probability 2 -new 2011

8/3/2019 Probability 2 -New 2011

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Cha

pter

7PROBABILITY II

LEARNING OBJECTIVES

Understand and use the concept of probability

of an event.

LEARNING OBJECTIVESChapter

7

Understand and use the concept of probability

of the complement of an event.

Understand and use the concept of probabilityof combined event.

PROBABILITY II


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KEY TERMS

Probability Kebarangkalian Sample space Ruang sampel Equally likely outcomes Kesudahan sama boleh jadi

Expected number of outcomes Jangkaan bilangan kesudahan Complementary event Peristiwa pelengkap Combined event Peristiwa bergabung Intersection Persilangan Union Kesatuan Tree diagram Gambar rajah pokok


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7.1 THE PROBABILITY OF AN EVENT

The outcomes of an event with the same probability to

occur are known as equally likely outcomes

EXPERIMENT Tossing a fair coin

Equally likely outcomes : { heads, tails }

The possibilities of obtaining the pictorial face (heads)and the numerical face (tails) of a fair coin are equal

P (heads) = P (tails) =

7.1 a Determining the probability of an Event


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7.1 a Determining the sample space of an experiment with equally

likely outcomes

EXAMPLE

A spinner contains the letters B, U, Rand N. If James spinsthe spinner, the probability of getting any of the letters B, U, RorN

is the same. List the sample space.

Solution

Sample space, S = { B, U, R, N}

B

U

R

N

Sample space is the set of all possible outcomes of an experiment


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7.1 b Determining the probability of an event with equiprobable

sample space

EXAMPLE 1 24 366 169

Mary puts the above six cards in a box. If Mary picks a card randomly

from the box, find the probability of obtaining

(a) an odd number

(b) a prime number(c) a number less than 15

(d) a positive number


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1 24 366 169

Sample space, S = { 1, 6, 9, 16, 24, 36 }n (S) = 6

J = the event of picking an odd number(a)

J = { 1, 9 }

n (J) = 2

P (J) = n (J) = 2 = 1

n (S) 6 3


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1 24 366 169

Sample space, S = { 1, 6, 9, 16, 24, 36 }n (S) = 6

K = the event of picking a prime number(b)

K = { }

n (K) = 0

P (K) = n (K) = 0 = 0

n (S) 6

It means the event of picking a prime

number will not happen


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1 24 366 169

Sample space, S = { 1, 6, 9, 16, 24, 36 }n (S) = 6

L = the event of picking a number less than 15(b)

L = { 1, 6, 9 }

n (L) = 3

P (L) = n (L) = 3= 1

n (S) 6 2


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1 24 366 169

Sample space, S = { 1, 6, 9, 16, 24, 36 }n (S) = 6

M = the event of picking a positive number(d)

M = { 1, 6, 9, 16, 24, 36 }

n (M) = 6

P (M) = n (M) = 6 = 1

n (S) 6

It means the event of picking a positive

number is sure to happen


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7.1 c Solving problems involving probability of an event

EXAM

PLE

A factory produces light bulbs for the domestic market. The quality

control will randomly pick 2500 bulbs daily to check on the quality.

On the average, 75 units of them are faulty. Find the probability thatany one bulb picked is faulty.

A = the event of picking a faulty bulb

n (A) = 75

P (A) = n (A) = 75 = 3

n (S) 2500 100


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7.1THE PROBABILITY FOR THE COMPLEMENT OF

AN EVENT

The complement of an event A is the set ofall outcomes in the sample space that are

not included in the outcomes of event A.

We write as A


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AN EVENT

7.2 a Expressing the complement of an event in wordsand set notation

A bag contains cards where each is labelled with letter from the word

PROBABILITY. One card is picked from the bag. State the complement of

(a)C, where C is the event of picking a consonant,

(b) V, where V is the event of picking a vowel I ,

in words and I set notation.P

Example


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AN EVENT


Sample space, S = {P,R,O,B,A,B,I,L,I,T,Y}

(a) C = {P,R,B,B,L,T, Y}

The complement of C, C is the set of vowels in the word PROBABILITY.

