Probability Theory: Probabilistic Model of an Experiment & Sample-point Approach

Outline Definitions Approaches Axioms Sample-Point Approach

1. Sets and Probability1.3 Probabilistic Model of an Experiment

1.4 Sample-Point Approach in Calculating Probability

Ruben A. Idoy, Jr.

Introduction to Probability Theory(Math 181)

21 June 2012


Outline

1 Definitions

2 Approaches of Probability Values

3 Axioms of Probability

4 Sample-Point Approach on Calculating ProbabilityStepsExamples


Definitions

experiment - the process of making an observation.


Definitions


An experiment can result in one, and only one, of a set of distinctlydifferent observable outcomes.


Definitions


An experiment can result in one, and only one, of a set of distinctlydifferent observable outcomes.

We are interested in experiments that generate outcomes which vary inrandom manner and cannot be predicted with certainty.


Definitions


sample space - denoted by S (or Ω in some books), is a set of pointscorresponding to all distinctly different possible outcomes of anexperiment. Each point corresponds to a particular single outcome.


Definitions



sample point - a single point in a sample space, S


Definitions


Discrete sample space - one that contains a finite number orcountable infinity of sample points.

Continuous sample space - has an infinite number of samplepoints.


Definitions


Discrete sample space - one that contains a finite number orcountable infinity of sample points.Continuous sample space - has an infinite number of samplepoints.


Definitions

event - any subset of the sample space, S. It can also be viewed as acollection of sample points.


Definitions


Example: Die-tossing Experiment

A: observe an odd number (A = 1, 3, 5),B: observe a number less than 5 (B = 1, 2, 3, 4),C: observe a 2 or a 3 (C = 2, 3),E1: observe a 1 (E1 = 1),E6: observe a 6 (E6 = 6)


Definitions



A: observe an odd number (A = 1, 3, 5),

B: observe a number less than 5 (B = 1, 2, 3, 4),C: observe a 2 or a 3 (C = 2, 3),E1: observe a 1 (E1 = 1),E6: observe a 6 (E6 = 6)


Definitions



A: observe an odd number (A = 1, 3, 5),B: observe a number less than 5 (B = 1, 2, 3, 4),

C: observe a 2 or a 3 (C = 2, 3),E1: observe a 1 (E1 = 1),E6: observe a 6 (E6 = 6)


Definitions



A: observe an odd number (A = 1, 3, 5),B: observe a number less than 5 (B = 1, 2, 3, 4),C: observe a 2 or a 3 (C = 2, 3),

E1: observe a 1 (E1 = 1),E6: observe a 6 (E6 = 6)


Definitions



A: observe an odd number (A = 1, 3, 5),B: observe a number less than 5 (B = 1, 2, 3, 4),C: observe a 2 or a 3 (C = 2, 3),E1: observe a 1 (E1 = 1),

E6: observe a 6 (E6 = 6)


Definitions





Definitions




Each of these 5 events is a specific collection of sample points.


Definitions


A simple event is one that contains a single sample point. Wemay refer to simple events as events that cannot be decomposed.


Definitions



Probability - a numerical measure of the chance of the occurrence ofan event.


Definitions



Probability - a numerical measure of the chance of the occurrence ofan event.

The final step in constructing a probabilistic model for an experimentwith a discrete sample space is to attach a probability to each sampleevent.


Approaches to the Assignment of Probability Values

Relative Frequency or A Posteriori Approach

The probability value is the relative frequency of the occurrence ofthe event over a long-run experiment (over a large number ofrepetitions of the experiment).

Classical, Theoretical or A Priori Approach

Probability value us based on an experimental model with certainassumptions

Subjective Approach

The researcher assigns probability according to his knowledge orexperience on the occurrence of the event. There is no objective wayof prediction of the occurrence of the event under this approach.





