Western Cape Education Department
WORKED SOLUTIONS GRADE 12 PROBABILITY
Example 1:
There are 5  4  6  120 different courses.
Example 2:
There were 20  20  20 10 10 10  8 000 000 different number plates on the old format and
20  20 10 10  20  20  16 000 000 (twice as many) using the new format.
Example 3:
8
There are 2  256 routes through the maze.
Conclusion from examples 1, 2, 3
If successive choices are made from m1 , m2 , m3 ,...mn options, then the total number of combined
options is the product: m1  m2  m3  ...  mn .
This is called the fundamental counting principle.
Example 4:
Nkosi can combine his clothes in 10  5  3  150 different ways.
Example 5:
There are 26 + 10 = 36 options for each of the 4 characters, so
The number of possible codes is 36  36  36  36  36 4  1 679 616
Example 6:
6.1.1 3 2 1  6 ways
6.1.2 4  3  2 1  24 ways 6.1.3 5  4  3 2 1  120 ways
6.1.4 m   m  1   m  2  ... 3 2 1  m !(called m factorial).
8  7  6  5  1 680
12 1110  9  8  95 040
6.2.1
10  9  8  720
6.2.4
m   m 1   m  2  ...  m  n  1 called the number of permutations of n from m:
6.2.2
6.2.3
m
Pn
Example 7:
There are 16 flags.
7.1
16 15 14 13  ...  3  2 1  16!  2.092278989 1013
(a calculator displays the best answer in the number of digits available, in scientific notation)
The actual answer is 20 922 789 888 000 which differs by 2 000 from the calculator answer!
In a number so large this difference is insignificant.
7.2
The number of permutations of a winner and a runner up is 16 15  240 .
Comment
Intuitive logic is to be recommended, even if you use your calculator to calculate the final
answer. Grasping at formulae without understanding what you are doing is not
recommended.
Example 8:
8.1
Total number of 3 digit codes = 5  4  3  60
This could also have been obtained using the 5 P3 button, but the direct method is easier!
8.2
By the fundamental counting principle: total number of 3 digit codes = 5  5  5  5 3  125
Example 9:
9.1.1 The number of arrangements is 4  3  2 1  4! 24
9.1.2 The number of arrangements is 12: LEEK, LEKE, LKEE, KEEL, KELE, KLEE, EEKL, EELK, EKEL,
EKLE, ELEK and ELKE. This is
9.1.2
4!
 12 .
2!
There are only 4 arrangements: LLLU, LLUL, LULL and ULLL. This is
4!
4
3!
1
9.1.3
Where there are 2 Ls and 2 Us, there are 6 ‘words’ that can be formed:
LULU, LUUL, LLUU, ULLU, ULUL and UULL. This number is
4!
6
2 ! 2 !
In general, the formula for the number of different ways that n items can be arranged,
where r1 of the n are identical items of one type, r2 of the n are identical items of a second
type, r3 of the n are identical items of a third type and so on… is
n!
r1 !  r2 !  r3 ! ...
Example 10
10.1 The number of ‘tunes’ is the seven different notes in order.
By the fundamental counting principle: 7  6  5  4  3  3  2  1  7 ! 5 040 . Notice that
once the first note has been ‘chosen’ from the 7 notes, there remain only 6 notes from
which to choose the second note in the ‘tune’, and then only 5 notes remain from which to
choose the third note… and so on.
10.2 The number of ‘tunes’ is 77 = 823 543. This time, there is a choice of all 7 notes
for every note in the ‘tune’. This means that doh, doh, doh, doh, doh, doh, doh is a ‘tune’, so
is doh, fah, me, re, doh, so, doh.
10.3 The number of ‘tunes’ is 1 7  7  7  7  7 1  16807 as there is only one way
of choosing the two dohs at the start and end of the ‘tune’.
10.4 There are 7 notes. Doh is repeated 3 times, me is repeated 2 times and so is
repeated 2 times, so the number of ‘tunes’ is
7!
 210 as repeating notes in a tune
3!  2!  2!
is the same as repeating letters in a word.
Example 11:
8!
 1 680
3!  2! 2!
11.1
There are 8 letters: 3 Us, 2Ns and 2Ds. So number of different words =
11.2
If we take one of the Ns for the first letter, there are 7 letters left, of which 3 are
Us and 2 are Ds. Hence the number of words =
11.3
7!
 420
3!  2 !
If we take both Ns for the first and last letters, there are 6 letters left, of which 3
are Us, and 2 are Ds. Hence the Number of words =
6!
 60
3!  2 !
Example 12:
12.1
4!  24
12.4
6!
 180
2! 2!
12.2
12.5
4!
 12
2!
9!
 45 360
2! 2! 2!
12.3
12.6
6!
 360
2!
9!
 30 240
3! 2!
Example 13:
Let the event that the word randomly generated starts with ‘H’ and ends with ‘Y’ be E. The sample
space S is all the possible permutations of the letters of the word HISTORY.
2
The number of ways the 7 letters of the word can be arranged is n(S) = 7 P7  7 !
To calculate the number of ways the event E can occur, we need to remember that the ‘H’ and ‘Y’
are fixed, so we have H _ _ _ _ _ Y. Hence n(E) = 5 P5  5!
The probability that a randomly generated word will start with ‘H’ and end with ‘Y’ is thus:
P(E)=
n( E ) 5! 1
 
