# Probability

```Business Statistics:
A Decision-Making Approach
7th Edition
Chapter 4
Using Probability and
Probability Distributions
Chap 4-1
Important Terms
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Random Variable - Represents a possible numerical value
from a random event and it can vary from trial to trial
Probability – the chance that an uncertain event will occur
Experiment – a process that produces outcomes for
uncertain events
Sample Space (or event) – the collection of all possible
experimental outcomes
4-2
Basic Rule of Probability
Individual Values
0 ≤ P(Ei) ≤ 1
For any event Ei
The value of a probability is
between 0 and 1
0 = no chance of occurring
1 = 100% change of occurring
Sum of All Values
k
 P(e )  1
i1
i
where:
k = Number of individual
outcomes in the
sample space
ei = ith individual outcome
The sum of the probabilities of all
the outcomes in a sample space
must = 1 or 100%
4-3
Simple probability

The probability of an event is the number of favorable
outcomes divided by the total number of possible
equally-likely outcomes.


This assumes the outcomes are all equally weighted.
Textbook: classical probability
Chap 4-4
Visualizing Events (or sample space)

Contingency Tables
Ace

Not Ace
Total
Black
2
24
26
Red
2
24
26
Total
4
48
52
Tree Diagrams
2
Sample
Space
Full Deck
of 52 Cards
Sample
Space
24
2
24
Chap 4-5
Experimental Outcome Example
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A automobile consultant records fuel type and vehicle
type for a sample of vehicles
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2 Fuel types: Gasoline, Diesel
3 Vehicle types: Truck, Car, SUV
6 possible experimental outcomes:
e1
Gasoline, Truck
e2
Gasoline, Car
e3
Gasoline, SUV
e4
Diesel, Truck
e5
Diesel, Car
e6
Diesel, SUV
e1
Car
e2
e3
e4
Car
e5
e6
Chap 4-6
Simple probability

What is the probability that a card drawn at
random from a deck of cards will be an ace?
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52 cards in the deck and 4 are aces
The probability is ????
Each card represents a possible outcome  52
possible outcomes.
Chap 4-7
Simple probability
There are 36
possible outcomes
when a pair of dice
is thrown.
Chap 4-8
Simple probability

Calculate the probability that the sum of the two
dice will be equal to 5?
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Four of the outcomes have a total of 5: (1,4; 2,3;
3,2; 4,1)
Probability of the two dice adding up to 5 is ????.
Chap 4-9
Simple probability

Calculate the probability that the sum of the
two dice will be equal to 12?
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Only one (6,6)
????
Chap 4-10
Simple Probability
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The probability of event A is denoted by P(A)
Example: Suppose a coin is flipped 3 times. What is the
probability of getting two tails and one head?
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The sample space consists of 8 possible outcomes.
S = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}
the probability of getting any particular outcome is 1/8.
Getting two tails and one head: A = {TTH, THT, HTT}
P(A) = ????
Chap 4-11
Essential Concepts and
Rules of Probability
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Independent event
Dependent event
Joint and Conditional Probabilities
Mutually Exclusive Event
Complement Rule
Multiplication Rule
Chap 4-12
Independent vs. Dependent Events
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Independent: Occurrence of one does not
influence the probability of
occurrence of the other
Dependent: Occurrence of one affects the
probability of the other
Chap 4-13
Examples
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Independent Events
A = heads on one flip of fair coin
B = heads on second flip of same coin
Result of second flip does not depend on the result of
the first flip.
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Dependent Events
X = rain forecasted on the news
Y = take umbrella to work
Probability of the second event is affected by the
occurrence of the first event
Chap 4-14
Probability Types
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Simple (Marginal) probability
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Joint probability
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Involves only a single random variable, the outcome of which is
uncertain
Involves two or more random variables, in which the outcome of
all is uncertain
Conditional probability (will be discussed soon!)
Chap 4-15
Exercise 4-22-A (page 158)
Answer for questions using “Pivot Table”
Chap 4-16
Exercise 4-22-B (page 158)
Chap 4-17
Practice

Exercise 4-23: try questions.
Electrical Mechanical
Total
Lincoln
28
39
67
Tyler
64
69
133
Total
92
108
200
Chap 4-18
Conditional Probability: ex 1
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A conditional probability is the probability of an event
given that another event has occurred.
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Involves two or more random variables, in which the outcome of
at least one is known
Conditional probability for any two events A , B:
P(Aand B)
P(A| B) 
P(B)
where P(B)  0
Notation
Chap 4-19
Conditional Probability: ex 1
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Of the cars on a used car lot, 70% have air
conditioning (AC) and 40% have a CD player
(CD). 20% of the cars have both.
