Probability
Grade 11 CAPS
Mathematics
Series
Outcomes for this Topic
In this topic we will :
 Revise Grade 10 Probability.
Unit 1
 Focus on dependent and independent events.
Unit 2
 Solve probability problems using Venn Diagrams.
Unit 3
Unit 1
Revision
of Grade 10
Probability
Grade11 CAPS
Mathematics
Series
Outcomes for Unit1
In this unit we will revise :
 Terminology related to probability.
 Theoretical probability.
 The Probability Identity : P  A  B  P  A  P B   P  A  B 
 Mutually exclusive events.
 Complementary events.
What is Probability?

Probability
is the mathematics of chance that enables us to predict
the likelihood of uncertain occurrences and tells us the relative frequency
with which we expect an event to occur.
Examples :
1. The probability that it will rain in PE today is 0,3 30%.
70
 70%.
100
1
3. The probability of winning the Lotto is
 0, 000 007%.
13 983 816
2. The probability of contracting the flu virus is
Probability can be reported as a fraction, decimal, percentage or ratio.
The greater the probability, the more likely the event is to occur.
Conversely, the smaller the probability, the less likely the event
is to occur. In the examples there is a strong probability that you
may contract the virus, a reasonable chance of rain in PE today,
but almost no chance of you winning the Lotto.
Terminology linked to probability
 To study probability in a mathematically precise way, we need special
terminology and notation.
 An experiment is the act of making an observation or taking a measurement
e.g. toss a fair coin and record whether the side up is heads or tails.
 An outcome is one of the possible things that can occur as a result of an
experiment e.g. H (heads) and T (tails) are the two possible outcomes in
the coin-tossing experiment.
 T h e s e t o f all p o s s i b l e o u t c o m e s is c a l l e d t h e s a m p l e s p a c e .
In the coin-tossing experiment S 
H ;T
is t h e s a m p l e s p a c e .
A n e v e n t is a n y s u b s e t o f t h e s a m p l e s p a c e .
Tossing a single coin in an experiment
Experiment 1 : Toss a coin, and record after each trial whether
the side up is heads or tails.
Two possible outcomes H or T.
Hence the sample space S  H;T and n S   2.
There are 2 2  4 events and since an event is simply a
subset of the sample space the events in this experiment
are , H, T and H ;T.
Events in this experiment can be described in the following way:
Event
Description of event
It is not always possible to give
H
T
S  H;T
  
Getting heads
descriptions of the different events.
Getting tails
Getting either heads or tails
Getting neither heads nor tails
Roll a pair of dice in an experiment
Experiment 2 : Roll a pair six-sided dice and record, respectively,the
number and number of dots showing on the top faces of the die after each trial.
For this experiment, the sample space consist
of 36 ordered pairs of outcomes and 236 events.
The 36 ordered pairs can be illustrated graphically.
Note :
5;3 & 3;5
Are different outcomes
Many more such pairs!
Experiment: Draw a card from a standard pack of cards
Experiment 3 : A card is drawn at random from a standard deck
of playing cards.
See illustration of :
Drawing a face card,
as one of these events.
For this experiment, the sample space consist
of 52 outcomes and 252  4 503 599 627 370 496
events.
Theoretical Probability

Probability is a real number in the closed interval 0;1
that describes how likely it is that an event will occur.
A probability of 0 means that an event will never occur (impossible).
 A probability of 1 means that an event will always occur (certain).

A probability of 0,5 means that an event will occur half the time.
 Define theoretical probability of an event as:
n E 
Number
of
outcomes
in
E
P E 

Number of outcomes in S n S 
P    0 and P S   1

 P    P E   P S 
0  P E  1
 

Theoretical Probability from single coin-tossing experiment
Event
Description of event
E 1  H
Getting heads
E 2  T
E 3  S   H ;T 
E4     
 P  getting heads   P  E
Getting tails
Getting either heads or tails
Getting neither heads nor tails
  nn ES1  
 
