Group Project 2 Sta108 Dah Siap Edit.docx

PERMUTATIONS AND COMBINATIONS

STA108: STATISTICS AND PROBABILITY

PREPARED BY: NOR SYAZWANI BINTI SHAHNUN (2016975079) NUR AIN NAZURAH BINTI MOHD NOOR (2016789305) SITI SURAINI BINTI AHMAD (2016107743) NUR SYUHADA BINTI AHMAD TERMIZI (2016726471) DIPLOMA IN SCIENCE FACULTY OF APPLIED SCIENCE DISTRIBUTED TO: M ADAM FADZILAH ABDOL RAZAK 1

CONTENT ACKNOWLEDGEMENT TABLE OF CONTENTS

CHAPTER 1: INTRODUCTION

PAGE

1.1 Introduction of experiment……………………………………….4 1.1.1

Probability Concepts and Rules……………………4

1.1.2

Procedures of the experiment……………………..,4

1.2 Background of experiment……………………………………….,4 1.3 Objectives of experiment………………………………………….,4

CHAPTER 2: METHODOLOGY 2.1 Suggestions of Probability Concepts and Rules…………………………………………………………….. 5 2.1.1 Methods………………………………………………. 5 2.1.2 Rules………………………………………………….. 5

CHAPTER 3: RESULTS 3.1 Tree Diagram………………………………………………………6 3.2 Probability Distribution……………………………………………7 - 8 3.3 Problem Solving…………………………………………………...9 - 14 3.2.1 Question for Tree Diagram…………………………….9 - 14 3.4 Justification ………………………………………………………..14 CHAPTER 4: CONCLUSION 4.1 Report Summary………………………………………………….,15

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ACKNOWLEDGEMENT

Firstly, we are very grateful because we managed to complete the second group project of Basic Statistics and Probability (STA108) within the time required. This project really help us in generating critical thinking using probability concepts rules toward the experiment given. Besides that, we really had a great teamwork while completing the project. This assignment cannot be completed without the effort and co-operation from our group members, Nor Syazwani Binti Shahnun, Siti Suraini Binti Ahmad, Nur Ain Nazurah Binti Mohd Noor and Nur Syuhada Binti Ahmad Termizi. We also sincerely thank our STA108: Basic Statistics and Probability lecturer Puan Fadzilah Binti Abdol Razak for the guidance and encouragement in finishing this project and also teaching us in this course.

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CHAPTER 1: INTRODUCTION 1.1 Introduction of experiment 1.1.1 Probability Concepts and Rules -0≤P(X=x)≤1 -ΣP(X=x)= 1.1.2 Procedures of the experiment The experiment was done by using a probability simulator -

Number of trials = 10 times

-

Value of X : 

{1, 2, 3}



{Green, Yellow, Red}

1.2 Background of experiment Humaira has been traveling to a country in Europe. In Europe there are a variety of mugs that attract Humaira. The mugs are hard to find in Malaysia because it is made from volcanic clay. The mugs are the best to be remembered as souvenirs when returning to Malaysia, she has brought some souvenirs which is 10 exclusive mugs from Europe. There are 10 colored mugs that consists of 4 green, 2 yellow and 4 red. Humaira kept the mugs in the cupboard because her friends could not come to her house on that day. In the next morning, Humaira’s mother took out three mugs at randomly to be used.

1.3 Objectives i.

To make an experiment based on the probability concept and probability distribution.

ii.

To solve the problems based on the probability and rules.

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CHAPTER 2: METHODOLOGY 2.1 Suggestion of Probability Concept and Rules. 2.1.1 Methods  Basic probability concepts 0 ≤ P( X = x ) ≤ 1  Discrete random variable ΣP( X = x ) = 1

2.1.2 Rules

The values of the probability must be 0 t0 1: 0 ≤ P( X = x ) ≤ 1

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CHAPTER 3: RESULTS 3.1 Tree Diagram

