(c) 2007 IUPUI SPEA K300 (4392)

Probability

Likelihood (chance) that an event occursClassical interpretation of probability: all

outcomes in the sample space are equally likely to occur (random sampling)

Empirical probability: conduct actual experiments to get the likelihood

Subjective probability: ask professors, friends, mom, etc.

(c) 2007 IUPUI SPEA K300 (4392)

Basic Concepts

Probability experiment: a chance process that leads to well-defined results (outcomes)

Outcome: the distinct possible result of a single trial of a probability experiment

Sample space: the set of possible outcomes

Event: identified with certain of outcomes

(c) 2007 IUPUI SPEA K300 (4392)

Sample space

Example 4-2 on page 180Sample space: 52 outcomesEvent “Queen”: 4Event “Heart”: 12Event “King Spade”: 1

Example 4-4Sample space: 8Event “Exactly two boys”: 3

(c) 2007 IUPUI SPEA K300 (4392)

Tossing a Coin

Tossing a coin once: head (H) or tail (T)Tossing two times: HH, HT, TH, TTTossing three times: HHH, HHT, HTH,

HTT, THH, THT, TTH, TTT 2 X 2 X 2

(c) 2007 IUPUI SPEA K300 (4392)

Tree Diagram

H: head, T: tail

T

T

H

T

H

H

HH

HT

TH

TT

(c) 2007 IUPUI SPEA K300 (4392)

Rolling a Die

Rolling once: 1, 2, 3, 4, 5, 6 Rolling twice: (1, 1), (1,2)…

(2, 1), (2, 2), …(6,6)6^2 Rolling three times: (1,1,1),

(1,1,2)… (1,2,1)… (1,6,6), (2,1,1)…(2,1,2)…(2,6,6), (3,1,1)…(6,6,6,)6^3

Rolling four times: How to get the sample space?

(c) 2007 IUPUI SPEA K300 (4392)

Combination

Selecting r distinct objects out of n objects regardless of order at a time

Example: select two students for awards among 5 students

N factorial: n! = n X (n-1) X (n-2) X … 1 0! = 1

)!(!

!

rnr

nCrn

10)123)(12(

12345

!3!2

!5

)!25(!2

!525

C

(c) 2007 IUPUI SPEA K300 (4392)

Permutation

An arrangement of n objects in a specific order using r objects at a time.

Taking r ordered objects out of n objects at a time. Selecting one student for $10K award and another

for $5K award among 5 students.

)!(

!

rn

nPrn

20123

12345

!3

!5

)!25(

!525

P

(c) 2007 IUPUI SPEA K300 (4392)

Classical Probability

P(E) is the probability that the event E occurs; expected (not actualized) likelihood

The number of outcomes of event E, NE, divided by the number of total outcomes in the sample space, N.

sapcesample

event

N

NEP E

_#

#)(

13

1

52

4)( QueenP

5

2

10

4)( AwardP

(c) 2007 IUPUI SPEA K300 (4392)

Probability Rules

P(E) is a number between 0 and 1Probability zero, P(E)=0, means the

event will not occur.Probability 1, P(E)=1, means only the

event occurs all the times.Sum of the probabilities of all outcomes

in the sample space is 1

(c) 2007 IUPUI SPEA K300 (4392)

Complementary Events

the set of outcomes in the sample space that are not included in the outcome of E

P(Ē) = 1 - P(E) P(E) = 1 - P(Ē) P(E) + P(Ē) = 1 P(Ē) P(E)

(c) 2007 IUPUI SPEA K300 (4392)

Empirical Probability

Is your quarter really fair? Hmm… I guess the probability of head is larger than ½ for some reason.

How about your die? Do all 1 through 6 have the equal chance of 1/6 to be selected?

How can you check that?

(c) 2007 IUPUI SPEA K300 (4392)

Addition Rule

Probability that event A or B occurs P(A or B) = P(A) + P(B) – P (A and B) P(A U B) = P(A) + P(B) – P (A ∩ B)

P(Nurse or Male)=P(N)+P(M)-P(N and M), Figure 4-5, p.198. Question 15, p.200.