1
1  50%
2

n E2 

1
 P  getting tails   P  E 2  
  50%
n S 
2


 P  getting either heads or tails   P  E   nnES3  2  1  100%
3

2

n E4   0  0  0 %

 P  getting neither heads nor tails   P  E 4   n  S 
2

Theoretical Probability from rolling a single die experiment
S  1; 2; 3; 4; 5; 6  n  S   6
P getting a number smaller than 3 P  getting a six 
 P E 

where E  1;2
 P E 
n E 2 1
 
n S  6 3
P getting aneven number
 P E 

n S 

1
6
P getting a number greater than six
where E  2; 4;6  P E 
n E  3 1

 
n S  6 2
n E 
where E  6

where E    
n E  0
 0
n S  6
Theoretical Probability from drawing a card experiment
 n S   52
P  drawing a diamond 

n E 
n S 
13 1


52 4
P drawing either an ace or a king
n E  8 2

 
n S  52 13
P drawing a diamond facecard

n E  3

n S  52
P  drawing a face card 

n E 
n S 

12
3

52 13
Number of Events in Sample Space
Sample
Space
S  
S  a
S  a;b
S  a;b;c
Events
Number of Events

,a
,a,b,a;b

a,b,c
a;b,a;c,b;c
a;b;c
1  2 0  2 nS 
11  2  21  2nS 
1 2 1  4  22  2nS 
1 3  3 1  8  23  2nS 

a,b,c,d
S  a;b;c; d a;b,a;c,a;d,b;c,b;d,c;d
a;b;c,a;b;d,a;c; d,b;c;d
S
1 4  6  4 1  16  2nS 
Tutorial 1: Theoretical Probabilities
Roll a pair six-sided dice and record, respectively, the number
and number of dots showing on the top faces of the die after each trial.
Determine the theoretical probability
for each of the following events:
1. Event A : The sum of the dice is equal to 7
2. Event B : The sum of the dice is greater than 5
3. Event C : At least one die must show a 4 faceup.
PAUSE
• Do Tutorial 1
• Then View Solutions
Tutorial 1: Problem 1: Suggested Solution
1.
Determine the theoretical probability for event A.
Event A : The sum of the dice is equal to 7
A  1,6  ;  2, 5  ; 3, 4  ; 4, 3; 5, 2  ; 6,1  n  A  6
P  A

n A
n S 
6
36
1

6

Tutorial 1: Problem 2: Suggested Solution
2.
Determine the theoretical probability for event B.
Event B : The sum of the dice is greater than 5
n B  26
P B

n  B
n S 
26
36
13

18

Tutorial 1: Problem 3: Suggested Solution
3.
Determine the theoretical probability for event C.
Event C : At least one die must show a 4 faceup.
n C   11
P C 

n C 
n S 
11

36
Venn diagram representation of events and sample space
Consider the following three sets of events linked to the sample
space S  outcomes when rolling a single die  1;2;3;4;5;6.
1.
A  getting an even number  2;4;6
B  getting a factor of 6  1;2;3;6
2.
A  getting an even number  2;4;6
B  getting a odd prime number  3;5
3.
A  getting an even number  2;4;6
B  getting a number  1  2;3;4;5;6
Example 1: Identity P  A  B  P  A  P B   P  A  B 
1. Consider the example where:
S  outcomes when rolling a single die and the events
A  getting even number  2;4;6 and
B  getting a factor of 6  1;2;3;6
Show that P  A  B  P  A  P B   P  A  B  holds for this example:
S  1;2;3; 4;5;6   n  S

6
A  2;4;6  n  A  3
B  1;2;3;6  n B   4
A  B  1;2;3;4;6  n  A  B   5
A  B  2;6  n  A  B   2
3 4 2 5
 P  A  P B   P  A  B       P  A  B 
6 6 6 6
P A  B  
n  A  B 5

n S 
6
P  A  n  A  3
n S  6
P B 
n B
n S 
P A  B  

4
6
n  A  B 2

n S 
6
Example 2: Identity P  A  B  P  A  P B   P  A  B 
1. Consider the example where:
S  outcomes when rolling a single die and the events
A  getting even number  2;4;6 and
B  getting an odd prime number  3;5
Show that P  A  B  P  A  P B   P  A  B  holds for this example:
S  1;2;3; 4;5;6   n  S

