 Likelihood (chance) that an event occurs
 Classical 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)

 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)

 Example 4-2 on page 180
 Sample space: 52 outcomes

Event “Queen”: 4

Event “Heart”: 12

Event “King Spade”: 1
 Example 4-4
 Sample space: 8

Event “Exactly two boys”: 3
(c) 2007 IUPUI SPEA K300 (4392)

 Tossing a coin once: head (H) or tail (T)
 Tossing two times: HH, HT, TH, TT
 Tossing three times: HHH, HHT, HTH,
HTT, THH, THT, TTH, TTT  2 X 2 X 2
(c) 2007 IUPUI SPEA K300 (4392)

 H: head, T: tail
H
T
(c) 2007 IUPUI SPEA K300 (4392)
H
T
H
T
HH
HT
TH
TT



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)

 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 (n2) X … 1
 0! = 1 n
C r
 r !
( n n !
 r )!
5
C
2

2 !
( 5
5 !

2 )!

5 !
2 !
3 !

5

( 2
4

3


1 )( 3

2

1
2

1 )

10
(c) 2007 IUPUI SPEA K300 (4392)

 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.
n
P r

( n n !
 r )!
5
P
2

5 !
( 5

2 )!

5
3 !
!

5

4

3

2

1
3

2

1

20
(c) 2007 IUPUI SPEA K300 (4392)

 P(E) is the probability that the event E occurs; expected (not actualized) likelihood
 The number of outcomes of event E, N
E
, divided by the number of total outcomes in the sample space, N.
P ( E )

N
E
N

# event
# sample _ sapce
P ( Queen )

4
52

1
13
P ( Award )

4
10

2
5
(c) 2007 IUPUI SPEA K300 (4392)

 P(E) is a number between 0 and 1
 Probability 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)




 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)

 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)



 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) P(A and B) P(B)
(c) 2007 IUPUI SPEA K300 (4392)



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)



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)

 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)





1.
2.
How to put an elephant into a refrigerator?
Open the door
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)

 Example 4-25, pp.206-207
 What if no replacement?
 Suppose a blue ball is selected in the 1 st trial
 P(blue) is 2/10 in the 1 st trial
P(Blue)
P(Red)
P(White)
1 st Trial
2/10
3/10
5/10
2 nd
1/9
3/9
5/9
Trial
(c) 2007 IUPUI SPEA K300 (4392)



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.
1 st Trial 2 nd Trial
P(Blue)
P(Red)
P(White)
2/10
3/10
5/10
1/9
3/9
5/9
(c) 2007 IUPUI SPEA K300 (4392)



 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) P(A and B) P(B)
(c) 2007 IUPUI SPEA K300 (4392)






 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)
Male
Female
Yes
32
8
40
No
18
42
60
Total
50
50
100
(c) 2007 IUPUI SPEA K300 (4392)



With Replacement :

P(W|B)=P(W ∩ B)/P(B)=
[2/10*5/10]/[2/10]=5/10=P(W)
 Events white (2 nd trial) and blue (1 st 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 (2 nd trial) and blue (1 st trial) are dependent
Event blue in the 1 st trial influences the probability of event white in the 2 nd trial.
(c) 2007 IUPUI SPEA K300 (4392)

 P(oppose|freshman)=[27/80]/[50/80]=27/50
 P(sophomore|favor)=[23/80]/[38/80]=23/38
 P(No opinion|sophomore)=?
 P(Favor | freshman)=?
Favor Oppose No opinion Total
Freshman 15
Sophomore 23
27
5
8
2
50
30
Total 38 32 10 80
(c) 2007 IUPUI SPEA K300 (4392)

 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
(c) 2007 IUPUI SPEA K300 (4392)