What is the probability that a car has a CD
player, given that it has AC (use the table on
the next slide)?
i.e., we want to find P(CD | AC)
Chap 4-20
Conditional Probability: ex 1
What is the probability that a car has a CD player, given
that it has AC ?
CD
No CD
Total
AC
.2
.5
.7
No AC
.2
.1
.3
Total
.4
.6
1.0
P(CDand AC) .2
P(CD| AC) 
  .2857
P(AC)
.7
Chap 4-21
Conditional Probability: ex 2

What is the probability that the total of two
dice will be greater than 8 given that the first
die is a 6?

This can be computed by considering only outcomes
for which the first die is a 6. Then, determine the
proportion of these outcomes that total more than 8.
Chap 4-22
Conditional Probability: ex 2
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There are 6 outcomes
for which the first die
is a 6.
And of these, there
are four that total
more than 8.
(6,3; 6,4; 6,5; 6,6)
Chap 4-23
Conditional Probability: ex 2
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6 outcomes for which the first die is a 6
There are four that total more than 8
The probability of a total greater than 8 given that the
first die is 6 is 4/6 = 2/3.
More formally, this probability can be written as:
p(total>8 | Die 1 = 6) = 2/3(6,3; 6,4; 6,5; 6,6).
Conditional probability using “Probability Type
Example” Excel file
Chap 4-24
Mutually Exclusive Events
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If A occurs, then B cannot occur

A and B have no common elements
A
Black
Cards
B
Red
Cards
A card cannot be
Black and Red at
the same time.
Chap 4-25
Mutually Exclusive Events
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If events A and B are mutually exclusive, then the
probability of A or B is p(A or B) = p(A) + p(B)
What is the probability of rolling a die and getting either
a 1 or a 6?
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impossible to get both a 1 and a 6
p(1 or 6) = p(1) + p(6) = 1/6 + 1/6 = 1/3
Chap 4-26
Not Mutually Exclusive Events
■
If the events are not mutually exclusive,
P(A or B) = P(A)+ P(B) – P(A and B)
A
+
B
=
A
B
P(A or B) = P(A) + P(B) - P(A and B)
Overlap: joint probability
Chap 4-27
Not Mutually Exclusive Events
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What is the probability that a card will be either an ace
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p(ace) = 4/52 and p(spade) = 13/52
The only way both can be drawn is to draw the ace of spades.
There is only one ace of spades: p(ace and spade) = 1/52 .
The probability of an ace or a spade can be:
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4/52 + 13/52 - 1/52 = 16/52 = 4/13
Chap 4-28
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Suppose we have two events and want to know
the probability that either event occurs.
Mutually exclusive:
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P(A or B) = P(A) + P(B)
Not Mutually exclusive:
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P(A or B) = P(A)+ P(B) – P(A and B)
Chap 4-29
Complement Rule
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The complement of an event E is the collection of all
possible elementary events not contained in event E.
The complement of event E is represented by E.
Complement Rule:
P(E)  1  P(E)
Or,
E
P(E)  P(E)  1
If there is a 40% ( E ) chance it will rain,
there is a 60% ( E ) chance it won’t rain!
E
Chap 4-30
Rule of Multiplication
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The rule of multiplication applies to the situation
when we want to know the probability that both
events (event A and event B) occur.
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Independent: P(A ∩ B) = P(A) P(B)
Dependent: P(A ∩ B) = P(A) P(B|A)
Chap 4-31
Independent Events
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If A and B are independent, then the probability that
events A and B both occur is: p(A and B) = p(A) x
p(B)
What is the probability that a coin will come up with
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Two events must occur: a head on the first toss and a head
on the second toss.
The probability of each event is 1/2
the probability of both events is: 1/2 x 1/2 = 1/4.
Chap 4-32
Independent Events
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What is the probability that the first card is the
ace of clubs (put it back: replace) and the
second card is a club (any club)?
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The probability of the first event is 1/52 (only one ace
of club)
The probability of the second event is 13/52 = 1/4
(composed of clubs)
Answer is 1/52 x 1/4 = 1/208
Chap 4-33
Dependent Events
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If A and B are not independent, then the
probability of A and B both occur is:
p(A and B) = p(A) x p(B|A)
where p(B|A) is the conditional probability of B
given A.
Chap 4-34
Dependent Events
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If someone draws a card at random from a deck and
then, without replacing the first card, draws a second
card, what is the probability that both cards will be
aces?
4 of the 52 cards are aces
First: p(A) = 4/52 = 1/13.
 Of the 51 remaining cards, 3 are aces. So, p(B|A)
= 3/51 = 1/17
 Probability of A and B is: 1/13 x 1/17 = 1/221.
Chap 4-35
Practice
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Page 180
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Exercise 4-26
Exercise 4-30
Chap 4-36
Summary
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Probability Rules