1. Chapter 6
Probability
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.

2. 6.1 Assigning probabilities to Events
Random experiment
a random experiment is a process or course of action, whose outcome is uncertain.
Examples
Experiment Outcomes
Flip a coin Heads and Tails
Record a statistics test marks Numbers between 0 and 100
Measure the time to assemble numbers from zero and above a computer
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.

3. 6.1 Assigning probabilities to Events
Repeating the same random experiment, may result in different outcomes, therefore, the best we can do is consider the probability that a certain outcome occurs.
To determine the probabilities, we need to define and list all possible outcomes first.
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4. Sample Space Determining the outcomes.
Build an exhaustive list of all possible outcomes.
Make sure the listed outcomes are mutually exclusive.
A list of outcomes that meets the two conditions above is called a sample space.
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5. Sample Space: S = {O1, O2,…,Ok}
a sample space of a random experiment
is a list of all possible outcomes of the
experiment. The outcomes must be
mutually exclusive and exhaustive.
Simple Events
The individual outcomes are called simple events.
Simple events cannot be further decomposed
into constituent outcomes.
Event
An event is any collection
of one or more simple events
Our objective is to
determine P(A), the
probability that event A will occur.
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6. Assigning Probabilities
Given a sample space S={O1,O2,…,Ok}, the following characteristics for the probability P(Oi) of the simple event Oi must hold:
Probability of an event: P(A), the probability of event A is the sum of the probabilities assigned to the simple events contained in A.
Ex. A={ O1 , O2, O3}, P(A)=P(O1)+P(O2)+ P(O3)
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7. Example The weights in pounds of a sample of 25 workers:
164, 148, 137, 157, 173, 156, 177, 172, 169, 165, 145, 168, 163, 162, 174, 152, 156, 168, 154, 151, 174, 146, 134, 140, and 171. .
If 25 workers are population, the relative frequency distribution can be regarded as the probability distribution.
What is the probability of randomly selecting a worker whose weight is larger than 160 pounds?
What is the probability of randomly selecting a worker with W<=180?
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8. Approaches to Assigning Probabilities…
There are three ways to assign a probability, P(Oi), to an outcome, Oi, namely:
Classical approach: make certain assumptions (such as equally likely, independence) about situation.
The assumption should make sense.
Relative frequency: assigning probabilities based on experimentation or historical data.
Assume that historical data is equal to the population.
Subjective approach: Assigning probabilities based on the assignor’s judgment.
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9. Classical Approach…
If an experiment has n possible outcomes, this method would assign a probability of 1/n to each outcome.
Experiment: Rolling a die
Sample Space: S = {1, 2, 3, 4, 5, 6}
Probabilities: Each sample point has a 1/6 chance of occurring.
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10. Relative Frequency Approach…
Bits & Bytes Computer Shop tracks the number of desktop computer systems it sells over a month (30 days):
For example,
10 days out of 30
2 desktops were sold.
From this we can construct
the probabilities of an event
(i.e. the # of desktop sold on a given day)…
Desktops Sold
# of Days 1 2 10 3 12 4 5
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11. Relative Frequency Approach…
Desktops Sold
# of Days 1
1/30 = .03 2
2/30 = .07 10
10/30 = .33 3 12
12/30 = .40 4 5
5/30 = .17
Total 30
∑ = 1.00
“There is a 40% chance Bits & Bytes will sell 3 desktops on any given day”
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12. Relative Frequency Approach
Insurance company uses the relative frequency approach.
What is the underlying assumption about the relative frequency approach in the case of the last example?
Past experience of 1 month will continue.
The relative frequency approach may not be accurate.
The past data about accidents may not be applied to the future.
The past data is not population.
Why should we use the relative frequency approach?
There is no other good alternative.
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13. Subjective Approach… Who will win, Hong Man Choi vs Mike Tyson?
Hong Man Choi will win against Tyson with 40% of probability.
