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Basic Probability

Chap 4-1

Learning Objectives
In this chapter, you learn:




Basic probability concepts
Conditional probability
To use Bayes’ Theorem to revise probabilities

Chap 4-2

Basic Probability Concepts


Probability – the chance that an uncertain event
will occur (always between 0 and 1)



Impossible Event – an event that has no
chance of occurring (probability = 0)



Certain Event – an event that is sure to occur
(probability = 1)

Chap 4-3

Assessing Probability
There are three approaches to assessing the
probability of an uncertain event:
1
...
empirical probability
probability of occurrence =

number of ways the event can occur
total number of elementary outcomes

3
...
191
total number of people
439

Chap 4-6

Subjective probability






Subjective probability may differ from person to person
 A media development team assigns a 60%
probability of success to its new ad campaign
...



Simple event





Joint event





An event described by a single characteristic
e
...
, A day in January from all days in 2013
An event described by two or more characteristics
e
...
A day in January that is also a Wednesday from all days in 2013

Complement of an event A (denoted A’)



All events that are not part of event A
e
...
, All days from 2013 that are not in January

Chap 4-8

Sample Space
The Sample Space is the collection of all
possible events
e
...
All 6 faces of a die:

e
...
All 52 cards of a bridge deck:

Chap 4-9

Organizing & Visualizing Events


Contingency Tables -- For All Days in 2013
Jan
...

Not Wed
...


5

47

52

27

286

313

32

333

365

Decision Trees
5

Sample
Space

Total

All Days

27

In 2013

47

Total
Number
Of
Sample
Space
Outcomes

286
Chap 4-10

Definition: Simple Probability


Simple Probability refers to the probability of a
simple event
...
P(Jan
...
P(Wed
...


Wed
...

Total

Not Jan
...
) = 52 / 365

P(Jan
...




ex
...
and Wed
...
P(Not Jan
...
)
Jan
...

Not Wed
...


Total

5

47

52

27

286

313

32

333

365

P(Not Jan
...
)
= 286 / 365

P(Jan
...
) = 5 / 365
Chap 4-12

Mutually Exclusive Events


Mutually exclusive events


Events that cannot occur simultaneously

Example: Randomly choosing a day from 2013
A = day in January; B = day in February


Events A and B are mutually exclusive

Chap 4-13

Collectively Exhaustive Events


Collectively exhaustive events



One of the events must occur
The set of events covers the entire sample space

Example: Randomly choose a day from 2013
A = Weekday; B = Weekend;
C = January; D = Spring;




Events A, B, C and D are collectively exhaustive
(but not mutually exclusive – a weekday can be in
January or in Spring)
Events A and B are collectively exhaustive and
also mutually exclusive
Chap 4-14

Computing Joint and
Marginal Probabilities


The probability of a joint event, A and B:
number of outcomes satisfying A and B
P( A and B) =
total number of elementary outcomes



Computing a marginal (or simple) probability:
P(A) = P(A and B1 ) + P(A and B 2 ) +  + P(A and Bk )


Where B1, B2, …, Bk are k mutually exclusive and collectively
exhaustive events
Chap 4-15

Joint Probability Example
P(Jan
...
)
=

number of days that are in Jan
...

5
=
total number of days in 2013
365

Jan
...

Not Wed
...


Total

5

47

52

27

286

313

32

333

365

Chap 4-16

Marginal Probability Example
P(Wed
...
and Wed
...
and Wed
...

Wed
...

Total

Not Jan
...
5

The sum of the probabilities of all
mutually exclusive and collectively
exhaustive events is 1
P(A) + P(B) + P(C) = 1
If A, B, and C are mutually exclusive and
collectively exhaustive

Certain

0

Impossible
Chap 4-19

General Addition Rule
General Addition Rule:
P(A or B) = P(A) + P(B) - P(A and B)
If A and B are mutually exclusive, then
P(A and B) = 0, so the rule can be simplified:
P(A or B) = P(A) + P(B)
For mutually exclusive events A and B
Chap 4-20

General Addition Rule Example
P(Jan
...
) = P(Jan
...
) - P(Jan
...
)
= 32/365 + 52/365 - 5/365 = 79/365

Jan
...