C = {O,A,I,I}

in words

in set notations


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AN EVENT


Sample space, S = {P,R,O,B,A,B,I,L,I,T,Y}

(b) V = {I,I}

The complement of V, V is the set of all letters

in the word PROBABILITY except the vowel I.

v = {P,R,O,B,A,B,L,T,Y}

in words

in set notations


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AN EVENT

7.2 b Determining the probability of the complement of an event

1. If P(A) is the probability that event A occurs, then P(A) is the

probability that event A does not occur

2. It can be shown that the relationship between P(A) and P(A) is

P(A) = 1 - P(A)


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AN EVENT


EXAM

PLE

A dice is thrown.

S = {1,2,3,4,5,6}. n (S) = 6

A = event of getting the number 6. n (A) = 1Thus, P(A) =

n (A)

n (S)

1

6=

A = event of not getting the number 6.

A = {1,2,3,4,5}. n (A) = 5

Thus, P(A) = n (A)n (S)

56

=

It can be seen that P(A) = 1 P (A)


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AN EVENT


EXAMPLE

A bag contains balls of different colours. The probability of

drawing a red ball from the bag is 5 .12

(a) Find the probability of drawing a ball which is not red.

(b) If there are 14 balls which are not red, find the total

number of the balls in the bag.


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AN EVENT


solution

(a) A = drawing a red ball

A = drawing a ball which is not red

P(A) =5

12

Thus, P(A) =1 - 5

12=

7

12

(b) n (A) = 14. P(A) =n (A)

n (S)

7

12 =14

n (S)

n (S) = 14 x 12

7= 24


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AN EVENT


EXAM

PLE

Two fair dice are rolled. Find the probability that

(a)both dice show prime numbers,

(b)not both dice show prime numbers(c) both dice do not show prime numbers


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AN EVENT


solution

The sample space can be represented in the graph below.

X X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X X

X

X

X

X

X

X

1 2 3 4 5 6

1

2

3

4

5

6

Dice x

Dicey

= A

(a) n (S) = 36

A = event of both dice showing

prime numbers.

n (A) = 9

Thus, P(A) =n (A)

n (S)

9

36=

= 14


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AN EVENT


solution


X X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X X

X

X

X

X

X

X

1 2 3 4 5 6

1

2

3

4

5

6

Dice x

Dicey

= A

(b) A = event of not both dice

showing prime numbers.

Thus, P(A) =

OR P(A) =1

4

=3

4

1 -

27

36

=3

4


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AN EVENT


solution


X X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X X

X

X

X

X

X

X

1 2 3 4 5 6

1

2

3

4

5

6

Dice x

Dicey

= B

(c) B = event of both dice notshowing prime numbers.

P(B) =n (B)

n (S)

9

36=

=1

4


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7.3 THE PROBABILITY FOR THE COMBINED EVENT

There are two types of combinations, namely combination ofeventAoreventB and a combination ofeventA andB.

When we roll a fair dice, the possible outcomes are getting either 1,2,3,4,5 or 6.

Each of these outcomes is equally likely to happen.

Therefore, the sample space,

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

If A is the event of getting an even number,

then, A = { 2,4,6 }

If B is the event of getting a prime number,

then, B = { 2,3,5 }

This can be represented by using a Venn

diagram

S

.1

.6

.3.4

.5.2

A B


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If X is the event of getting anevennumber or a primenumber, then, X is acombination ofeventA oreventB.

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

Hence, X = A U B.

The shaded region, setX istheunion ofset

A andsetB.

This can be represented by using the Venn

diagram as shown.

X = { 2,3,4,5,6 },

S

.1

.6

.4

A B

.3

.5.2


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If Y is the event of getting anevennumber anda primenumber, then, Y is acombination ofeventA andeventB.

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

Therefore, the events X and Y are called combinedevents.

Hence, Y = A B.

The shaded region, setY isthe intersection

ofsetA andsetB.

This can be represented by using the Venn

diagram as shown.

Y = { 2 },

S

.1

.6

.4

A B

.3

.5.2


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7.3 THE PROBABILITY FOR THE COMBINED EVENT7.3 a Listing the outcomes for events (a) A or B ( AU B ) (b) A and B

( A B )

Experiment : Roll a fair dice and toss a fair coin. List all the possible outcomes.

1. Complete the table

Single

dice1 2 3 4 5 6

CoinHead (H)

Tails (T)


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( A B )


2. Let A be the event of getting a Heads.

(H,1) (H,2) (H,3) (H,4) (H,5) (H,6)

(H,1) (H,3) (H,5) (T,1) (T,3) (T,5)

A = { , , , , , }

Let B be the event of getting an odd number.