6
A  2;4;6  n  A  3
B  3;5  n B   2
A  B  2;3;4;5;6  n  A  B   5
A  B      n  A  B   0
3 2 0 5
 P  A  P B   P  A  B       P  A  B 
6 6 6 6
P A  B  
n  A  B 5

n S 
6
P  A  n  A  3
n S  6
P B  n B  2
n S  6
P A  B  
n  A  B 0

n S 
6
Tutorial 2: Identity P  A  B  P  A  P B   P  A  B 
Consider the example where:
S  outcomes when rolling a single die and the events
A  getting an even number  2;4;6 and
B  getting a number  1  2;3;4;5;6
1. Illustrate the events and sample space by means
of an appropriate Venn diagram.
2. Confirm that the Identity
P  A  B   P  A  P B   P  A  B 
holds for this example.
PAUSE
• Do Tutorial 2
• Then View Solutions
Tutorial 2: Problems 1 and 2: Suggested solutions
1. Appropriate Venn Diagram:
Given:
S  outcomes when rolling a single die and the events
A  getting an even number  2;4;6 and
B  getting a number  1  2;3;4;5;6
2. Show that P  A  B  P  A  P B   P  A  B  holds for this example:
S 
1; 2 ; 3 ; 4 ; 5 ; 6  
n S

6
A  2;4;6  n  A  3
B  2;3;4;5;6  n B   5
A  B  B  2;3;4;5;6  n  A  B   5
A  B  A  2;4;6  n  A  B   3
3 5 3 5
 P  A  P B   P  A  B       P  A  B 
6 6 6 6
P  A 
n  A 3

n S  6
P B  n B  5
n S  6
P A  B  
P A  B  
n  A  B
n S 

5
 P B 
6
n  A  B 3
  P  A
n S 
6
Mutually exclusive events: A  B  
Definition :
Two events are called mutually exclusive if they
cannot occur at the same time.
Alternatively two event sets, A and B, are mutually
exclusive if they cannot have any elements in common,
which implies that A  B   or P  A  B   0.
In addition: n  A  B   n  A  n B   P  A  B   P  A  P B 
A  B    n  A  B  0
Example of two mutually exclusive events :
Sample space S  outcomes when rolling a single die  1;2;3; 4;5;6
 A  getting an even number  2;4;6
Events: 
 B  getting a odd prime number  3;5
In addition: P  A  P  B  
3
6

2
6

5
6
 P A  B 
Complementary events: A  B   and A  B  S
Definition :
Two events, A and B, are complementary events
if and only if
1. A  B  and
2. A  B  S
Note :
1
2
A  A
3
1  P S   P  A  A
A  A S
 P  A  P  A P  A  A
 P  A  P  A
 A'  B  S  A
4 P  A  1  P  A
In addition:  B '  A S  B

 A and B is a partitioning of S
Example of two complementary events :
Sample space S  rolling a die  1;2;3; 4;5;6 
 A   getting an even number   2;4;6
Events: 
 B  A  getting a odd number   1;3;5
or P  A 1  P  A
Unit 2
Dependent and
Independent
Events
Grade11 CAPS
Mathematics
Series
Outcomes for Unit 2
In this Unit we will focus on :
 Identification of independent an d dependent events.
 Calculate the probability of dependent and independentevents.
 Product Rule for independent events : P  A  B  P  A P B 
P  A  B
 Conditional Probability Formula : P B A 
P  A
 Conditional Probability formula for two independent events :
P  A  B P  A   P  B 
P B A 
 P B 

P  A
P  A
Independent Events and Product Rule
Two events are said to be independent if the result of the second
event is not affected by the result of the first event.
Example 1 : A white ball, red ball and yellow ball are placed in a box.
In an experiment a ball is randomly drawn, its color recorded
and it is returned to the box. A second ball is drawn and its
color recorded. What are the probability of drawing a red ball
and then a white ball?
If A and B are independent events
Use a contingency table :
then P  A and B  P  A  B   P  A P B 
First
Draw
n S   3 3  9
W
R
Y
WW
RW
YW
WR
RR
YR
3 1

9 3
WY
RY
YY
B  white second  WW; RW ;YW  P B  
A  red first  RW; RR; RY  P  A 
C  first red then white  RW  P C  
1  P C  P A P B
     