“Subjective approach relies on the specialist’s hunch.
E.g. weather forecasting’s “P.O.P.”
“Probability of Precipitation (Rain)” (or P.O.P.) is basically a subjective probability based on past observations combined with current weather conditions.
POP 60% – based on current conditions, there is a 60% chance of rain (say).
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14. Example
Flip a coin
What is the probability of Head? ½
The box includes 5 blue balls, 3 red balls, 2 green balls.
What is the probability of randomly selecting a blue ball? 5/10 =1/2
What is the probability of randomly selecting a red ball? 3/10
What is the probability of randomly selecting a red ball or a blue ball?
What is the probability of randomly selecting a blue, red or green ball?
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15. 6.2 Joint, Marginal, and Conditional Probability
There are 6 yellow shirts, 4 green shirts and 10 blue shirts.
What is the probability of randomly selecting a yellow shirt? 6/20
Color
Frequency
Relative Frequency
Yellow shirts 6
6/20
Green shirts 4
4/20
Blue shirts 10
10/20
Total 20 1
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16. Two different sizes: Large and small.
Frequency and Relative Frequency of shirts based on size and color
Color\Size
Large
Small
Total
Yellow
3 (3/20)
6 (6/20)
Green
1 (1/20)
4 (4/20)
Blue
10 (10/20)
8 (8/20)
12 (12/20)
20 (20/20)
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17. Joint probability and marginal probability
Color\Size
Large
Small
Margin
Yellow
3/20
6/20
Green
1/20
4/20
Blue
10/20
8/20
12/20 1
When population is classified by two or more dimensions, relative frequency is joint probability.
When population is classified by one dimension, relative frequency is called as marginal probability.
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18. Joint probability and marginal probability
Color\Size
Large
Small
Margin
Yellow
3/20
6/20
Green
1/20
4/20
Blue
10/20
8/20
12/20 1
What is the probability of selecting a shirt that is yellow and large? Intersection
What is the probability of selecting a shirt that is large or yellow? Union
What is the probability of selecting a shirt that is blue? Marginal probability
What is the probability of selecting a shirt that is large? Marginal Probability
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.

19. Conditional Probability
Color\Size
Large
Small
Margin
Yellow 3 6
Green 1 4
Blue 10 8 12 20
If we know that shirts are large, what is the sample space?
If we know that shirts are large, then how many shirts are yellow? 3
If we know that shirts are large, then what is the probability of selecting a shirt that is yellow? Conditional Probability. P(Yellow|Large) = 3/8
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20. Conditional Probability
Color
Large
Yellow 3
Green 1
Blue 4
Total 8
If we know that shirts are large, then what is the probability of selecting a shirt that is yellow?
If we know that shirts are large, what is the probability of selecting a shirt that is green?
Color\Size
Large
Small
Margin
Blue 4 6 10
If we know that shirts are blue, then what is the probability of randomly selecting a small shirt? 6/10 =3/5
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21. Conditional Probability
Color\Size
Large
Small
Total
Yellow
3 (3/20)
6 (6/20)
Green
1 (1/20)
4 (4/20)
Blue
10 (10/20)
8 (8/20)
12 (12/20)
20 (20/20)
If we know that shirts are small, what is the probability of randomly selecting a green shirt? P(green|small) =3/12
Prob(Green and Small) = 3/20, Prob(Small) = 12/20
Prob(green|small) = Prob(Green and Small)/Prob(Small) = (3/20)/(12/20)
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22. 6.2 Joint, Marginal, and Conditional Probability
There are several types of combinations and relationships between events:
Intersection of events
Union of events
Dependent and independent events
Complement event
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23. Intersection
Example 6.1
A potential investor examined the relationship between the performance of mutual funds and the school the fund manager earned his/her MBA from.
The following table describes the joint probabilities.