B = { , , , , , }


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( A B )


3. List the outcomes of obtaining a Heads or an odd number.( A or B )

(H,1) (H,2) (H,3)

(H,1) (H,3) (H,5)

(T,1) (T,3) (T,5) A U B = { , , , , , }

4. List the outcomes of obtaining a Heads and odd number . ( A and B )

A B = { , , }


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H

T

H

T

H

T

H

T

H

T

H

T

5. Based on the outcomes obtained, complete the following diagram.

Dice Coin

(2,H)

Outcomes

(2,T)

(3,H)

(3,T)

(4,H)

(4,T)

(5,H)

(5,T)

(6,H)

(6,T)

(1,H)

(1,T)

Tree Diagram


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7.3 THE PROBABILITY FOR THE COMBINED EVENT7.3 b Finding the probability by listing the outcomes of

the combined event

Steps in finding the probability of combined events

1. Find the sample space,S

, and n (S

)2. List all the outcomes of event A U B or A B and n (A U B) or n (A B)

3. Find the probability using the formula

(a)P(A or B) = P(A U B) = n (A U B)

n (S

)(a)P(A and B) = P(A B) = n (A B)

n (S)

1. If A B = ,

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

T I P sT I P s

2. If A B ,P(A or B)

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


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7.3 b Finding the probability by listing the outcomes ofthe combined event

EXA

MPLE

Box M contains three cards labelled 1,2 and 3. Box N contains a red

button and a blue button. If one randomly takes out a card from box M

and a button from box N, find the probability of obtaining

(a) an even number or a blue button.

(b) an even number and a blue button


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the combined event

solution

Sample space, S = { (1,R), (2,R), (3,R), (1,B), (2,B), (3, B) }

n (S) = 6

A = { (2,R), (2,B) }

Let A = event of getting an even number

Let B = event of getting a blue button

B = { (1,B), (2,B), (3,B) }

(a) The combined event of getting an even number or blue button

A U B = { (2,R), (2,B), (1,B), (3, B) } n (A U B) = 4

So, P(A U B) = n (A U B) = 4 = 2

n (S) 6 3


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the combined event

solutio

n

Sample space, S = { (1,R), (2,R), (3,R), (1,B), (2,B), (3, B) }

n (S) = 6

A = { (2,R), (2,B) }

Let A = event of getting an even number

Let B = event of getting a blue button

B = { (1,B), (2,B), (3,B) }

(b) The combined event of getting an even number and blue button

A B = { (2,B) } n (A B) = 1

So, P(A B) = n (A B) = 1

n (S) 6


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7.3 THE PROBABILITY FOR THE COMBINED EVENT7.3 c Calculating the probability of a combined event involving the

sum of probabilities

1. If A B = ,P(A or B) = P(A) + P(B)

2. If A B ,P(A or B)

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

A B = A B

S SA B A B


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sum of probabilities

EXAMPLE 3 24 366 169

Mary puts the above six cards in a box. If Mary picks a card randomly

from the box, find the probability of obtaining

(a) an odd number or an even number.

(b) an odd number or a multiple of 3.


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3 24 366 169

Sample space, S = { 3, 6, 9,16, 24, 36 } n (S) = 6

A = the event of picking an odd number(a)

A = { 3, 9 } n (A) = 2

Hence, P(A U B) = n (A U B) = 6 = 1

n (S) 6

B = the event of picking an even number

B = { 6,16, 24, 36 } n (B) = 4

Event of choosing an odd number or an even number = A U B

= { 3, 6, 9, 16, 24, 36 }

n (A U B) = 6


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3 24 366 169

Sample space, S = { 3, 6, 9,16, 24, 36 } n (S) = 6

P(A U B) = P(A) + P(B)(a)

P(A U B) = n (A) + n (B)

n (S) n (S)

= 2 + 4

6 6

= 66

= 1


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3 24 366 169

Sample space, S = { 3, 6, 9,16, 24, 36 } n (S) = 6

C = the event of picking an odd number(b)

C = { 3, 9 } n (C) = 2

Hence, P(A U B) = n (A U B) = 5

n (S) 6

D = the event of picking a multiple of 3

D = { 3, 6, 9, 24, 36 } n (D) = 5

Event of choosing an odd number or a multiple of 3 = A U B

= { 3, 6, 9, 24, 36 }

n (A U B) = 5


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3 24 366 169

Sample space, S = { 3, 6, 9,16, 24, 36 } n (S) = 6

P(C U D) = P(C) + P(D) P(C D)

(b)

P(C U D) = n (C) + n (D) n (C D)

n (S) n (S) n (S)

= 2 + 5 - 2

6 6 6

= 5

6

C D = { 3, 9 }n (C D) = 2


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product of probabilities

EXAM

PLE

Two fair dice are rolled. Find the probability that

both dice show prime numbers.


