chapter 7 probability 1

8/3/2019 Chapter 7 Probability 1

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7.0 PROBABILTY 1

Created By: Mohd Said B Tegoh


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KEY TERMS

o Probability Kebarangkalian

o Sample space Ruang Sampel

o Outcome Kesudahano Experiment Eksperimen

o Reasoning Penaakulan

o Event Peristiwa

o Element Unsuro Trial Percubaan

o Occurrence - Berlakunya


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7.1 Sample Space

7.1 a Possible Outcomes of

an Experiment

o An experimentis a process or an action

in making an observation to obtain the

required results

o An outcome of an experiment is a possibleresultthat can be obtained from the

experiment


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7.1 Sample Space

Exam

ple In a box, there arered,blue and green

Marbles. A marble is drawn at random

from the box. Determine whether each ofthe following outcomes is a possible

outcome.

(a)A red marble is drawn

(b)A green marble is drawn(c)A yellow marble is drawn

(d)A blue marble is drawn


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7.1 Sample Space

Exam


marbles. A marble is drawn at random


outcome.

(a) A red marble is drawn

So

lution

(a)Drawing a red marble is a possible

outcome


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7.1 Sample Space

Exam




outcome.

So

lution(b) A green marble is drawn(b) Drawing a green marble is a possible

outcome


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7.1 Sample Space

7.1 b Listing All Possible Outcomes

Examp

le

An experiment is carried out by tossingA coin. List all possible outcomes.

So

lutionWhen a coin is tossed, the possible

possible outcomes are heads andtales.

heads-the factorial face of the coin

tales-the numerical face of the coin


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7.1 Sample Space

Exam




outcome.

So

lution(a) A blue marble is drawn(a)Drawing a blue marble is a possible

outcome


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x x

x


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Experiment: Roll a fair dice and toss a fair coin.

List all the possible outcomes.

1

2

3

4

5

6

H

T

H

T

H

T

H

T

H

T

H

T

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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7.1 Sample Space

A sample spaceis the set of all thepossible outcomes of an experiment

7.1 c Determining the Sample Space


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7.1 Sample Space

A letter is chosen from the wordHARMONY. Write the sample space, S,

using set notation.

Example

SolutionSample space, S = { H,A,R,M,O,N,Y }


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7.1 Sample Space

ExampleA spinner contains the lettersB,U,R and N. If James spins

the spinner, list the sample space,

S, using set notation

B

U

R

N

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

Solution


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7.2 Events

When a specific condition is given, we

can list theelements of a sample space

which satisfy the given condition

7.2 a Elements with Satisfy Given

Conditions


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7.2 Events

A two-digit number which is not more than20 is chosen at random. List theelements

of the sample space which satisfy each of

the following conditions

(a)A prime number is chosen

(b) A perfect square is chosen

Example


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7.2 Events

A two-digit number which is not more than 20 is

Chosen at random. List theelements of the

sample space which satisfy each of the followingconditions

(a) A prime number is chosen

SolutionS= { 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 }

(a) { 11,13,17,19 }

11 13 17 19

Two- digit prime number s which are less than 20


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7.2 Events

A two-digit number which is not more than 20 is

Chosen at random. List theelements of the

sample space which satisfy each of the followingconditions

(a) A perfect square is chosen

SolutionS= { 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 }

(b) { 16 }

16

Two- digit perfect square which is less than 20


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7.2 Events

An eventis a set of outcomes whichsatisfy a specific condition and it is

a subsetof thesample space.

7.2 b Events for Sample Space


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7.2 Events

ExampleA coin and die are thrown simultaneously. The

events A and B are defined as follows.A = Event of obtaining heads from the coin and

an even number from the die

B = Event of obtaining a tails from the coin and

a number less than 3 from the die

(a) List sample space, S

(b) List theelements of

(i) theevent A

(ii) theevent B


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7.2 Events

Solu

tion(a) Construct a table as shown below to help us

list theelements of the sample space

Single

dice1 2 3 4 5 6

Coin

Head (H)

Tails (T)

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

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

S = { (H,1), (H,2), (H,3), (H,4), (H,5), (H,6),

(T,1), (T,2), (T,3), (T,4), (T,5), (T,6) }


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7.2 Events

So

lution(b) List theelements of event A

S = { (H,1), (H,2), (H,3), (H,4), (H,5), (H,6),

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

A = { (H,2), (H,4), (H,6) }

Set of outcome ofevent A: obtaining a heads from the

coin and even number from the die


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7.2 Events

So

lution(b) List theelements of event B

S = { (H,1), (H,2), (H,3), (H,4), (H,5), (H,6),

(T,1), (T,2), (T,3), (T,4), (T,5), (T,6) }(T,1) (T,2)

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

Set of outcome ofevent B: obtaining a tails from the

coin and a number less than 3 from the die


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7.2 Events

o If R and R S, then theevent R is

possiblefor the sample space, S.

