B) = f / N = 1/4 = 0.25 = P(C)

That is, probability of getting a diamond, given that the card

selected is a king.

2b) Therefore,

P(B&C) = (0.077)(0.25) = 0.019

Where: 0.077 is computed using the (f/N); and 0.25 is calculated

as shown in (2a)




MUTUALLY EXCLUSIVE V/S INDEPENDENT

When the occurrence of one event has NO effect on the probability of

occurrence of another event the two events are independent. That is,

P(A&B) = P(A)P(B)

When two events cannot occur simultaneously, they are mutually exclusive.

That is, P(A&B) = 0.

NOTE: If two events are independent they cannot be mutually exclusive,

and vice versa.

Example: Independent Events

Consider a fair coin and a fair six-sided die. Consider two events: A: Obtaining heads B: Rolling a 6. Then we can reasonably assume that events A and B are independent, because the outcome of one does not affect the outcome of the other. The probability that both A and B occur is: P(A&B) = P(A)P(B) = (1/2)(1/6) = 1/12. Since this value is not zero, then events A and B cannot be mutually exclusive.

Example: Mutually Exclusive Events

Consider a fair six-sided die as before, only in addition to the numbers 1 through 6 on each face, we have the property that the even-numbered faces are colored red, and the odd-numbered faces are colored green. Consider the two events: A: Rolling a green face, and B: Rolling a 6. Then, P(A) = 1/2 (two colors – red and green); and P(B) = 1/6 A and B cannot simultaneously occur, since rolling a 6 means the face is red, and rolling a green face means the number showing is odd. Therefore: P(A&B) = 0.




BAYES' RULE (OPTIONAL) -- SKIP IN STA308

Exhaustive Events

Events A1, A2, ..., Ak are said to be exhaustive if at least one of the

events must occur when the experiment is performed.

Example 1:

Picking a student in the calculus class example implies that either

A1 or A2 is true. (A1 & A2 are exhaustive and mutually exclusive

events. That is, when the experiment is performed exactly one of

the events must occur).

Example 2:

In another experiment of selecting a card from a deck of 52 cards

the following events are defined:

A. A card selected is not a face card.

B. A card selected is a face card.

C. A card selected is a diamond

Events A, B, & C are exhaustive.

Events A & B are exhaustive and mutually exclusive.

Events (AUB) & C are not mutually exclusive because

event C is a subset of (AUB).




The Rule Of Total Probability

If events A1, A2, A3, ..., Ak are mutually exclusive and exhaustive,

then for any event B,

P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) + ... +

P(Ak)P(B|Ak).

Bayes' Rule

If events A1, A2, A3, ..., Ak are mutually exclusive and exhaustive,

then for any event B,

P(Ai|B) = P(Ai)P(B|Ai)

P(A1)P(B|A1) + P(A2)P(B|A2) + ...

+ P(Ak)P(B|Ak)

Where: Ai is any one of the events A1, A2, ..., Ak

Other Definitions:

Prior Probability: The probability of an event.

Posterior Probability: Conditional probability.




Example

The National Center for Health Statistics provides information on

suicides by sex and method used. The information published said

there were 30,904 total suicides in the US. of which 24,226 were

males and the remaining were females. The following table gives

the conditional probability of a male using a particular method

and a female using a particular method.

Methods Male (M) Female(F)

Poisoning (P) 0.145 0.377

Hanging (H) 0.155 0.127

Firearms (G) 0.641 0.395

Other (O) 0.059 0.101

Solution Given: Total suicides = 30,904 Male suicides = 24,226 Female suicides = 6,678 Therefore: P(M) = 24226/30904 = 0.784 P(F) = 6678/30904 = 0.216




Q: Determine the probability that a firearm was used for the

suicide. That is, what is P(G) ?

A: P(G) = P(M&G) + P(F&G)

= P(M)P(G|M) + P(F)P(G|F)

= (0.784)(0.641) + (0.216)(0.395)

= 0.58786

Q: What is the prior probability that the person is female?

A: P(F) = 0.216

Q: What is the posterior probability that the person was a female

given that a firearm was used?

A: P(F|G) = P(F&G)

P(G)

= P(F)P(G|F)

P(G)

= (0.216 * 0.395) / 0.587

= 0.145




COUNTING RULES

The Fundamental Counting Rule

Suppose there are r actions to be performed in a definite order.

If there are m1 possibilities for the first action, m2 for the second

action, etc. Then the total number of possibilities is:

m1 m2 mr

Factorials

Let k be a positive integer. Then the product of the first k positive

integers is called k factorial and is denoted as:

k! = k(k - 1) 2 1

Examples:

3! = 3 2 1 = 6

0! = 1

1! = 1




Permutations

The number of possible permutations of r objects from a

collection of m objects is given by:

(m)r = m!

(m - r)!

Note: (m)m = m!

Example:

A zip code consists of five digits.

a) How many possible zip codes are there? b) How many possible zip codes are there in which no digit

appears more than once.

Solution Each location can get any of the 10 numbers selected one at a

time. That is, (m)r = (10)1 = 10.

There are five such locations, thus there are:

10 10 10 10 10 = 100,000 possible zip codes.

If no digit appears more than once then there are:

(m)r = (10)5

= 10 9 8 7 6

= 30,240 possible zip codes




Combinations

The number of possible combinations of r objects from a

collection of m objects.

m

r = m!

r!(m - r)!

Example:

How many samples of size 5 are possible from a population of size

70?

Solution

The # of samples of size 5 from a population of 70 is:

70

5 = 70!

5!65! = 12,103,014

Probability - NKD Group · 2013. 1. 29. · Probability of an event is always between 0 & 1. 2. An impossible event has a probability of 0. 3. ... Each joint event is also mutually

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