Chapter 7 Form 5 Probability Tanpa Jawapan Try This

CHAPTER 7 Probability ADDITIONAL MATHEMATICS FORM 5 MODUL SKOR A

Date Checked by Good Well Done Keep Trying Do Correction | 1

ANALYSIS OF THE SPM PAPERS 2004 - 2011 Additional Mathematics Paper 1 & 2

TOPIC NUMBER OF QUESTIONS

2004 2005 2006 2007 2008 2009 2010 2011

FORM 5 (Paper 1)

07 Probability 1 1 1 0 1 0 1 1

FORM 5 (Paper 2)

06 Probability 0 0 0 0 0 0 0 0

ADDITONAL MATHEMATICS NOTE

1 Probability of an Event

Sample space, S is a set of all possible outcomes of an experiment.

Example :

(a) When a dice is rolled, all possible outcomes are 1, 2, 3, 4, 5 and 6

S = {1, 2, 3, 4, 5, 6}

n(S) = 6

(b) When a coin is tossed, all possible outcomes are Head H and Tail T

S = {H, T}

n(S) = 2

Event is the set of outcomes which satisfy certain conditions.

Example :

(a) A is the event that an odd number is obtained when a dice is

rolled.

A = {1, 3, 5}

n(A) = 3

(b) B is the event that an even number is obtained when a dice is

rolled.

B = {2, 4, 6}

n(B) = 3

CHAPTER

7 Probability



The probability of an event A happens, P(A) is given by

P(A) = )(

)(

Sn

An

where n(A) = number of outcomes of event A,

n(S) = number of possible outcomes of sample space and

0 P(A) 1

If P(A) = 1, then event A is sure to happen.

If P(A) = 0, then event A will never happen.

If A is the complement of event A, then n(A) = n(S) n(A)

Dividing by n(S), gives

)(

)(

)(

)(

)(

)'(

Sn

An

Sn

Sn

Sn

An

P(A) = 1 P(A)

Therefore, the probability of the complement of event A is

P(A) = 1 P(A)

Example :

A fair dice is rolled. Calculate the probability of obtaining an odd

number.

Solution

When a dice is rolled, all possible outcomes are 1, 2, 3, 4, 5 and 6

Sample space, S = {1, 2, 3, 4, 5, 6}

n(S) = 6

If A is the event of obtaining odd number

A = {1, 3, 5}

n(A) = 3

Therefore, the probability of obtaining an odd number, P(A) = )(

)(

Sn

An

= 6

3

= 2

1



A box contains 6 blue pens, 2 red pens and 7 black pens. A pen is drawn

at random from the box. Calculate the probability of getting a red pen.

Strategy

Determine the number of possible outcomes of sample space, n(S) and

number of outcomes of getting a red pen.

Use formulae P(A) = )(

)(

Sn

An to calculate the probability of getting a red pen.

Solution :

Number of possible outcomes of sample space, n(S)

= Total number of pens

=

Let R = event of getting a red pen

R = {all red pens}

n(R) =

Therefore, the probability of getting a red pen, P(R) = )(

)(

Sn

Rn

=

Try This 3

A box contains 5 red marbles and k green marbles. If a marble is picked at

random from the box, the probability of obtaining a red marble is 4

1. Find

the value of k.

Strategy

Determine the number of possible outcomes of sample space, n(S) and

number of outcomes of getting a red marble.

Use formulae P(A) = )(

)(

Sn

An to find the value of k.

1 TRY THIS

2 TRY THIS



Solution :

Number of possible outcomes of sample space, n(S)

= Total number of marbles

=

Let R = event of getting a red marble

R = {all red marbles}

n(R) =

Given, the probability of obtaining a red marble is 4

1

Therefore, the probability of getting a red marble,

P(R) = )(

)(

Sn

Rn

4

1 =

k5

5

4

1(5 + k) = 5

4

5 +

4

1k = 5

4

1k = 5

4

5

4

1k =

4

15

k = 4

15

4

1

k = 15



2 Probability of Two Events

For two events A and B which are not mutually exclusive (tidak saling

eksklusif),

(a) the union of two events, A and B (the event of A or B happens)

are written as A B.

(b) the intersection of two events, A and B (the event of A and B

happens) are written as A B.

A B means A or B or both

A B means A and B occur together

For two events A and B which are not mutually exclusive,

P(A B) = P(A) + P(B) P(A B)

Example :

An experiment is carried out by randomly choosing a card from a set

of card numbered 1 to 8, as shown in the diagram. Calculate the

probability that the card chosen has a number less than 5 or an even

number.

