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PROBABILITY
Introduction



Every trial or experiment has one or more possible
outcomes
An event, denoted by capital letter is a collection of
one or more of those outcomes.
For example, in tossing a die, event may be defined
as getting odd numbers which consist of three
possible outcomes.




The probability denoted by of an event A is a
measure of the likelihood or chance of the event
occurring in any one trial or experiment.
The value for P(A) is between 0 and 1.
P(A)=0 mean there are no chance for an event to
occur (impossible event)
P(A)=1 means an event that is sure to occur
(certain event).
Venns Diagram
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

Sample spaces, S contains all possible outcomes of
an experiment.
Event is a subset of the sample space.
For example, event A is a subset of S, .
S
A B
C



The union of event A and B, which is denoted as A  B
is the event of all elements that belong to A or B or
both.
The intersection of event A and B, which is denoted
by A  B , is the event of all elements that belong to
both A and B.
Event A and C have no outcomes in common, they
are said to be mutually exclusive.
Classical definition of probability

If S is a sample space, event A is a subset of S, and
the experiment has an equal probability of
accurance, then the probability of A can be obtain
by:
Number outcome in A n( A)
P( A) 

Number outcome in S n( S )
Example 1
Karim has a set of eight cards numbered 1 to 8. A
card is drawn randomly from the set of cards. Find
the probability that the number drawn is
(a) 8
(b) Not 8

Solution:
(a) If E is the event that a number 8 is drawn and S is
the sample space, then S={1,2,3,4,5,6,7,8} ; E={8}
; E={1,2,3,4,5,6,7}
n( E ) 1
P( E ) 

n( S ) 8
(b)
n( E ) 7
P( E ) 

n( S ) 8
1 7
Note that 1  P( E )  1    P( E )
8 8
Example 2
Two fair dice are thrown. Determine
I.
the sample space of the experiment.
II.
the elements of event A if the outcomes of both
dice thrown are showing the same number of dot.
III.
the elements of event B if the first thrown giving a
greater number of dot than the second thrown.
IV.
probability of event A, P(A) and event B, P(B)

If A and B are two events from an experiment with
the conditions that P( A)  0 and P( B)  0, then
P  A  B   P  A  P  B   P  A  B 
Where A  B is the event that A occurs or B occurs
or both events A and B occurs while A  B is the
event that both A and B occur together
S
A
B
a-d

d
b-d
Let say n(A)=a, n(B)=b, n(S)=n and n(A  B)=d
n A  B
P  A  B 
nS 
a  d   d  b  d 


n
abd
n
a b d
  
n n n
 P  A  P  B   P  A  B 

Example 3
In a class consisting of 25 students, 10 are girls
and15 are boys. 4 of the girls and 8 of the boys
wear glasses. If a student is selected at random,
what is the probability that the student selected is a
girl or wearing a glasses?
Solution
If A is event that selected student is a girl and B is
event that the selected student wear glasses.
n  A  10
P  A 

n  S  25
n  B  12
P B  

n  S  25
n  A  B 4
P  A  B 

n S
25
P  A  B   P  A   P B   P  A  B 
10 12 4



25 25 25
18

25
18
Probability that the student selected is a girl or wearing glasses is
25
Probability Of Mutually Exclusive Event
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

If event A can occurs or even B can occurs but not
both event A and B can occurs, then two even A and
B are said to be mutually exclusive.
This mean that if A and B are mutually exclusive
event, then if A occurs, B cannot occur and if B
occurs, A cannot occur.
For two mutually exclusive event A and B,
A B  
P  A  B   0 and
P  A  B   P  A   P B 
S
A
B
Probability of two mutually exclusive event
Example
Five graduates of equal ability apply for vacancy.
Only one applicant will be successful. The applicant
are Ali, Bakri, Chandran, David and Eng Kok. Find
the probability that
(a) Bakri will be successful
(b) Bakri or david will be successful
Solution
(a) Each applicant has an equal chance of getting the
job and selections are mutually exclusive because
only one will be chosen.
S  Ali, Bakri, Chandran, David, Eng Kok , n  S   5
1
Therefore, P  Bakri    0.2
5
(b) P(Bakri or David)  P  Bakri   P  David 
1 1 2
  
5 5 5
Probability If Three Events
The result of probability for two event
P  A  B   P  A   P B   P  A  B 
can be extended to three event as follows:
If A, B and C are three events from sample space S, then,
P  A  B  C   P  A   P B   P  C   P  A  B   P  A  C  
P B  C   P  A  B  C 

Usually, problems involving probability of three
events can be solved with the aid of Venn diagram.
S
A
B
A B
A BC
AC
BC
C
P  A  B  C   P  A   P B   P  C   P  A  B   P  A  C  
P B  C   P  A  B  C 
Tree Diagram


When a trial or an experiment of interest consist of
a sequence of several stages, it is convenient to
represent these with tree diagram.
A tree diagram highlights all the possible outcomes
and systematically works out the corresponding
probabilities.
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
Example 2.6
A coin is tossed three times. If A is the event that a
head occurs on each of the first two tossed, B is the
event that a tail occurs on the third toss and C is the
event that exactly two tails occur in the three tosses.
Show a tree diagram for this experiment and list
the possible outcomes of the sample space, S and
events A, B and C. Find the probability for events A,
B and C to occur.
Answer for exercise
1. Let A be a event that a student like to read
megazine AA and B the event that a student like to
read megazine BB.
P A  B  0.05
P(A)=0.8 P  A  B  0.1


Put these information into Venn diagram.
S
A
B
0.70 0.10
0.15
0.05
P  B   0.15  0.10
 0.25
a)Probability that the student selected like to read
megazine BB is 0.25
b)P(student like to read AA or BB but not both)
=0.70+0.15
=0.85
2.
(a)
P E  M  P E   P  M  P E  M
0.65  0.4  P  M   0.15
P  M   0.65  0.4  0.15
 0.4
(b)
P E  S   P E   P S   P E  S 
0.8  0.4  P  S   0.2
P  S   0.6
P E  M  S   P E   P M  P S   P E  M
P  E  S   P  M  S 
P  E  M  S 
0.95  0.4  0.4  0.6  0.15  0.2  0.2
P  E  M  S 
P  E  M  S   0.1
(c)
P  not a member of any society 
 1 P  E  M  S 
 1 0.95
 0.05
3.
The sample space S={3 black balls, 4 white balls, 9
yellow balls}
n(S)=16
Let A: event that black balls is chosen
B: event ball is chosen
(a) P  A   n  A   3
n S
16
(b) P  A  B   P  A   P B  because A and B are mutually exclusive 
n  A  n B 


n S n S
3
9 3



16 16 4
4
(a) Let A:Event that Aru be a champion
B:Event that Bakri be a champion
C:Event that Chan be a champion
Since only one person will be a champion, event A,B
and C are mutually exclusive. Thus
P  A  B  C   P  A   P B   P  C 
 0.2  0.3  0.1
 0.6
Probability that one of them will be a champion is 0.6


(b) P A  B  C  1 P  A  B  C 
 1 0.6
 0.4
Probability that none of them will be a champion is
0.4