9
3 1

9 3
Independent Events another example
Example 2 : A packet contains 3 red paperclips, 4 green paperclips
and 5 blue paperclips. One paperclip is taken from the packet, the
color recorded and then placed back in packet. Another paperclip
is taken from the packet and the color recorded.What is the probability
that the first paperclip is red and the second paperclip is blue?
Solution : Because the first paper clip is replaced, the sample space
of 12 paperclips does not change from the first to the second event.
Events are independent
 P  red then blue   P  red   P  blue 
P red then blue 
3

5
12 12

15
144

5
48
Probability for Dependent Events :Contingency Table and Rule
Example 3 : A bag contains three marbles : one red, one blue and one green.
If we draw two marbles from the bag, one after the other, without replacing
the first one drawn, what is the probability that the first marble is red and the
second green?
Solve using ContigencyTable
If a result of one event is affected by the
result of another event the events are
First
Draw
said to be dependent.
Drawing two marbles implies that sample space is
RR; RB; RG; BR; BB; BG;GR;GB; GG.
R
B
G
RR
BR
GR
RB
BB
GB
RG
BG
GG
Condition, first red, changes sample space A  first must be red  RR;RB; RG
to RR;RB; RG.
 P first red then green 
 A  B  RG
1P B A
 
3
Not Examinable in CAPS
B  second must be green  RG; BG;GG
1
P  A  B 9
1 9 1
Then

    P B A.
3
P  A
9 3 3
9
Probability of Dependent Events : Using Tree Diagrams
Example 4 : When tossing three fair coins, what is the probability of
getting two tails given that the first coin came up heads?
Solve using Tree Diagram
Tossing three coins implies that sample space is :
HHH; HHT; HTH; HTT;THH ;THT;TTH ;TTT .
Condition, first coin head, changes sample space
to HHH; HHT; HTH ; HTT .
Only one outcome, HTT , fits the event description.
Conditional probability of getting two tails
given that the first of the three coins is head is
1
.
4
1
P B A  P getting two tails once a head occured 
4
Not Examinable in CAPS
Conditional Probability
Let A and B be events in a sample space S where P  B   0.
The conditional probability that event A occurs, once event
B occured, denoted by P  A B , is defined as
P  A  B
P A B  
.
P B
If A and B are independent events, then we know that
P  A  B  P  A   P B 
Conditional probability for two independent events
can be redefined using the previous relationship :
P B A 
P  A  B
P  A

P  A P B
P  A
 P B 
Not Examinable in CAPS
Probability for Dependent Events : Using Conditional Probability Formula
Example 4 : When tossing three fair coins, what is the probability of
getting two tails given that the first coin came up heads?
Know: P B A  P getting two tails once a head occured 
1
4
Tree Diagram
Alternative Solution:
Tossing three coins implies that sample space is :
HHH ; HHT ; HTH ; HTT ;THH ;THT ;TTH ;TTT .
A  first head  HHH; HHT; HTH ; HTT 
B  two tails  HTT;THT ;TTH 
 A  B  HTT 
P B A  P getting two tails once a head occured 
P B A 
1
8

4
8

1

8
8 4

P  A  B
P  A
1
4
Not Examinable in CAPS
Dependent Events : Using Venn Diagram
Example 4 : When tossing three fair coins, what is the probability of
getting two tails given that the first coin came up heads?
Tossing three coins implies that sample space is :
S  HHH; HHT; HTH ; HTT ;THH ;THT ;TTH ;TTT .
Not Examinable in CAPS
Let A be the event of getting exactly two tails:
A  HTT;THT ;TTH .
Let B be the event of getting head with first coin:
B  HHH; HHT; HTH; HTT 
 A B  HTT 
 P A B  
1
4
Condition of head first limits us to set B (shaded) :
Only one of these four outcomes contains two tails (HTT ).
Venn Diagrams to illustrate definition of ConditionalProbability
Not Examinable in CAPS
A sample space S of equally likely outcomes is
shown in the next Venndiagram.
P A  6
n S   14
   14
P  A  B 2 7 2