A B

S

A B

S

1 2 3 4

5 6 7 8



Solution

Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8}

n(S) = 8

Let A = event of card chosen has a number less than 5

A = {1, 2, 3, 4}

n(A) = 4

Let B = event of card chosen has an even number

B = {2, 4, 6, 8}

n(B) = 4

A B = {1, 2, 3, 4, 6, 8}

n(A B) = 6

A B = {2, 4}

n(A B) = 2

The probability that the card chosen has a number less than 5 or an

even number is given by P(A B)

Method 1 (Set)

P(A B) = )(

)(

Sn

BAn

= 8

6

= 4

3

Method 2 (Formula)

P(A B) = P(A) + P(B) P(A B)

= )(

)(

Sn

An +

)(

)(

Sn

Bn

)(

)(

Sn

BAn

= 8

4 +

8

4

8

2

= 4

3



A sample space of an experiment is given by S = {a, b, c, d, e, f, g, h, I, j}.

Events A and B are defined as follows :

A = { a, b, c, d}

B = {c, d, e, f, g, h}

Find (a) P(A)

(b) P(B)

(c) P(A or B)

(d) P(A and B)

Solution

(a) S = {a, b, c, d, e, f, g, h, I, j}

n(S) =

A = {a, b, c, d}

n(A) =

P(A) = )(

)(

Sn

An

= 10

4

= 5

2

(b) S = {a, b, c, d, e, f, g, h, I, j}

n(S) =

B = {c, d, e, f, g, h}

n(B) =

P(B) = )(

)(

Sn

Bn

= 10

6

= 5

3

1 TRY THIS



(c) P(A or B) = P(A B)

A = { a, b, c, d}

B = {c, d, e, f, g, h}

A B =

n(A B) =

A B =

n(A B) =

P(A B) = )(

)(

Sn

BAn or P(A B) = P(A) + P(B) P(A B)

= 10

8 =

)(

)(

Sn

An +

)(

)(

Sn

Bn

)(

)(

Sn

BAn

= 5

4 =

5

2 +

5

3

10

2

= 5

4

(d) P(A and B) = P(A B)

A = { a, b, c, d}

B = {c, d, e, f, g, h}

A B =

n(A B) =

P(A B) = )(

)(

Sn

BAn

= 10

2

= 5

1



3 Probability of Mutually Exclusive Events

Two events, A and B are said to be mutually exclusive (saling

eksklusif) if they cannot happen at the same time.

In general,

Two events A and B are mutually exclusive if and only if A B = or

A B = { }

P(A B) = 0

Therefore, for two events A and B which are mutually exclusive,

P(A B) = P(A) + P(B)

Example :

A number is chosen at random from a set S = {x : 1 x 20, x is an

integer }. What is the probability of getting an odd number or

multiples of 4?

Strategy

Determine the number of possible outcomes of sample space, n(S),

number of outcomes of getting an odd number [ Let n(A) ], number

of outcomes of getting multiples of 4 [ Let n(B) ], number of outcomes

of getting odd number and multiples of 4 [ Let n(A B) ].

If n(A B) = 0 (mutually exclusive event) then use

P(A B) = P(A) + P(B)

Solution

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

n(S) = 20

Let A = event of getting an odd number

A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

n(A) = 10

A B

S



Let B = event of getting multiples of 4

B = {4, 8, 12, 16, 20}

n(B) = 5

A B = { }

n(A B) = 0

Since A B = { }, therefore A and B are mutually exclusive events.

The probability of getting an odd number or multiples of 4 is given

by

P(A B) = P(A) + P(B)

= )(

)(

Sn

An +

)(

)(

Sn

Bn

= 20

10 +

20

5

= 4

3

Given that A and B are two mutually exclusive events where P(A) = 3

1 and

P(B) = 5

2. Find (a) P(A)

(b) P(B) (c) P(A B)

(d) P(A B)

Solution

(a) P(A) = 1 P(A)

= 1 3

1

= 3

2

(b) P(B) = 1 P(B)

= 1 5

2

= 5

3

1 TRY THIS



(c) P(A B) =

(d) Since A and B are mutually exclusive events

P(A B) = P(A) + P(B)

= 3

1 +

5

2

= 15

11

A and B are two events in which P(A B) = 0.8 and P(A) = 0.55. Find P(B) if

A and B are mutually exclusive.