7.2 c Determining whether an Event

is Possible for a Sample Space

o If R = , then theevent R is notpossiblefor the sample space, S.


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ExampleTwo dice are thrown simultaneously. Theevents T, Q

and R are defined as follows.

T = Event such that the sum of the numbers from the two

dice is not more than 6

Q = Event such that the product of numbers from the two

dice is a prime numberR = Event such that the sum of the numbers from the two

dice is more than 12

(a) List the sample space, S.(b)Determine whether each of the following events is

possible for the sample space, S.

(i) Event T

(ii) Event Q

(iii) Event R


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(a) Construct a table as shown below to list theelements

of the sample space

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

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

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


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Solution

Two dice are thrown simultaneously. Theevents T, Q


T = Event such that the sum of the numbers from the two

dice is not more than 6

(a) Determine whether each of the following events is


(i) Event T

T = { (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4),

(3,1), (3,2), (5,1) }

o Since T and T S, then theevent T is



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Q = Event such that the product of numbers from the two

dice is a prime number

(b) Determine whether each of the following events is


(ii) Event Q

SolutionQ = { (1,2), (1,3), (1,5), (2,1), (3,1), (5,1) }

o Since Q and Q S, then theevent Q is



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R = Event such that the sum of the numbers from the twodice is more than 12

(b) Determine whether each of the following events is


(iii) Event R

SolutionR =

o Since R = , theevent R is not possiblefor the

sample space, S.


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7.3 Probability of an Event

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

P(A) = Number of times of the occurrence of event A

Number of trials

where 0 P(A) 1

7.3 a Probability of an Event from Big

Enough Number of Trials


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

P(A) = Number of outcomes in event A

Number of outcomes in the sample space

= n(A)

n(S)

where 0 P(A) 1


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

P(A) = n(A)

n(S)

o If P(A) = 0, then theevent A will certainly not occur

o If P(A) = 1, then theevent A is certainly occur


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E E L

Example

Three cards, as shown in the above diagram, are put

in into a box. A card is drawn at random from the box

and their letter is recorded. Then the card is put back

Into the box before another card is drawn at random.

This process is repeated 240 times and the results arerecorded in a table as shown below.

Outcome E L

Number of occurrences 162 78

Based on the above table, calculate the probability

that card with

(a) the letter E is drawn

(b) the letter L is drawn


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Outcome E L

Number of occurrences

Based on the above table, calculate the probabilitythat card with

(a) the letter E is drawn

So

lutio

nProbability that a card with the letter E is drawn= P(E)

Number of times the letter E is drawn=

Number of trials

162 78

=162

240

=27

40

240


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Outcome E L

Number of occurrences

Based on the above table, calculate the probabilitythat card with

(a) the letter L is drawn

So

lutio

nProbability that a card with the letter E is drawn= P(E)

Number of times the letter L is drawn=

Number of trials

78162

=78

240

=13

40

240


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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

Examp

le


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1 24 366 169

So

lutionSample space, S = { 1, 6, 9, 16, 24, 36 }

n (S) = 6

J = theevent 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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So

lutio

nSample space, S = { 1, 6, 9, 16, 24, 36 }

n (S) = 6

K = theevent 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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So


n (S) = 6

L = theevent 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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So


n (S) = 6

M = theevent of picking a positive number(d)

M = { 1, 6, 9, 16, 24, 36 }n (M) = 6

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

n (S) 6It means theevent of picking a positive

number is sure to happen


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A factory produces light bulbs for the domestic market.

Thequality control will randomly pick 2500 bulbs daily to

check on thequality. On the average, 75 units of them are

faulty. Find the probability that any one bulb picked is faulty.