Mutual fund outperform the market
Mutual fund doesn’t outperform the market
Top 20 MBA program
.11
.29
Not top 20 MBA program
.06
.54
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24. Intersection Example 6.1 – continued
The joint probability of [mutual fund outperform…] and […from a top 20 …] = .11
The joint probability of [mutual fund outperform…] and […not from a top 20 …] = .06
P(A1 and B1)
Mutual fund outperforms the market
(B1)
Mutual fund doesn’t outperform the market (B2)
Top 20 MBA program (A1)
.11
.29
Not top 20 MBA program (A2)
.06
.54
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.

25. Intersection Example 6.1 – continued
The joint probability of [mutual fund outperform…] and […from a top 20 …] = .11
The joint probability of [mutual fund outperform…] and […not from a top 20 …] = .06
P(A1 and B1)
P(A2 and B1)
Mutual fund outperforms the market
(B1)
Mutual fund doesn’t outperform the market (B2)
Top 20 MBA program (A1)
.11
.29
Not top 20 MBA program (A2)
.06
.54
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26. Mutually Exclusive
If P(A and B) = 0, then A and B are mutually exclusive.
A and Ac are mutually exclusive.
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27. Marginal Probability
These probabilities are computed by adding across rows and down columns.
Mutual fund outperforms the market (B1)
Mutual fund doesn’t outperform the market (B2)
Marginal Prob.
P(Ai)
Top 20 MBA program (A1)
Not top 20 MBA program (A2)
Marginal Probability P(Bj)
P(A1 and B1)+ P(A1 and B2) = P(A1)
P(A2 and B1)+ P(A2 and B2) = P(A2)
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28. Marginal Probability
These probabilities are computed by adding across rows and down columns
Mutual fund outperforms the market (B1)
Mutual fund doesn’t outperform the market (B2)
Marginal Prob.
P(Ai)
Top 20 MBA program (A1)
.11
.29
.40
Not top 20 MBA program (A2)
.06
.54
.60
Marginal Probability P(Bj) + = + =
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29. Marginal Probability
These probabilities are computed by adding across rows and down columns
Mutual fund outperforms the market (B1)
Mutual fund doesn’t outperform the market (B2)
Marginal Prob.
P(Ai)
Top 20 MBA program (A1)
.40
Not top 20 MBA program (A2)
.60
Marginal Probability P(Bj)
P(A1 and B1) +
P(A2 and B1 =
P(B1)
P(A1 and B2) +
P(A2 and B2 =
P(B2)
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.

30. Marginal Probability
These probabilities are computed by adding across rows and down columns
Mutual fund outperforms the market (B1)
Mutual fund doesn’t outperform the market (B2)
Marginal Prob.
P(Ai)
Top 20 MBA program (A1)
.11
.29
.40
Not top 20 MBA program (A2)
.06
.54
.60
Marginal Probability P(Bj)
.17
.83 + +
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31. Conditional Probability
Example 6.2 (Example 6.1 – continued)
Find the conditional probability that a randomly selected fund is managed by a “Top 20 MBA Program graduate”, given that it did not outperform the market.
Solution
P(A1|B2) = P(A1 and B2) = .29 = P(B2)
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32. Conditional Probability
Example 6.2
Find the conditional probability that a randomly selected fund is managed by a “Top 20 MBA Program graduate”, given that it did not outperform the market.
Solution
P(A1|B2) =
P(A1 and B2) P(B2) =.29/.83 = .3494
New information
reduces the relevant sample space to the 83% of event B2.
Mutual fund outperforms the market (B1)
Mutual fund doesn’t outperform the market (B2)
Marginal Prob.
P(Ai)
Top 20 MBA program (A1)
.11
.29
.40
Not top 20 MBA program (A2)
.06
.54
.60
Marginal Probability P(Bj)
.17
.83
.29
.83
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33. Conditional Probability
Before the new information becomes available we have P(A1) = 0.40
After the new information (B2) becomes available P(A1) changes to P(A1 given B2) = .3494
Since the occurrence of B2 has changed the probability of A1, the two event (A1 , B2 ) are related and are called “dependent events”.