Solution

Since A and B are mutually exclusive events

P(A B) = P(A) + P(B)

0.8 = 0.55 + P(B)

P(B) = 0.25

A bag contains 6 blue chips, 5 green chips and 4 yellow chips. A chip is

selected at random from the bag. Find the probability of selecting

(a) a blue chip

(b) a green chip

(c) either a blue chip or a green chip

Solution

(a) Number of possible outcomes of sample space, n(S) =

Let B = event of selecting a blue chip

B = {all blue chips}

n(B) =

The probability of selecting a blue chip, P(B) = )(

)(

Sn

Bn

= 15

6

= 5

2

2 TRY THIS

3 TRY THIS



(b) Number of possible outcomes of sample space, n(S) =

Let G = event of selecting a green chip

G = {all green chips}

n(G) =

The probability of selecting a green chip, P(G) = )(

)(

Sn

Gn

= 15

5

= 3

1

(c) The probability of selecting either a blue chip or a green chip is given

by P(B G)

Since we cannot get a blue chip and a green chip at the same time

when a chip is selected at random from the bag, therefore the event

B and G are mutually exclusive events.

P(B G) = P(B) + P(G)

= 5

2 +

3

1

= 15

11

The probability that Zhao Ming will win the 100m butterfly swimming

competition is 5

2 while the probability that Muthu will win is

10

1. Find the

probability that

(a) either Zhao Ming or Muthu wins

(b) someone else wins

3 TRY THIS



Solution

(a) Let Z = event that Zhao Ming will win the 100m butterfly swimming

competition

P(Z) =

Let M = event that Muthu will win the 100m butterfly swimming

competition

P(M) =

The probability that either Zhao Ming or Muthu wins is given by

P(Z M)

Since Zhao Ming and Muthu cannot win at the same time in the 100m

butterfly swimming competition, therefore the event Z and M are

mutually exclusive events.

P(Z M) = P(Z) + P(M)

= 5

2 +

10

1

= 2

1

(b) Someone else wins means there is another person who win in the

competition (beside Zhao Ming and Muthu)

OR

Someone else wins means either Zhao Ming or Muthu do not win

(Z U M)

The probability that someone else wins is given by :

1 Probability of either Zhao Ming or Muthu wins

1 P(Z M) = 1 2

1

= 2

1

Z M

S



The probabilities of three teams P, Q and R winning a basketball

championship are 4

1,

6

1 and

12

1. Calculate the probability that

(a) either P or Q wins

(b) neither Q nor R wins

Solution

(a) Let P = event that team P winning a basketball championship

P(P) =

Let Q = event that team Q winning a basketball championship

P(Q) =

Let R = event that team R winning a basketball championship

P(R) =

The probability that either P or Q wins is given by P(P Q)

Since P and Q cannot win at the same time in the basketball

championship, therefore the event P and Q are mutually exclusive

events.

P(P Q) = P(P) + P(Q)

= 4

1 +

6

1

= 12

5

(b) neither Q nor R wins means either Q or R do not win (Q R)

Let Q = event that team Q winning a basketball championship

P(Q) =

Let R = event that team R winning a basketball championship

P(R) =

4 TRY THIS



Probability that either Q or R win = P(Q R)

= P(Q) + P(R)

= 6

1 +

12

1

= 4

1

Probability that neither Q nor R wins (Probability either Q or R do not

win) is given by :

1 Probability either Q or R wins = 1 P(Q R)

= 1 4

1

= 4

3

A bag contains 3 whites balls, 4 blacks balls and 5 red balls. A ball is drawn

from the bag. Find the probability that the ball is

(a) either black or red

(b) neither white nor black

Solution

(a) Number of possible outcomes of sample space, n(S) =

Let W = event that the ball drawn is white

n(W) =

Let B = event that the ball drawn is black

n(B) =

Let R = event that the ball drawn is red

n(R) =

The probability that the ball is either black or red is given by P(B R)

Q R

S

5 TRY THIS



Since we cannot get a black ball and a red ball at the same time

when a ball is drawn from the bag, therefore the event B and R are

mutually exclusive events.

P(B R) = P(B) + P(R)

= )(

)(

Sn

Bn +

)(

)(

Sn

Rn

= 12

4 +

12

5

= 4

3

(b) neither white nor black ball is drawn means either white or black is not

drawn (W B)

Number of possible outcomes of sample space, n(S) =

Let W = event that the ball drawn is white

n(W) =

Let B = event that the ball drawn is black

n(B) =

The probability that the ball is either white or black is given by

P(W B) = P(W) + P(B)

= )(

)(

Sn

Wn +

)(

)(

Sn

Bn

= 12

3 +

12

4

= 12

7

The probability that the ball is neither white nor black (The probability that

the ball either white or black is not drawn) is given by :

1 Probability that the ball is white or black = 1 P(A B)

= 1 12

7

= 12

5



4 Probability of Independent Events

Two events A and B are said to be independent events (peristiwa tak

bersandar) if the outcomes of event A are not influenced (tidak

dipengaruhi) by the outcomes of event B and vice versa (sebaliknya).