P A B 

 

n
A

6



P B
14 14 7
7

 P B 

14
n B  7

Condition
2
n  A  B   2 

P  A  B   14
B shaded
P  A  B 2
6 2
P B A 

 
P A
14 14 6
Condition
A shaded
Tutorial 3: Dependent and Independent Events
A golfer can select any of four makes of golf balls from his bag.
PAUSE
• Do Tutorial 3
• Then View Solutions
We will lable the four balls as T , M , F and P.
The golfer selects randomly two balls from his bag.
What will be the probability in the following cases:
1. Two balls are drawn with replacement
and the first must be P and the second T.
Use Contingency Table and Tree Diagram.
2. Two balls are drawn without replacement
and the first must be P and the second T.
Use Venn Diagram and Conditional Formula.
Tutorial 3: Problem 1: Solution using a Contingency Table
A golfer selects two golf balls from his bag with replacement.
1. What is probability that the first is a P and the second a T ?
n S   4 4  16
A  select a Pinnacle first  PT; PM ; PF; PP
 n  A  4 and P  A 
First Ball
Selected
T
M
F
P
TT
MT
FT
PT
TM
MM
FM
PM
TF
MF
FF
PF
TP
MP
FP
PP
4 1

16 4
B  select a Titleist second  TT; MT; FT; PT
n B  4 and P B  
A and B are independent events
 P  A  B   P  A   P B 
4
16

1
4
P first a Pinacle then a Titleist 
1 1 1
 P  A  P  B    
4 4 16
Tutorial 3: Problem 1: Solution using a Tree Diagram
A golfer selects two golf balls from his bag with replacement.
1. What is probability of
firstly P and secondly T ?
n S   4 4  16
Follow blue branches:
1 1 1
 P first P and then T    
4 4 16
A   select a Pinnacle first    PT ; P M ; P F ; PP
 n  A   4 and P  A  
4
1

16 4
B   select a Titleist second   TT; MT ; FT ; PT 
 n  B   4 and P  B  
4
1 A and B are indepe ndent events

16 4  P  A  B   P  A   P  B 
P  first a Pinacle then a Titleist 
 P  A  P  B  
1
4

1
4

1
16
Tutorial 3: Problem 2: Solution using a Venn Diagram
A golfer selects two golf balls from his bag without replacement.
2.
W h a t is probability of
firstly P an d secondly T ?
S  select two balls  TT;TM ;TF;TP; MT; MM ; MF; MP; FT; FM ; FF; FP; PT; PM ; PF; PP.
Let A  select a Pinnacle first  PT; PM ; PF; PP
 A B  PT

Let B  select a Titleist second  TT; MT; FT; PT
First Ball
Selected
T
M
F
P
TT
MT
FT
PT
TM
MM
FM
PM
TF
MF
FF
PF
TP
MP
FP
PP
1
 P B A  
Condition of Pinacle first limits us to the set A (shaded).
4
Only one of these four outcomes contains a Titleist as a second ball.
Not Examinable in CAPS
Tutorial 3: Problem 2: Conditional Probability Formula
A golfer selects two golf balls from his bag without replacement.
2.
What is probability of
firstly P and secondly T ?
Know : P B A  P  select a Titleist once a Pinacle was selected  
1
4
S  select two balls  TT;TM ;TF;TP; MT; MM ; MF; MP; FT; FM ; FF; FP; PT; PM ; PF; PP.
Let A  select a Pinnacle first  PT; PM ; PF; PP
 A B  PT

Let B  select a Titleist second  TT; MT; FT; PT
P select a Titleist once a Pinacle was selected
 P B A 