Not Wed
...


Total

5

47

52

27

286

313

32

333

365

Don’t count
the five
Wednesdays
in January
twice!

Chap 4-21

Computing Conditional Probabilities


A conditional probability is the probability of one
event, given that another event has occurred:

P(A and B)
P(A | B) =
P(B)

The conditional
probability of A given
that B has occurred

P(A and B)
P(B | A) =
P(A)

The conditional
probability of B given
that A has occurred

Where P(A and B) = joint probability of A and B
P(A) = marginal or simple probability of A
P(B) = marginal or simple probability of B
Chap 4-22

Conditional Probability Example




Of the cars on a used car lot, 70% have air
conditioning (AC) and 40% have a GPS
...

What is the probability that a car has a GPS,
given that it has AC ?
i
...
, we want to find P(GPS | AC)

Chap 4-23

Conditional Probability Example
(continued)


Of the cars on a used car lot, 70% have air conditioning
(AC) and 40% have a GPS and
20% of the cars have both
...
2

0
...
7

No AC

0
...
1

0
...
4

0
...
0

P(GPS and AC) 0
...
2857
P(AC)
0
...
Of these,
20% have a GPS
...
57%
...
2

0
...
7

No AC

0
...
1

0
...
4

0
...
0

P(GPS and AC) 0
...
2857
P(AC)
0
...
2

...
5

...
2

P(AC and GPS’) = 0
...
2

...
1

...
2

P(AC’ and GPS’) = 0
...
2

...
2

...
2

P(GPS and AC’) = 0
...
5

...
1

...
5

P(GPS’ and AC’) = 0
...




Developed by Thomas Bayes in the 18th
Century
...


Chap 4-31

Bayes’ Theorem

P(A | B i )P(B i )
P(B i | A) =
P(A | B 1 )P(B 1 ) + P(A | B 2 )P(B 2 ) + ⋅ ⋅ ⋅ + P(A | B k )P(B k )



where:
Bi = ith event of k mutually exclusive and collectively
exhaustive events
A = new event that might impact P(Bi)

Chap 4-32

Bayes’ Theorem Example


A drilling company has estimated a 40%
chance of striking oil for their new well
...
Historically, 60% of successful
wells have had detailed tests, and 20% of
unsuccessful wells have had detailed tests
...
4 , P(U) = 0
...
6



(prior probabilities)

P(D|U) = 0
...
6)(0
...
6)(0
...
2)(0
...
24
=
= 0
...
24 + 0
...
667
Chap 4-35

Bayes’ Theorem Example
(continued)


Given the detailed test, the revised probability
of a successful well has risen to 0
...
4

Event

Prior
Prob
...


Joint
Prob
...


S (successful)

0
...
6

(0
...
6) = 0
...
24/0
...
667

U (unsuccessful)

0
...
2

(0
...
2) = 0
...
12/0
...
333

Sum = 0
...




In this topic, you learn various counting
rules for such situations
...
There are 3 parks, 4 restaurants, and 6 movie
choices
...
How many
different ways can these books be placed on the shelf?
Answer: 5! = (5)(4)(3)(2)(1) = 120 different possibilities

Counting Rules - 5

Counting Rules
(continued)


Counting Rule 4:


Permutations: The number of ways of arranging X
objects selected from n objects in order is

n!
n Px =
(n − X)!


Example:




You have five books and are going to put three on a
bookshelf
...
How many different combinations are there, ignoring
the order in which they are selected?
Answer:

n

Cx =

n!
5!
120
=
=
= 10
X!(n − X)! 3! (5 − 3)! (6)(2)

different possibilities

Counting Rules - 7



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