Example

A = theevent of picking a faulty bulb

n (A) = 75

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

n (S) 2500 100

Solution


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7.3 b Calculating theExpected Number of

Times an Event Will Occur

If the probability of en event A and the

number of trials are given, then the

expected number of times theevent will

occur

= P(A) x Number of trials


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Exam

pleIn the month of December, 5000 cars were sold. If the

probability that a Proton cars were sold is 1 , calculate

5the number of Proton cars that were sold in that month.

So

lutionP(a Proton car is sold) =

Number of Proton cars sold =

Total number of cars sold

Number of Proton cars sold =

5000

Number of Proton cars sold =@

5

1

5

1

5

1

5005

1x

1000!


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7.3 c Predicting the Occurrence of an

Outcome

An event that

certainly will

not occur

An event which

low possibility

of occurring

0 0.1 0.5 0.9 1

An event withequal possibility

of occurring

An event with

high possibility

of occurring

An event thatcertainly will

occur


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0

0.1

0.5

0.9

1


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Bed

Dustbin

Exam

ple

Rank the following home products from

the most useful (1) to the least useful (4)

The diagram below shows a contest

organised by a shopping centre


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(a) Calculate the total number of the possible arrangements.

(b) Hence, state the probability that an entry will win.

(c) Madam Fong would like to try her luck in this contest

by sending two entries with different arrangements.Has she got a big chance of winning? Explain why.

(d) Puan Zaitun would also like to try her luck in this

contest by sending 20 entries with difference

arrangements. Has she got a big chance of winning?

Explain why?

Example


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(a) Calculate the total number of the possible arrangements.

SolutionLet ; 1 - Bed, 2 Table lamp, 3 Electric kettle, 4 - Dustbin

The possible arrangements are as follow

1234 1243 1324 1342 1423 1432

2134 2143 2314 2341 2413 2431

3124 3142 3214 3241 3412 3421

4123 4132 4213 4231 4312 4321

Hence, the total number of the possible arrangements

is 24


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(b) Hence, state the probability that an entry will win

SolutionThe probability that an entry will win =

24

1

The total number of the possible

arrangements is 24


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(c) Madam Fong would like to try her luck in this contest

by sending two entries with different arrangements.

Has she got a big chance of winning? Explain why.

SolutionLet F = Event that Madam Fong will win by sending 2

entries with different arrangements

P (F) =

24

2

=

12

1

Since, thevalue of P(F) below 1 and quite close to 0,

2

the chance of Madam Chong winning is very small.


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(d) Puan Zaitun would also like to try her luck in this

contest by sending 20 entries with difference

arrangements. Has she got a big chance of winning?

Explain why?

SolutionLet Z = Event that Puan Zaitun will win by sending 20

entries with different arrangements

P (F) =

24

20

=6

5

Since, thevalue of P(Z) is way above 1 and quite close to 1,

2

the chance of Puan Zaitun winning is great.


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7.3 d Solving Problems

A number is chosen at random from a set of positive

integers from 10 to 20 (inclusive of 10 and 20).

Calculate the probability that a prime number is

chosen.

Example


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A number is chosen at random from a set of positive

integers from 10 to 20 (inclusive of 10 and 20). Calculate

the probability that a prime number is chosen.

SolutionLet

R = Event that a prime number is chosenS = Sample space

S = { 10,11,12,13,14,15,16,17,18,19,20 }

R = { 11,13,17,19 }

@)(

)()(

Sn

RnRP !

11

4!


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Given that the probability of choosing a man at random

from a group of tourists is .11

4

If there are 28 women in the group, find the totalnumber of tourists in the group.

Example



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y = number of man group

y + 28 = number of tourists

11

4

28!

y

y

112411 ! yy

1127 !y

16!y

Number of tourists = 16 + 28 = 44

Solution


from a group of tourists is .11

4

If there are28 women in the group, find the total

number of tourists in the group.


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In a group of 80 students, 50 are boys. A further 10

girls then join the group. If the student is chosen at

random from the group, calculate the probability that

the student is a girl.

Example


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In a group of80 students,50 are boys. A further 10

girls then join the group. If the student is chosen at

random from the group, calculate the probability that

the student is a girl.

SolutionLet

B = Event that a boy is chosen

G = Event that girl is chosen

S = Sample space

n(S) =

n(S) =

n(G) =

80 + 10

50

( 80 50 )

= 90

+ 10 = 40

)(

)(

)( Sn

GnGP !

90

40!

9

4!

chapter 7 probability 1

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