P(A and B)P(A) P(B)

(c) 2007 IUPUI SPEA K300 (4392)

Mutually Exclusive Events

P(A U B) = P(A) + P(B) - 0 P (A ∩ B) = 0 P(Monday or Sunday)=P(Monday)+P(Sunday)-0

P(A) P(B)

(c) 2007 IUPUI SPEA K300 (4392)

Multiplication Rule

Probability that both events A and B occur P(A ∩ B) = P(A) X P(B) Example 4-24, p.206:

P(queen and ace) = P(queen) X P(ace) = 4/52 X 4/52

Example 4-25: P(blue and white)=P(blue) X P(white) = 2/10

X 5/10 What if event A and B are related?

(c) 2007 IUPUI SPEA K300 (4392)

Statistical Independence

Occurrence of an event does not change the probability that other events occur.

Occurrence of one measurement in a variable should be independent of the occurrence of others.

Drawing a card with/without replacement. With replacement->independent (Ex. 4-25) Without replacement->dependent

How do we know if two events are statistically independent?

(c) 2007 IUPUI SPEA K300 (4392)

Examples

How to put an elephant into a refrigerator? 1. Open the door

2. Put an elephant into the refrigerator

3. Close the door Now, how to put an hippo into the

refrigerator? What makes a difference? Question 1, p.215

(c) 2007 IUPUI SPEA K300 (4392)

Statistical Dependence 1

Example 4-25, pp.206-207 What if no replacement? Suppose a blue ball is selected in the 1st trial P(blue) is 2/10 in the 1st trial

1st Trial 2nd Trial

P(Blue) 2/10 1/9

P(Red) 3/10 3/9

P(White) 5/10 5/9

(c) 2007 IUPUI SPEA K300 (4392)

Statistical Dependence 2

P(blue then white)=2/10 X 5/10 w/o replacement P(blue then white)=2/10 X 5/9 w/ replacement 5/9: probability that event B (white ball)

occurs given event A (blue) already occurred. Figure 4-6. p. 210.

1st Trial 2nd Trial

P(Blue) 2/10 1/9

P(Red) 3/10 3/9

P(White) 5/10 5/9

(c) 2007 IUPUI SPEA K300 (4392)

Conditional Probability

P(B|A) is the probability that event B occurs after event A has already occurred.

P(B|A)=P(A ∩ B) / P(A) P(A ∩ B)= P(A) X P(B|A) in case of statistical dependence

P(A and B)P(A) P(B)

(c) 2007 IUPUI SPEA K300 (4392)

Statistical Independence, again

Events A and B are statistically independent, if and only If P(B|A)=P(B) or P(A|B)=P(A)

Example 4-34, p.211 P(Yes|Female)=P(Female ∩ Yes) / P(Female) =

[8/100]/[50/100]= 8/50 ≠ 40/100 Events Female and Yes are not independent P(A ∩ B)= P(A) X P(B|A)=[50/100]*[8/50]=8/100 P(A ∩ B)= P(A) X P(B) in case of statistical independence

because P(B|A)=P(B)

Yes No Total

Male 32 18 50

Female 8 42 50

40 60 100

(c) 2007 IUPUI SPEA K300 (4392)

Examples: Example 4-25, p207

With Replacement: P(W|B)=P(W ∩ B)/P(B)=

[2/10*5/10]/[2/10]=5/10=P(W) Events white (2nd trial) and blue (1st trial) are

independent Without Replacement:

P(W|B)=P(W ∩ B)/P(B)=[2/10*5/9]/[2/10] =5/10 ≠ P(W)

Events white (2nd trial) and blue (1st trial) are dependent

Event blue in the 1st trial influences the probability of event white in the 2nd trial.

(c) 2007 IUPUI SPEA K300 (4392)

Examples: Question 34, p216

P(oppose|freshman)=[27/80]/[50/80]=27/50P(sophomore|favor)=[23/80]/[38/80]=23/38P(No opinion|sophomore)=?P(Favor | freshman)=?

Favor Oppose No opinion TotalFreshman 15 27 8 50Sophomore 23 5 2 30

Total 38 32 10 80

(c) 2007 IUPUI SPEA K300 (4392)

Summary

Addition: probability that event A or B occurs P(A U B) = P(A) + P(B) – P (A ∩ B)P (A ∩ B) =0 if mutually exclusive

Multiplication: probability that both events A and B occur P(A ∩ B) = P(A) X P(B|A) P(B|A)=P(B) if statistically independent