In general,

If A and B are independent, then

P(A B) = P(A) x P(B)

Given A, B and C are three independent events, the probability of

event A, B and C occurring (berlaku) is

P(A B C) = P(A) x P(B) x P(C)

Problems involving two or three independent events can also be

solved by using a tree diagram

Example 1 :

Box Red Balls Yellow Balls

A 4 5

B 7 3

The table shows two boxes A and B which contained red and yellow

balls. A ball is drawn at random from each box, find the probability

that

(a) a red ball is drawn from box A and a yellow ball is drawn from

box B.

(b) both balls drawn are yellow.

(c) both balls drawn are of different colour.



Solution

(a) Number of possible outcomes of sample space for box A,

n(S) = 4 + 5

= 9

Let R = event of a red ball is drawn from box A

n(R) = 4

Number of possible outcomes of sample space for box B,

n(S) = 7 + 3

= 10

Let Y = event of a yellow ball is drawn from box B

n(Y) = 3

Probability that a red ball is drawn from box A,

P(R) = )(

)(

Sn

Rn

= 9

4

Probability that a yellow ball is drawn from box B,

P(Y) = )(

)(

Sn

Yn

= 10

3

The outcomes of event A is not influenced by the outcomes of

event B, therefore A and B are independent events.

The probability that a red ball is drawn from box A and a yellow

ball is drawn from box B,

P(R Y) = P(R) x P(Y)

= 9

4 x

10

3

= 15

2



Tree Diagram

Box A Box B Outcomes Probability

10

7 R RR

9

4 x

10

7 =

45

14

9

4 R

10

3 Y RY

9

4 x

10

3 =

15

2

10

7 R YR

9

5 x

10

7 =

18

7

9

5 Y

10

3 Y YY

9

5 x

10

3 =

6

1

(b) The probability that both balls drawn are yellow,

P(Y Y) = 9

5 x

10

3 =

6

1

(c) The probability that both balls drawn are of different colour,

P[ (R Y) P(Y R) ] = 9

4 x

10

3 +

9

5 x

10

7

= 15

2 +

18

7

= 90

47

Example 2 :

The probability that Amin and Zaki win in a table tennis match are 3

1

and 4

3 respectively. Find the probability that

(a) both of them win (b) either one of them win (kedua-duanya menang) (salah seorang antara mereka

menang)



Solution

Let A = event that Amin win in a table tennis match

P(A) = 3

1

Let Z = event that Zaki win in a table tennis match

P(Z) = 4

3

(a) The probability that both of them win = P(Amin and Zaki win)

P(A Z) = P(A) x P(Z)

= 3

1 x

4

3

= 4

1

(b) The probability that either one of them win

= P(Amin win and Zaki does not win) +

P(Amin does not win and Zaki win)

Let A = event that Amin win in a table tennis match

P(A) = 3

1

A = event that Amin does not win in a table tennis match

P(A) = 1 3

1 =

3

2

Let Z = event that Zaki win in a table tennis match

P(Z) = 4

3

Z = event that Zaki does not win in a table tennis match

P(Z) = 1 4

3 =

4

1

The probability that either one of them win

= P(Amin win and Zaki does not win) +

P(Amin does not win and Zaki win)

= P(A Z) + P(A Z) = P(A) x (P(Z) + P(A) x P(Z)



= 3

1 x

4

1 +

3

2 x

4

3

= 12

1 +

2

1

= 12

7

Example 3 :

The probability that Selina is chosen as a school traffic warden is 3

1

while the probability that Anne is chosen is 12

7. Find the probability

that

(a) neither of them is chosen as a school traffic warden

(kedua-dua mereka tidak dipilih sebagai warden trafik sekolah)

(b) only one of them is chosen as a school traffic warden

(hanya seorang daripada mereka dipilih sebagai warden trafik

sekolah)

Solution

(a) The probability that neither of them is chosen as a school traffic

warden

= P(Selina is not chosen and Anne is not chosen)