1

4
16 16

P  A  B
P A
1
16

16
4

1
4
Not Examinable in CAPS
Unit 3
Solving
Probability
Problems
using
Venn
Diagrams
Grade11 CAPS
Mathematics
Series
Outcomes for Unit 3
In this unit we will:
 Revision from Grade10 :The use of Venn diagrams
to solve probability problems limited to two events.
 Use Venn diagrams to solve probability problems
for three events.
EXAMPLE 1
7.3
A school organised a c a mp for their 103 G r ad e 1 2 learners.
T h e learners wer e asked to indicate their food preferences for the camp.
T h e y h ad to choose f r om chicken  C  , vegetables  V
 and
fish
 F .
T h e following information w a s collected:
2 learners d o n ot eat chicken, fish o r vegetables
2 learners o n l y eat chick en
5 learners eat on l y vegetables
3 learners eat on l y fish
21 learners d o n ot eat fish
66 learners eat chick en a n d fish 7 5 learners eat vegetables a n d fish
Let the n u mb e r of learners w h o eat chicken, vegetables a n d fish b e x.
1.
D r a w an appropriate Ve n n diagram to represent the information.
2.
Calculate x.
3.
Calculate the probability that a learner, chosen at random:
(a) Eats only chicken an d fish, an d n o vegetables.
b
Eats any T W O of the given food choices.
7 
2 
2 
2 
Understanding a VENN DIAGRAM
C or V or F   C V  F   e  a  b  x  d  h  k
C or F   C  F   e  a  x  b  k  d
C or V   C V   e  a  x  b  h  d
V or F   V  F   k  b  x  d  h  a
C and V and F   C V  F   x
C and F   C  F   x b
C and V   C V   x  a
V and F   V  F   x d
not C   C /  k  d  h  m
only C  e
not V   V /  e  b  k  m
only V   h
only F   k
not F   F /  e  a  h  m
Representation using a VENN DIAGRAM
7.3.1 Solution:
A school organised a camp for their 103 learners
2 learners do not eat chicken or fish or vegetables
5 learners eat only vegetables
2 learners only eat chicken
3 learners eat only fish
66 learners eat chicken andfish
75 learners eat vegetables and fish
21 learners do not eat fish
Learners who eat chicken, vegetables and fish be x
7.3.2
2 12  5  x  66  x  75  x  3  2  103
165  x  103  x  62 x  62
Representation using a VENN DIAGRAM
7.3.3 Calculate the probability that a learner:
a Eats only chicken and fish, and no vegetables.
b Eats any TWO of the given food choices.
Solution:
7.3.3 a 
7.3.3 b
4
103
0,0388
4
12 13 29
  
or 0, 2816 or 28,16%
103 103 103 103
Determine Probability using Venn Diagrams if A  B  
S  1;2;3;4;5;6  n S   6
S  outcomes when rolling a single die and the events
1. Consider the example where:
A  getting even number  2;4;6 and
B  getting a factor of 6  1;2;3;6
P E   n E 
n S 
Use appropriate Venn Diagram to determine:
P  A 
1. P A
2. P B
3. P  A  B 
and P B  
6
P A  B  
5
4
6
and P  A  B  
P  A
5. P  A P not A  P A
6. P B   P not B   P B 
A  B   P A B  
6
3
and P  B
A 1;3;5  B  4;5
6
2
A  B  1;2;3;4;6
A  B  2;6
6

P  A B   P  AB
8 P  A  B P  A only
1

P
A

B




9
P
A

B

P
B
only
 
 

A  2;4;6  B  1;2;3;6
2
6
4. P  A  B 
7. P
3
and P  A B  
6
4
6
2
6
 A  B  1;3;4;5
A B 1;3;4;5
A  B 4
A B  1;3
Determine Probability using Venn Diagrams if A B  
1. Consider the example where:
S  outcomes when rolling a single die and the events
A  getting even number  2;4;6 and
B  getting an odd prime number  3;5
Use the Venn Diagram to determine the probabilities of the given events:
A  2;4;6  n  A  3
B  3;5  n B   2
S  1;2;3;4;5;6  n S   6
A  B  2;3;4;5;6  n  A  B   5
A  B      n  A  B   0
P  A
n A 3

n S 
P AB 
n AB 5

n S 
6
P A  n A  3
n S  6

nB 2
P B 

n S 

1
n S  6
A  B  S  n A  B  6P  A B   P S  
n S 
A  1;3;5  n A  3
 P  A  P B   P  A  B  
6
3 2 0 5
    P A  B 
6 6 6 6
n AB 0
P AB 

n S 
6
Determine Probabilty using Venn Diagrams if A  B
Given:
S  outcomes when rolling a single die and the events
A  getting an even number  2;4;6 and
B  getting a number  1  2;3;4;5;6
Use the Venn Diagrams to determine the probabilities of the given events:
S  1,2,3,4,5,6  n S   6
A  2;4;6  n  A  3
B  2;3;4;5;6  n B   5
A  B  B  2;3;4;5;6  n  A  B   5
A  B  A  2;4;6  n  A  B   3
3 5 3 5
 P  A  P B   P  A  B       P  A  B 
6 6 6 6
P  A 
P B 
n A
n S 
n B 


