Probability
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Transcript of Probability
(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!5
25
C
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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.
)!(!rnnPrn
2012312345
!3!5
)!25(!5
25
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.
sapcesampleevent
NNEP E
_##)(
131
524)( QueenP
52
104)( 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 refrigerator3. 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 TrialP(Blue) 2/10 1/9P(Red) 3/10 3/9P(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 TrialP(Blue) 2/10 1/9P(Red) 3/10 3/9P(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 TotalMale 32 18 50Female 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