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solution


X X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X X

X

X

X

X

X

X

1 2 3 4 5 6

1

2

3

4

5

6

Dice x

Dicey

= A

(a) n (S) = 36

A = event of both dice showing

prime numbers.

n (A) = 9

Thus, P(A) =n (A)

n (S)

9

36=

= 14




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SOLUT

ION

n (S) = 6A = event dice x showing prime numbers

B = event dice y showing prime numbers

n (A) = 3

P(A) = n (A)n (S)

36

=

=1

2

n (B) = 3 n (S) = 6

P(B) = n (B)n (S)

36

=

=1

2The event both dice showing prime numbers = A B

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


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7.3 d Solving problem involving probability of combined event

EXAMP

LE

Ramesh and Sazali will choose either to register at school A or

School B. The probability of Ramesh and Sazali choosing school A

is 1/4 and 2/3 respectively. What is the probability that

(a) Ramesh will choose school B(b) Ramesh will choose school A and Sazali will choose school B

(c) Ramesh and Sazali will choose the same school


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solution

School A

School B

School B

School A

School B

School A

1/4

1/3

2/3

1/3

2/3

3/4

RASA

RASB

RBSA

RBSB

Ramesh Sazali Outcomes


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solution

(a) Probability that Ramesh will choose school B is

P(RB) = 1 P(RA)

= 1 1/4

= 3/4

(b) Probability that Ramesh will choose school A and Sazali will

choose school B is

P(RA and SB) = P(RA x SB)

= 1/4 x 1/3= 1/12


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solution

(c) Probability that Ramesh and Sazali will choose the same

school is

P[(RA and SA)]

= (1/4 x 2/3) + (3/4 x 1/3)

= 2/12 + 3/12

or P[(RB and SB)]

= P[(RA x SA)] + P[(RB x SB)]

= 5/12


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Table 2 shows the probability of James and Florence passing

a Mathematics and a Science test. [5 marks]

Probability of passing

Student Mathematics Science

James 3/5 2/3

Florence 7/10 3/5

Table 2

Calculate the probability that

(a) Both James and Florence pass the Science test

(b) Florence passes one of two tests


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Answer:

2/3 x 3/5 = 2/5

7/10 x 2/5 + 3/5 x 3/10 = 23/50

K1

K2N1

Probability

Mathematics Science

Student Pass Fail Pass Fail

James 3/5 2/3

Florence 7/10 3/5

(a) Both James and Florence pass the Science test

(b) Florence passes one of two tests

N1

2/5 1/3

3/10 2/5


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Diagram 10 shows three numbered cards in box P and two cards labelled

with letters in box Q.

2 3 6 Y R

Box P Box Q

Diagram 10

A card is picked at random from box P and then a card is picked at random

from box Q.

By listing the sample of all the possible outcomes of the event, find the

probability that

a) a card with an even number and the card labelled y are picked,

b) a card with a number which is multiple of 3 or the card labelled R

are picked


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2

3 6 Y

R

P Q


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2

3

6

Y

R

P Q


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2

3

6 Y

R

P Q


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2 3

6 Y

R

P Q


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2 3

6

Y

R

P Q


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2

3

6

Y R

3

6

Y R

2

Y R


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Box P

2 3 6

BoxQ

Y

R

{ (2,Y),(2,R),(3,Y),(3,R),(6,Y),(6,R) }

a) a card with an even number and

the card labelled Y are picked

{(2,Y),(6,Y)}

2/6 = 1/3

b) a card with a number which is multiple 3 or the

card labelled R are picked

{(3,Y),(3,R),(6,Y),(6,R),(2,R)}

5/6

P1

K1

N1

K1

N1


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Diagram 4 shows ten labelled cards in two boxes

A 2 B C D 3 E 4 F G

Box P Box Q

Diagram 4

A card is picked at random from each of the boxes. By listing the outcomes,

find the probability that

a) both cards are labelled with a number

b) one card is labelled with a number and the other card is labelledwith a letter