Let S = event that Selina is chosen as a school traffic warden

P(S) = 3

1

S = event that Selina is not chosen as a school traffic warden

P(S) = 1 3

1 =

3

2

Let A = event that Anne is chosen as a school traffic warden

P(A) = 12

7

A = event that Anne is not chosen as a school traffic warden

P(A) = 1 12

7 =

12

5



The probability that neither of them is chosen as a school traffic

warden

= P(Selina is not chosen and Anne is not chosen)

= P(S A) = P(S) x P(A)

= 3

2 x

12

5

= 18

5

(b) The probability that only one of them is chosen as a school

traffic warden

= P(Selina is chosen and Anne is not chosen) +

P(Selina is not chosen and Anne is chosen)

Let S = event that Selina is chosen as a school traffic warden

P(S) = 3

1

S = event that Selina is not chosen as a school traffic warden

P(S) = 1 3

1 =

3

2

Let A = event that Anne is chosen as a school traffic warden

P(A) = 12

7

A = event that Anne is not chosen as a school traffic warden

P(A) = 1 12

7 =

12

5

The probability that only one of them is chosen as a school

traffic warden

= P(Selina is chosen and Anne is not chosen) +

P(Selina is not chosen and Anne is chosen)

= P(S A) + P(S A) = P(S) x (P(A) + P(S) x P(A)

= 3

1 x

12

5 +

3

2 x

12

7

= 36

5 +

18

7

= 36

19



SPM 2004 | Soalan 1 Probability of an event

SPM 2005 | Soalan 2 Probability of Independent Events

Ans : k = 9

Ans : k = 19/66

SPM 2015 SPM PRACTICE (PAPER 1)





Ans : (a) 2/15 (b) 7/15

Ans : (a) 1/6 (b) 29/60



SPM 2010 | Soalan 5 Probability of Mutually Exclusive Events

(a) (b)

SPM 2011 | Soalan 6 Probability of Two Events

(a) (b)

Ans : (a) 1/15 (b) 3/5

Ans : (a) 3/10 (b) 1/10



SPM CLONE 2004 | Soalan 1

A box contains 8 red marbles and k blue marbles. If a marbles is picked

randomly from the box , the probability of getting a blue marble is 9

5. Find

the value of k.

Ans : k = 10


There are 6 red chips, 5 green chips and 4 blue chips in a bag. Two chips

are drawn at random from the bag, one after the other without

replacement. Calculate the probability that both chips are of the same

colour.

Ans : 31/105

SPM CLONE 2006/ 2008 | Soalan 3

The probability for Azean , Dalilah and Nurul to qualify for the final of a

singing contest are 3

1,

5

3 and

7

3 respectively. Calculate the probability that

only one of them will qualify.

Ans : 44/105

SPM 2015 SPM CLONE (PAPER 1)




Sebuah kotak mengandungi 5 biji bola kuning dan n biji bola merah. Jika

sebiji bola dipilih secara rawak daripada kotak, kebarangkalian mendapat

bola merah ialah 8

n. Cari nilai n.

Ans : n = 3


Sebuah kotak mengandungi 4 keping kad. Kad itu tercatat dengan

nombor 1, 2, 3 dan 4 masing-masing. 2 keping kad dipilih secara rawak

daripada kotak. Cari kebarangkalian bahawa kedua-dua nombor itu ialah

nombor perdana.

Ans : 1/6 SPM CLONE 2008 | Soalan 6

Kebarangkalian Siti akan menang kedudukan pertama dalam larian 100m

ialah 5

2 manakala kebarangkalian Fatimah akan menang kedudukan

pertama ialah 4

1. Cari kebarangkalian bahawa

(a) kedua-dua orang gagal memenangi tempat pertama.

(b) hanya salah seorang menang tempat pertama.

Ans : (a) 9/20 (b) 9/20




Dalam sebuah beg terdapat beberapa biji bola pingpong yang berlainan

warna. Sebiji bola pingpong dipilih secara rawak daripada beg,

kebarangkalian mendapat bola pingpong hijau ialah 4

1 dan

kebarangkalian mendapat bola pingpong hijau atau kuning ialah 3

2. Cari

kebarangkalian bahawa

(a) bola pingpong kuning dipilih.

(b) bola pingpong hijau atau kuning tidak terpilih.

Ans : (a) 5/12 (b) 1/3


Ans : (a) 3/14 (b) 17/28




Ans : 83/253


Ans : (a) 1/15 (b) 2/5




Ans : (a) 11/20 (b) 1/4

SPM 2012 | Soalan 12

Ans : (a) 1/19 (b) 15/38



SPM 2013 | Soalan 13

Ans : (a) 5/8 (b) 1/3

Chapter 7 Form 5 Probability Tanpa Jawapan Try This

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