P (E) =number of times the event occurred

number of repetitions of the experiment



Subjective Approach








Subjective Approach



Axioms of Probability

For every event E in a sample space S, we assign a numerical valueP(E), known as the probability of E, such that:

1 P(E) > 0;2 P(S) = 1;3 If E1, E2, . . . form a sequence of pairwise mutually exclusive events

in S (Ei ∩ Ej = ∅, i , j), then

P(E1 ∪ E2 ∪ E3 ∪ · · · ) =∞∑

i=1

P(Ai)




1 P(E) > 0;

2 P(S) = 1;3 If E1, E2, . . . form a sequence of pairwise mutually exclusive events


P(E1 ∪ E2 ∪ E3 ∪ · · · ) =∞∑

i=1

P(Ai)




1 P(E) > 0;2 P(S) = 1;

3 If E1, E2, . . . form a sequence of pairwise mutually exclusive eventsin S (Ei ∩ Ej = ∅, i , j), then

P(E1 ∪ E2 ∪ E3 ∪ · · · ) =∞∑

i=1

P(Ai)




1 P(E) > 0;2 P(S) = 1;3 If E1, E2, . . . form a sequence of pairwise mutually exclusive events


P(E1 ∪ E2 ∪ E3 ∪ · · · ) =∞∑

i=1

P(Ai)


Example

Let A be the event of obtaining a number less than or equal to 3 intossing a die.

Find the probability of A if:

1 the die is fair;2 the die is biased such that an odd number is twice as likely to

occur as an even number.


Example


Find the probability of A if:1 the die is fair;

2 the die is biased such that an odd number is twice as likely tooccur as an even number.


Example


Find the probability of A if:1 the die is fair;2 the die is biased such that an odd number is twice as likely to

occur as an even number.


Example

Solution for [1]

First note that S = 1, 2, 3, 4, 5, 6. Since the die is fair, the probabilityfor each simple event is equal, say p. That is,

P(1) = P(2) = · · · = P(6) = p.

We further observe that

P(1) + P(2) + · · ·+ P(6) = 1.

Substituting p to each probability of the simple event, we get

p + p + p + p + p + p = 6p = 1.

Thus, p = 16 .


Example

Solution for [1]

The event A = 1, 2, 3, has therefore a probability:

P(A) = P(1) + P(2) + P(3) =16+

16+

16=

36


Example

Solution for [2]

The sample space of the experiment is still the set S = 1, 2, 3, 4, 5, 6.Let p be the probability of each even number to occur and 2p be theprobability of each odd number to occur. That is,

P(2) + P(4) + P(6) =pP(1) + P(3) + P(5) =2p

Substituting each probability to the simple event, we get

2p + p + 2p + p + 2p + p = 9p = 1.

Thus, p = 19 .


Example

Solution for [2]

The event A = 1, 2, 3, has therefore a probability:

P(A) = P(1) + P(2) + P(3) =29+

19+

29=

59

Not all problems dealing with probability of an event are solvable bysimply using the Axioms of Probability.

Thus, there are 2 ways or approaches known to calculate theProbability of an Event: the sample-point approach and theevent-composition method.


Steps

Sample-Point Approach on Calculating Probability

Steps:

1 Define the experiment.2 List the simple events associated with the experiment and test

each to make certain that they cannot be decomposed. Thisdefines the sample space, S.

3 Assign reasonable probabilities to the sample points in S,making certain that ∑

S

P(Ei) = 1

.4 Define the event of interest, E, as a specific collection of sample

points.5 Find P(E) by summing the probabilities of the sample points in E.


Steps


Steps:

1 Define the experiment.

2 List the simple events associated with the experiment and testeach to make certain that they cannot be decomposed. Thisdefines the sample space, S.


S

P(Ei) = 1




Steps


Steps:




S

P(Ei) = 1




Steps


Steps:




S

P(Ei) = 1

.

4 Define the event of interest, E, as a specific collection of samplepoints.

5 Find P(E) by summing the probabilities of the sample points in E.


Steps


Steps:




S

P(Ei) = 1


points.