3
6
5
n S  6
n  A  B 5
P A  B  
  P B 
n S 
6
P A  B  
n  A  B 3
  P  A
n S 
6
Modelling using an appropriate Venn Diagram
A conference was attended 370 educators.
Only Biology  B  , Accounting  A and Chemistry C 
educators were in attendance. Each of the 370 were
connected to at least one of the three subjects.
For the 370 educators the following were identified:
50 were both B and A but not C educators;
70 were B but neither A or C educators;
60 were A but neither B nor C educators;
40 were both A and C but not B educators;
50 were both B and C but not A educators and
80 were C but neither A or B educators.
1. Construct an appropriate Venn Diagram.
2. Symbolize given information.
n S   n  A  B  C   370
n  A  B  C  50 n
 B  A C 70 n  A
 B C  60 n  A 
B C   40 n  A B
 C   50 n  A B
C   80
Venn Diagrams in Problem Solving
3. Place appropriate cardinal number in each region.
4. How many at the conference were A, B and C educators?
Assume that:
n A  B  C   x
n S   n  A  B  C   370
n  A  B  C  50 n
 B  A C 70 n  A
 B C  60 n  A 
B C   40 n  A B
 C   50 n  A B
C   80
60  40  50  50  70  80  x  370
350  x 370
 x  20
 20 educators were A, B and C educators.
Using Venn Diagrams to solve a Probability Problem
Use the Venn Diagram to determine the
probability of the following conference events:
n  A
170 17
5. P Educator teaches Accounting 


n S  370 37
6. P  Educator teaches both B and C  
7. P Educator teaches A and C but not B 
8. P  Educator teaches A only  
n B  C 
n S 
70
7


370 37
n A  C  B
n S 
n  A  B C
n S 

40
4


370 37
60
6

370
37
Challenge : Formulate and solve a probability problem linked
to each of the seven areas in the Ve n n Diagram.
Tutorial 4: Using Venn Diagrams to solve Probability Problems
In a survey 100 people were questioned to find out how many
read the newspapers A, B and C. The survey revealed that 10
did not read any of the three newspapers; 50 read A; 44 read B;
20 read A and B; 9 read A and C; 6 read B and C and 5 read all
three papers.
1. Construct an appropriate Venn Diagram.
2. Symbolize given information.
PAUSE
• Do Tutorial 4
• Then View Solutions
3. Place the appropriate cardinal number in each of the eight regions.
4. Determine how many read newspaper C only?
5. Determine the probability represented by each of the eight regions.
Tutorial 4: Problems 1 and 2: Suggested Solutions
In a survey 100 people were questioned to find out how many
read the newspapers A, B and C. The survey revealed that 10
did not read any of the three newspapers; 50 read A; 44 read B;
20 read A and B; 9 read A and C; 6 read B and C and 5 read all
three papers.
1.
C o n s t r u c t a n a p p r o p r i a t e Ve n n D i a g r a m .
n S   100
2.
S y m b o l i z e g i v e n i n f o r ma t i o n .
n A  B  C   10


n  A  50
n B  44
n  A  B   20
n A  C   9
n B  C   6
n A  B  C   5
Tutorial 4: Problems 3 and 4: Suggested Solutions
3. Place the appropriate cardinal number in each of the eight regions.
4. Determine how many read newspaper C only?
n S   100
Assume that n  A B C   x


n A  B  C   10
n  A  50
n B  44
n  A  B   20
n A  C   9
15
26
n B  C   6
23
n A  B  C   5
5
4
1
26  4  5 15  23 1 x 10  100
x
10
84  x  100  x  16
16 read C only.
Tutorial 4: Problem 5: Suggested Solution
5.
Determin e the probability for each of the eight regions.
5
1

100 20
3

Region 2 : P  A  B  C 
20
Region 1 : P  A  B  C  
Region 3 : P  A  B C  
Region 4 : P  A  B C 
1
25
13
50
1

Region 5 : P  A B  C  
Region 8 :
100
P  A B C
23


R egion 6 : P  A  B  C  
 P  A  B  C 
100

10
1


100 10

Region 7 : P  A B C  
4
25
End of the Topic Slides on Probability
REMEMBER!
• Consult text-books and past papers and memos for
additional examples.
• Attempt as many as possible other similar examples
on your own.
• Compare your methods with those that were
discussed in these Topic slides.
• Repeat this procedure until you are confident.
• Do not forget: Practice makes perfect!