BU522: LECTURE 7
Chapter 6:
Elementary Probability
Objectives:
•Understand elementary probability concepts.
•Calculate the probability of events,
•Distinguish between mutually exclusive,
•Use general additional law for probabilities,
•dependent and independent events,
•Calculate conditional probabilities,
ILLUSTRATION – SBM Students
Sample
space
QMII
Bachelor
Accounting
Financial
Management
2
Macroeconomics
MATHS II
Events
Probability is the study of uncertainty and is concerned
with events happening by chance.
Probability is the numerical measure of the likelihood
that the event will occur.
The greater the probability the more likely the event
will occur.
It can be written as a fraction, decimal, percent, or
ratio.
Probability Scale
Value is between 0 and 1.
Suppose the sample space S of an experiment has n
1
Certain
outcomes given by:
S  {e1, e 2 , e3 ,....en }
To each outcome S, we assign a real
number, P(e), called the probability
of the outcome e,
Then,
.5
50/50
P(e1 )  P(e 2 )  P(e3 )  ...  P(en )  1
0
•Probability must be non negative
•Probability must NOT be greater than one.
Impossible
A probability experiment is an action through which
specific results (counts, measurements, or responses)
are obtained.
Such as Tossing a coin, throwing a Die,
The result of a single trial in a probability experiment is
an outcome.
The set of all possible outcomes of a probability
experiment is the sample space, denoted by the letter S.
Sample Space
This is a set consisting of all the possible outcomes.
Example 1:
Tossing of a coin: S = { H, T}
Example 2:
All 6 faces of a die: S = { 1 , 2 , 3 , 4 , 5 , 6 }
Example 3:
A survey is undertaken in which the gross annual income
of accountants in year 2005 is recorded. The figure
recorded is whether this income exceeded $100,000.00.
The sample space is:
S = {gross income  $100 000, gross income  $100 000}
Example 4:
A stockbroker makes a listing of all his stocks. For each
one he records whether he feels it will rise or fall in value in
the next 12 months. The sample space is:
S = {rise in value, fall in value}
Example 5:
Consider an experiment in which, for the sake of
simplicity, one die is black and the other is red. When
the dice are rolled, the set of outcomes consists of all
the different ways that the dice may come to rest. Find a
sample space for this experiment.
1
2
3
4
5
6
1 1,1 1,2 1,3 1,4 1,5 1,6
2 2,1 2,2 2,3 2,4 2,5 2,6
3 3,1 3,2 3,3 3,4 3,5 3,6
4 4,1 4,2 4,3 4,4 4,5 4,6
5 5,1 5,2 53 5,4 5,5 5,6
6 6,1 6,2 6,3 6,4 6,5 6,6
Example 6:
Describe a sample space for the experiment of selecting
one family from the set of all possible three-child families.
Solution:
This experiment can be depicted by the tree diagram as
follows:
Boy
Boy
Girl
Boy
Girl
Girl
Boy
BBB
Girl
BBG
Boy
BGB
Girl
BGG
Boy
GBB
Girl
Boy
GBG
Girl
GGG
GGB
Revision:
Formula:
n(E)
P(E) 
n(S)
Where n(E) is the number of
elements in the event and n(S) is
the number of elements in the
sample space.
Example 7:
The probability of picking a picture card from a pack of 52
playing cards is:
n (Picture cards) 12 3
P(Picture card) 


n (S)
52 13
Example 8:
What is the probability of obtaining a Head upon
tossing of a coin (NB. not the Tail). The event is Head
to show up.
Solution:
When tossing a coin, the are 2 possible outcomes
i.e. {H, T}..
Thus,
n(H) 1
P(H) 
  0.5
n(S) 2
Example 9:
On rolling a die, what is the probability that an odd
number will show?
Solution:
A die has 6 faces and three of them have odd numbers.
Therefore,
n (odd ) 3 1
P(odd ) 
 
n (S)
6 2
Complementary Events
The complements of an event are those outcomes of the
sample space for which the event does not occur.
Two events that are complements of each other are said
to be complementary.
The complement of event E is denoted by E or E
Properties of Probability:
0  P(E )  1
P(E)  P(E )  1
P(E)  1  P(E )
P(E )  1  P(E)
Example 10:
Upon tossing a die what is the probability of obtaining not a 5.
Solution:
n (5) 1
P(5) 

n (S) 6
n (5)
1 5
 P( Not a 5)  1 
 1 
n (S)
6 6
P(E)  P(E )  1
The given equation is particularly useful when calculating the
probability that, for example, at least one object in a random
sample has a particular characteristics.
Define the two events:
A = no one has the characteristics
B = at least one person has the characteristics.
If A and B are complementary events,
Then,
P(at least one object has the characteristics)
= 1 – P(no object has the characteristics)
Example 11:
A coin is tossed 6 times. Find the probability of obtaining at least 1
head.
Solution:
Define events A and B:
A = at least 1 head
B = no heads
Since the 6 tosses are independent,
1 1 1 1 1 1 1
P(B)       
2 2 2 2 2 2 64
Hence,
P(at least one head)  1  P(no heads)
1
 1
64
63