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Box P

A 2 B C

BoxQ

D

3

E

4

F

G

{ (A,D),(A,3),(A,E),(A,4), (A,F),(A,G),(2,D),(2,3), (2,E),(2,4),(2,F),(2,G),

(B,D),(B,3),(B,E),(B,4),(B,F),(B,G),(C,D),(C,3),(C,E),(C,4),(C,F),(C,G }

a) both cards are labelled with a

number

{(2,3),(2,4)}

2/24 = 1/12

b) one card is labelled with a number and the other

card is labelled with a letter

{(A,3),(A,4),(2,D),(2,E),(2,F),(2,G), (B,3),(B,4),

(C,3),(C,4)}

10/24 = 5/12

P2

K1

N1

K1

N1


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Diagram 9 shows five cards labelled with letters.

S M I L E

All these cards are put into a box. A two-letter code is to be

formed by using any two of these cards. Two cards are picked

at random, one after another, without replacement.

(a) List the sample space

(b) List all the outcomes of the events and find the probabilitythat

(i) the code begins with letter M,

(ii) the code consists of two vowels or two consonants.

Diagram 9


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S M I L E

s X


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S M I L E


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S

M

I

L E


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S

M I

L

E


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S

M I L

E


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S M

I

L

E

S

S

S

M S

M

M

M

I

L

E

I

I

I

I

S

M

L

E

E S

E

E

E

M

L

I

L S

L M

L I

L E

S M I L E


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S M I L E

S

M

I

L

E

(a) { (S,M),(S,I),(S,L),(S,E), (M,S),(M,I),(M,L),(M,E), (I,S),(I,M),

(I,L),(I,E),(L,S),(L,M),(L,I),(L,E),(E,S),(E,M),(E,I),(E,L)}

(S,M) (S,I) (S,L) (S,E)

(M,S) (M,I) (M,L) (M,E)

(I,S) (I,M) (I,L) (I,E)

(L,S) (L,M) (L,I) (L,E)

(E,S) (E,M) (E,I) (E,L)

S M I L E


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S M I L E

S

M

I

L

E

(S,M) (S,I) (S,L) (S,E)

(M,S) (M,I) (M,L) (M,E)

(I,S) (I,M) (I,L) (I,E)

(L,S) (L,M) (L,I) (L,E)

(E,S) (E,M) (E,I) (E,L)

(b) (i) { (M,S),(M,I),(M,L),(M,E) } 4

20

S M I L E


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S M I L E

S

M

I

L

E

(S,M) (S,I) (S,L) (S,E)

(M,S) (M,I) (M,L) (M,E)

(I,S) (I,M) (I,L) (I,E)

(L,S) (L,M) (L,I) (L,E)

(E,S) (E,M) (E,I) (E,L)

(b) (ii) { (S,M),(S,L),(M,S),(M,L),(I,E),(L,S),(L,M),(E,I) } 8

20

20032003 No 9


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Starting point

Taman Aman

Taman Sentosa

Other places

2003 No.9

Diagram 5 shows the route of a vehicle which carries a group of

volunteers. The group consists of 7 males and 5 females who aredropped off at random to sell flags at various points along routes.

a) If two volunteers are dropped off at Taman Aman, calculate theprobability that both are males.

b) The volunteers of different gender are dropped off at Taman

Aman. If two other volunteers are dropped off at Taman

Sentosa, calculate the probability that at least one of them is

female.

Diagram 5


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9.

Starting point

Taman Aman

Taman Sentosa

Other places

M F

7 5 M,M


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M

F

M

F

M

F

(M,M)

(M,F)

(F,M)

(F,F)

7/12

6/11

5/12

5/11

7/11

4/11

9. ( a ) 7/12 x 6/11=

7/22

K1N1

Answer:


72/73

9.

Starting point

Taman Aman

Taman Sentosa

Other places

M F

1 1

M F6 4

(M,F)

(F,M)(F,F)

(M M)


73/73

9. (b)

M

F

M

F

M

F

(M,M)

(M,F)

(F,M)

(F,F)

6/10

5/9

4/10

4/9

6/9

3/9

6/10 x 4/9 + 4/10 x 6/9 + 4/10 x 3/9

3

K23

Answer:

probability 2 -new 2011

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