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34. Independence Independent events
Two events A and B are said to be independent if P(A|B) = P(A and B)/P(B) =P(A) or P(B|A) = (B and A)/P(A) =P(B)
That is, the probability of one event is not affected by the occurrence of the other event.
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35. Dependent and independent events
Example 6.3 (Example 6.1 – continued)
We have already seen the dependency between A1 and B2.
Let us check A2 and B2.
P(B2) = .83
P(B2|A2)=P(B2 and A2)/P(A2) = .54/.60 = .90
Conclusion: A2 and B2 are dependent.
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36. Union
The union event of A and B is the event that occurs when either A or B or both occur.
It is denoted “A or B”.
Example 6.4 (Example 6.1 – continued) Calculating P(A or B))
Determine the probability that a randomly selected fund outperforms the market or a fund’s manager graduated from a top 20 MBA Program.
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37. Union of events
Union
Solution
Mutual fund outperforms the market (B1)
Mutual fund doesn’t outperform the market (B2)
Top 20 MBA program (A1)
.11
.29
Not top 20 MBA program (A2)
.06
.54
Comment:
P(A1 or B1) = 1 – P(A2 and B2)
= 1 – .54 = . 46
A1 or B1 occurs
whenever either:
A1 and B1 occurs,
A1 and B2 occurs,
A2 and B1 occurs.
P(A1 or B1) = P(A1 and B1) + P(A1 and B2) + P(A2 and B1) =
= .46
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38. 6.3 Probability Rules and Trees
We present more methods to determine the probability of the intersection and the union of two events.
Three rules assist us in determining the probability of complex events from the probability of simpler events.
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39. Complement event
Complement Rule
The complement of event A (denoted by AC) is the event that occurs when event A does not occur.
The probability of the complement event is calculated by
A and AC consist of all the simple events in the sample space. Therefore, P(A) + P(AC) = 1
P(AC) = 1 - P(A)
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40. Multiplication Rule For any two events A and B
When A and B are independent
P(A and B) = P(A)P(B|A)
= P(B)P(A|B)
P(A and B) = P(A)P(B)
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41. Multiplication Rule Example 6.5 Solution
What is the probability that two female students will be selected at random to participate in a certain research project, from a class of seven males and three female students?
Solution
Define the events: A – the first student selected is a female B – the second student selected is a female
P(A and B) = P(A)P(B|A) = (3/10)(2/9) = 6/90 = .067
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42. Multiplication Rule Example 6.6 Solution
What is the probability that a female student will be selected at random from a class of seven males and three female students, in each of the next two class meetings?
Solution
Define the events: A – the first student selected is a female B – the second student selected is a female
P(A and B) = P(A)P(B/A) = P(A) P(B)= (3/10)(3/10) = 9/100 = .09
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43. P(A or B) = P(A) + P(B) - P(A and B)
Addition Rule
For any two events A and B
P(A or B) = P(A) + P(B) - P(A and B)
P(A) =6/13
A or B + A
P(B) =5/13 _ B
P(A and B) =3/13
P(A or B) = 8/13
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44. Addition Rule A B B When A and B are mutually exclusive,
P(A or B) = P(A) + P(B) A B B
P(A and B) = 0
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45. Addition Rule
Example 6.7
The circulation departments of two newspapers in a large city report that 22% of the city’s households subscribe to the Sun, 35% subscribe to the Post, and 6% subscribe to both.
What proportion of the city’s household subscribe to either newspaper?
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46. Addition Rule Solution Define the following events:
The Addition and
Multiplication Rules
Addition Rule
Solution
Define the following events:
A = the household subscribe to the Sun
B = the household subscribe to the Post
Calculate the probability P(A or B) = P(A) + P(B) – P(A and B) = = .51
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47. Probability Trees Example 6.5 revisited (dependent events). Definition
Find the probability of selecting two female students (without replacement), if there are 3 female students in a class of 10.