64
Two events A and B are mutually exclusive if and only if:
P( A  B)  0
In a Venn diagram this means that event A is disjoint
from event B.
A
B
A and B are M.E.
A
B
A and B are not M.E.
The Addition Rule
The probability that at least one of the events A or B
will occur, P(A or B), is given by:
P( A or B)  P( A  B)  P( A)  P( B)  P( A  B)
If events A and B are mutually exclusive, then the addition
rule is simplified to:
P( A or B)  P( A  B)  P( A)  P( B)
This simplified rule can be extended to any number of
mutually exclusive events.
Example 12:
An accountant found that receivables fell into one of 4
categories.
A = paid on time
C = paid late
B = paid early
D = uncollectable,
Of a sample of 120 receivables, 35 were paid on time,
40 were paid early, 28 were paid late and the remainder
were uncollectable.
Find the probability that a receivable was paid early or
on time?
Solution:
The events A and B are mutually exclusive. We have
35
40
P(A) 
P(B) 
120
120
 P(A or B)  P(A)  P(B)
35
40


120 120
75

120
5

8
Example 13:
Two dice are rolled simultaneously. What is the probability that a
sum of 5 or 6 is shown.
Solution:
1
2
3
4
5
6
1 1,1 1,2 1,3 1,4 1,5 1,6
2 2,1 2,2 2,3 2,4 2,5 2,6
3 3,1 3,2 3,3 3,4 3,5 3,6
4 4,1 4,2 4,3 4,4 4,5 4,6
5 5,1 5,2 53 5,4 5,5 5,6
6 6,1 6,2 6,3 6,4 6,5 6,6
There is a total of 36 possible
outcomes.
4
P( sum 5) 
36
5
P ( sum 6) 
36
P( sum 5  sum 6)  0
 P( Sum 5 or Sum 6)
9 1
4
5




0
36 4
36 36
Example 14:
From a well shuffled pack of 52 cards, what is the
probability of drawing a Jack or a Heart?
Solution:
Let J be Jack and H be Heart
4
P( J )  ,
52
13
P( H )  ,
52
1
P( J  H )  ,
52
 P( J or H )  P( J )  P( H )  P( J  H )
4 13 1



52 52 52
16
4


52 13
Independent events
Two events A and B are independent if the occurrence of
one not alter the likelihood of the event occurring.
If two events A and B are independent, then this
relationship holds:
P(A and B)  P(A)  P(B)
Note:
This relationship can also be applied to more
independent events.
Conditional Probability
Conditional probability is the probability of an event
occurring, given that another event has already occurred.
The conditional probability of event B occurring, given that
event A has occurred, is denoted by P(B|A) and is read as
“probability of B, given A.”
We use conditional probability when two events occurring
in sequence are not independent. In other words, the fact
that the first event (event A) has occurred affects the
probability that the second event (event B) will occur.
Conditional Probability
Formula for Conditional Probability
P(B  A )
P( A  B )
P( A | B ) 
or P(B | A ) 
P(B)
P( A )
The formula above can also be rearrange to:
P(B  A)  P(B)  P(A B)
P(A  B)  P(A)  P(B A)
Example 15:
Table below shows the sex and age characteristics of a
population of 20 males and 30 females.
Age
Age under 30
Age 30 or over
Total
Sex
Male
8
12
20
Female
20
10
30
Total
28
22
50
A person is chosen at random from the population.
Let A = person is male, B = person is female, C = person is
aged under 30 and D = person is aged 30 and over.
Calculate the following probabilities.
a ).
20 2
P(A) 

50 5
b).
30 3
P(B) 

50 5
c).
28 14
P ( C) 

50 25
d ).
22 11
P ( D) 

50 25
e).
12
6
P(A and D) 

50 25
f ).
20 2
P(B and C) 

50 5
g ).
P(A and D)
P ( A D) 
P ( D)
12
6
50


22 11
50
h ).
20
P(C and B) 50 2
P(C B) 


30 3
P(B)
50
i).
P(A or C)  P(A)  P(C)  P(A  C)
20 28 8



50 50 50
40

50
4

5
The Multiplication Rule
For any two events A and B, the probability of events, A and B
occurring in a sequence (or simultaneously) is:
P( A and B)  P( A  B)  P( A)  P( B A)
If events A and B are independent, then the probability of two
events, A and B occurring in a sequence (or simultaneously) is:
P ( A and B )  P ( A  B )  P ( A)  P ( B )
This rule can extend to any number of independent events.
Example 16:
Two cards are drawn from a well shuffled deck of 52 cards.
First card is not replaced. Find the probability of drawing 2
Hearts.
Solution
Let A be the Second card is a Heart and let B be the first
card is a Heart.
P(B  A )
12
13
P( A | B ) 
P(B) 
, P( A | B ) 
P(B)
51
52
P(B  A)  P(B)  P(A
B)
1
13 12

 0.059


17
52 51
END