 0,023...
n( S ) 7 ! 42
You may be tempted to reason as follows:
The probability that the first letter is a ‘H’ is
1
and the probability that the last letter is an ‘Y’ is also
7
1
1 1
1
, so P(E) is  
 0,020...
7
7 7 49
This is incorrect as the assumption has been made that the two elementary events of drawing an ‘H’
and then drawing a ‘Y’ are independent, but they are not because once the ‘H’ has been drawn the
probability of drawing the ‘Y’ is greater since the sample space is now smaller.
With or without a tree diagram, the solution can be calculated as follows:
P(drawing the ‘H’) =
1
1
. Once the ‘H’ has been drawn, P(drawing the Y) = , hence
7
6
1 1
1
P(E)=  
7 6 42
Example 14
In this example, letters and digits can be repeated.
14.1 Let E be the event that a number plate starts with a ‘B’ and ends with a
five. We are looking for something like this: BRR 615.
Since 20 letters and 10 digits can be used, we calculate:
n( S )  20  20  20  10  10  10  8 000 000
n( E )  1  20  20  10  10  1  40 000
 P( E ) 
40 000
n( E )
1


 0,005
n( S ) 8 000 000 200
This can also be done using the following logic:
P(choosing a ‘B’) =
1
1
and P(choosing a five) = .
20
10
Since these are independent events,
P(E) = P(choosing a ‘B’ and a five)= P(choosing a ‘B’)  P(choosing a five)
=
14.2
1
1
1
 =
 5  10 3  0,005
20 10 200.
Let the event of exactly one B being chosen be A. This B could be the first or the
second or the third letter in the number plate. In each case, the other letter must not be a
‘B’ but one of the other 19 letters.
There are 10 10 10 ways of choosing the three digits and this number ( 1 000) must be
multiplied by the number of ways that there can be exactly one ‘B’.
This B can be the first letter or the second letter or the third letter.
There is 1 way of choosing a B for the first letter and 19 ways of choosing the other two
letters: (1  19  19) .
This must be added to the number of ways of choosing a letter other than B, then a B and
then not a B 19  1 19 and the number of ways of choosing a letter other than a B twice
and then a B 19  19  1 .
3
These numbers added together must be multiplied by 10 10 10 = 1 000.
n(A) = 10 10 10119 19  19 119  19 19 1  1 083 000
So:
 P( A) 
14.3
n A 1 083 000

 0,135 375
nS  8 000 000
Let the event of at least one 5 being chosen be C
This problem is best solved using the fact that P(C) = 1 - P( C  ) where C 
(the complement of C) is the event of not getting a single five.
P(C )  1  P(C )  1 
20  20  20  9  9  9
 0,271
8 000 000
Example 15:
STATISTICS has 10 letters. It consists of 3 Ss, 3 Ts and 2 Is, an A and a C. To find out how many
possible arrangements of these letters there are, we use the formula:
n( S ) 
10 !
(10 letters of which there are 3 Ss, 3 Ts and 2 Is)
3! 3! 2 !
= 50 400
If E is the event of the word starting and ending with the same letter, then
we are looking for S_ _ _ _ _ _ _ _ S or T_ _ _ _ _ _ _ _ T or I_ _ _ _ _ _ _ _ I
Then n(E) = number of words using the letters TATISTIC (excluding the two Ss at the
start and the end of the word) + the number of words using SATISICS (excluding the
two Ts at the start and end of the word) + the number of words using the letters
STATSTCS (excluding the two Is at the start and end of the word)

8!
8!
8!
(3 Ts and 2 Is) +
(3 Ss and 2 Is) +
(3 Ss and 3 Ts)
3!  2 !
3! 2 !
3! 3!
 7 840
So the probability of the rearranged letters starting and ending with the same letter is:
 P( E ) 
7 840
 0,15
50 400
Alternative method:
P(choosing an ‘S’) 
3
because 3 of the 10 letters are Ss.
10
The probability of choosing a second ‘S’ is not independent of whether or not an ‘S’ was chosen on
the first draw:
P(choosing a second ‘S’ once it is known that the first letter drawn is an ’S’) =
2
.
9
4
Similarly, as shown in the following tree diagram, the probabilities of the successive events, of
drawing one and then another ‘T’ and one and then another ‘I’, which are not independent, can be
calculated:
So P( E ) 
3 2 3 2 2 1
.  .  .  0,15
10 9 10 9 10 9
5