Definition
F1: The first student is a female.
F2: The second student is a female.
M1: The first student is a male.
M2: The second student is a male.
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48. Probability Trees Example 6.5 revisited (dependent events).
Find the probability of selecting two female students (without replacement), if there are 3 female students in a class of 10.
P(F1 and F2)=(3/10)(2/9)
P(F1 and M2)=(3/10)(7/9)
P(M1 and F2)=(7/10)(3/9)
P(M1 and M2)=(7/10)(6/9)
Joint probabilities
Second selection
P(F2|M1) = 3/9
P(F2|F1) = 2/9
P( M2|M1) = 6/9
P( M2|F1) = 7/9
First selection
P(F1) = 3/10
P( M1) = 7/10
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49. Probability Trees Example 6.6 – revisited (independent events)
Find the probability of selecting two female students (with replacement), if there are 3 female students in a class of 10.
P(F1 and F2)=(3/10)(3/10)
P(F1 and M2)=(3/10)(7/10)
P(M1 and F2)=(7/10)(3/10)
P(M1 and M2)=(7/10)(7/10)
= P(F2) =
= P(M2) =
Second
selection
P(F2|M1) = 3/10
P(F2|F1) = 3/10
P( M2|M1) =7/10
P( M2|F1) = 7/10
First
selection
P(F1) = 3/10
P( M1) = 7/10
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50. Probability Trees Example 6.8 (conditional probabilities)
The pass rate of first-time takers for the bar exam at a certain jurisdiction is 72%.
Of those who fail, 88% pass their second attempt.
Find the probability that a randomly selected law school graduate becomes a lawyer (candidates cannot take the exam more than twice).
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51. Probability Trees Solution First exam Second exam
P(Pass) = .72
P(Fail and Pass)= .
28(.88)=.2464
P(Fail and Fail) =
(.28)(.(12) = .0336
First
exam
P(Pass) = .72
P( Fail) = .28
Second
exam
P(Pass|Fail) = .88
P( Fail|Fail) = .12
P(Pass) = P(Pass on first exam) + P(Fail on first and Pass on second) = .9664
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52. Bayes’ Theorem for the Revision of Probability
In the 1700s, Thomas Bayes developed a way to revise the probability that a first event (A) occurred from information obtained from a second event (B).
Bayes’ Theorem: For two events A and B
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53. Bayes’ Law
Example 6.9
Medical tests can produce false-positive or false-negative results.
A particular test is found to perform as follows:
Correctly diagnose “Positive” 94% of the time.
Correctly diagnose “Negative” 98% of the time.
It is known that 4% of men in the general population suffer from the illness.
What is the probability that a man is suffering from the illness, if the test result were positive?
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54. Bayes’ Law Solution Define the following events
D = Has a disease
DC = Does not have the disease
PT = Positive test results
NT = Negative test results
Build a probability tree
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55. Bayes’ Law Solution – Continued The probabilities provided are:
P(D) = .04 P(DC) = .96
P(PT|D) = .94 P(NT|D)= .06
P(PT|DC) = .02 P(NT|DC) = .98
The probability to be determined is
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56. Bayes’ Law + D PT D PT|D PT|D PT|D PT D D PT|D D D| PT|D PT PT PT PT
P(D and PT)
=.0376
PT|D
P(PT|DC) = .02
P( NT|DC) = .98
P(PT|D) = .94
P( NT|D) = .06
P(PT) =.0568 +
P(DC) = .96
P(D) = .04
P(DC and PT)
=.0192
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57. Bayes’ Law Prior Likelihood probabilities probabilities
P(PT|DC) = .02
P( NT|DC) = .98
P(PT|D) = .94
P( NT|D) = .06
Posterior probabilities
P(DC) = .96
P(D) = .04
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