5 Find P(E) by summing the probabilities of the sample points in E.


Steps


Steps:




S

P(Ei) = 1




Examples

Example 1

Toss a coin 3 times and observe the top face. What is the probabilityof observing exactly 2 heads, assuming the coin is fair?

Solution

1 Experiment: Tossing a fair coin 3 times.2 List of simple events:

S = HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

3 Assignment of probability to each sample points:

P(Ei) =18

, i = 1, 2, . . . , 8.

4 Define event of interest: Let A be the event that 2 heads willappear after tossing the coin 3 times.

5 Find P(A):

P(A) = P(HHT) + P(HTH) + P(THH) =18+

18+

18=

38


Examples

Solution

1 Experiment: Tossing a fair coin 3 times.

2 List of simple events:



P(Ei) =18

, i = 1, 2, . . . , 8.


5 Find P(A):

P(A) = P(HHT) + P(HTH) + P(THH) =18+

18+

18=

38


Examples

Solution

1 Experiment: Tossing a fair coin 3 times.2 List of simple events:



P(Ei) =18

, i = 1, 2, . . . , 8.


5 Find P(A):

P(A) = P(HHT) + P(HTH) + P(THH) =18+

18+

18=

38


Examples

Example 2

Patients arriving at a hospital outpatient clinic can select any of threeservice counters. Physicians are randomly assigned to the stationsand the patients have no station preference. Three patients arrived atthe clinic and their selection is observed. Find the probability thateach station receives a patient.

Solution

1 Experiment: Assigning patients to service counters.2 Let (a, b, c) be the ordered triple where a, b, c ∈ 1, 2, 3. That is,

each patient could be assigned to any of the service counter 1,2and 3. Furthermore, |S| = 33 = 27.

3 Since each simple events are likely to occur, then

P(Ei) =1|S|

=1

27, ∀i = 1, 2, . . . , 27

4 Define event of interest: Let B be the event that each stationreceives a patient.

5 P(B) = P((1, 2, 3)) + P((1, 3, 2)) + · · ·+ P((3, 2, 1))= 1

27 + 127 + · · ·+ 1

27 = 627


Examples

Solution

1 Experiment: Assigning patients to service counters.

2 Let (a, b, c) be the ordered triple where a, b, c ∈ 1, 2, 3. That is,each patient could be assigned to any of the service counter 1,2and 3. Furthermore, |S| = 33 = 27.


P(Ei) =1|S|

=1

27, ∀i = 1, 2, . . . , 27


5 P(B) = P((1, 2, 3)) + P((1, 3, 2)) + · · ·+ P((3, 2, 1))= 1

27 + 127 + · · ·+ 1

27 = 627


Examples

Solution

1 Experiment: Assigning patients to service counters.2 Let (a, b, c) be the ordered triple where a, b, c ∈ 1, 2, 3. That is,

each patient could be assigned to any of the service counter 1,2and 3. Furthermore, |S| = 33 = 27.


P(Ei) =1|S|

=1

27, ∀i = 1, 2, . . . , 27


5 P(B) = P((1, 2, 3)) + P((1, 3, 2)) + · · ·+ P((3, 2, 1))= 1

27 + 127 + · · ·+ 1

27 = 627


Examples

Example 3

Four cards are drawn from a standard deck of 52 cards. What is theprobability that the cards drawn are:

1 of the same suit;2 of the same color;3 of the same type.


Examples

Assignment 1

Write your STEP-BY-STEP solution in a 1/2 sheet of yellow paper.

A box contains seven laptops. Unknown to the purchaser, three aredefective. Two of the seven are selected for thorough testing and thenclassified as defective or nondefective.

(i) Find the probability of the event A that the selection includes nodefective.

(ii) Find the probability of the event B that the selection includesexactly one defective.

Probability Theory: Probabilistic Model of an Experiment & Sample-point Approach

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