2/8 2/8 3/9

4/8 2/9

3/8 1/8 4/8 4/9

3/8 2/8 3/8

4/10

3/8 1/8 4/9 2/10

4/8 1/9

4/8 0/8

4/9

4/8 4/8 1/8 3/8

4/10 3/8 2/8 4/9

3/8 4/8 2/9

1/8 3/8

3/9 4/8 2/8 2/8 6

3.2 Probability distribution Element of sample space

Probability

X=x

GGG

4 3 2 1 × × = 10 9 8 30

1

GGY

4 3 2 1 × × = 10 9 8 30

2

GGR

4 3 4 1 × × = 10 9 8 15

2

GYG

4 2 3 1 × × = 10 9 8 30

2

GYY

4 2 1 1 × × = 10 9 8 90

1

GYR

4 2 4 2 × × = 10 9 8 45

1

GRG

4 4 3 1 × × = 10 9 8 15

2

GRY

4 4 2 2 × × = 10 9 8 45

1

GRR

4 4 3 1 × × = 10 9 8 15

1

YGG

2 4 3 1 × × = 10 9 8 30

2

YGY

2 4 1 1 × × = 10 9 8 90

1

YGR

2 4 4 2 × × = 10 9 8 45

1

YYG

2 1 4 1 × × = 10 9 8 90

1

YYY

2 1 0 × × =0 10 9 8

0

YYR

2 1 4 1 × × = 10 9 8 90

0

YRG

2 4 4 2 × × = 10 9 8 45

1

YRY

2 4 1 1 × × = 10 9 8 90

0

7

YRR

2 4 3 1 × × = 10 9 8 30

0

RGG

4 4 3 1 × × = 10 9 8 15

2

RGY

4 4 2 2 × × = 10 9 8 45

1

RGR

4 4 3 1 × × = 10 9 8 15

1

RYG

4 2 4 2 × × = 10 9 8 45

1

RYY

4 2 1 1 × × = 10 9 8 90

0

RYR

4 2 3 1 × × = 10 9 8 30

0

RRG

4 3 4 1 × × = 10 9 8 15

1

RRY

4 3 2 1 × × = 10 9 8 30

0

RRR

4 3 2 1 × × = 10 9 8 30

0

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3.3 Problem Solving 3.3.1 Question For Tree Diagram 1. At least three mugs are the same colors? P (three mugs the same colour) = P (GGG or YYY or RRR) = P (GGG) + P (YYY) + P (RRR) = (1/30) + (0) + (1/30) = 1/15 2. The mugs are of different colors? P (three mugs have different colour) = P (GYR or GRY or YGR or YRG or RGY or RYG) = P (GYR) + P (GRY) + P (YGR) + P (YRG) + P (RGY) + P (RYG) = (2/45) + (2/45) + (2/45) + (2/45) + (2/45) + (2/45) = 4/15 3. At least one of the mugs is green? P (at least one green mug) = P (GGG or GGY or GGR or GYG or GYY or GYR or GRG or GRY or GRR or YGG or YGY or YGR or YYG or YRG or RGG or RGY or RGR or RYG or RRG) = P (GGG) + P (GGY) + P (GGR) + P (GYG) + P (GYY) + P (GYR) + P (GRG) + P (GRY) + P (GRR) + P (YGG) + P (YGY) + P (YGR) + P (YYG) + P (YRG) + P (RGG) + P (RGY) + P (RGR) + P (RYG) + P (RRG) = (1/30) + (1/30) + (1/15) + (1/30) + (1/90) + (2/45) + (1/15) + (2/45) + (1/15) + (1/30) + (1/90) + (2/45) + (1/90) + (2/45) + (1/15) + (2/45) + (1/15) + (2/45) + (1/15) = 5/6

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OR 1 - P (YYY + YYR + YRY + YRR + RYY + RYR + RRY + RRR) 1 - [(0) + (1/90) + (1/90) + (1/30) + (1/90) + (1/30) + (1/30) + (1/30)] = 1 – (1/6) = 5/6

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The probability distribution function for a random variable 𝑋 is:

0, 1 , 6 1 , 𝑓(𝑥) = 2 3 , 10 1 , { 30

𝑥<0 0≤𝑥<1 1≤𝑥<2 2≤𝑥<3 𝑥≥3

Calculate the probability of P (1 ≤ 𝑥 < 3) =

1 3 + 6 10

=

7 15

Find P(𝑋 ≥ 1.5) = 𝑃 (𝑋 = 2) + 𝑃(𝑋 = 3) =

3 1 + 10 30

=

1 3

11

Find the cumulative probability distribution function 𝐹(𝑥) for 𝑋.

0, 1 , 6 2 𝐹(𝑥) = 3 , 29 , 30 1, {

𝑥<0 0≤𝑥<1 1≤𝑥<2 2≤𝑥<3 𝑥≥3

Calculate 𝐸(𝑋) and 𝐸(𝑋 2 )

2

𝐸(𝑋) = ∑ 𝑥. 𝑓(𝑥) 0

1 1 3 1 6 𝐸(𝑋) = [(0 × ) + (1 × ) + (2 × ) + (3 × )] = 6 2 10 30 5

2

𝐸(𝑋

2)

= ∑ 𝑥 2 . 𝑓(𝑥) 0

1 1 3 1 𝐸(𝑋 2 ) = [(02 × ) + (12 × ) + (22 × ) + (32 × )] = 2 6 2 10 30

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Calculate 𝐸(3𝑥 + 2) 𝐸(3𝑥 + 2) = 𝐸(3𝑥) + 𝐸(2) = 3 𝐸(𝑥) + 𝐸(2) 6 = (3 × ) + 2 5 =

28 5

Calculate 𝐸(4𝑋 2 + 𝑋 + 7) 𝐸(2𝑋 2 + 𝑋 + 7) = 𝐸(2𝑋 2 ) + 𝐸(𝑋) + 𝐸(7) = 4𝐸(𝑋 2 ) + 𝐸(𝑋) + 𝐸(7) 6 = 4(2) + + 7 5 =

81 5

Find i.

variance of 𝑋. 𝑉𝑎𝑟 (𝑋) = 𝐸(𝑋 2 ) − 𝐸(𝑋)2 =2−

= ii.

6 5

4 5

𝑣𝑎𝑟 (3𝑥 + 8) = 32 𝑣𝑎𝑟 (𝑋) + 0 =9 × =

4 5

36 5

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Assume 𝑋 = Green mug 𝑋=𝑥

0

1

2

3

∑ 𝑃 (𝑋 = 𝑥)

𝑃 (𝑋 = 𝑥)

1 6

1 2

3 10

1 30

1

E (𝑋)

0

1 2

6 10

1 10

6 5

E (𝑋 2 )

0

1 2

6 5

3 10

2

3.4 Justification Based on the theorem, the function can serve as probability distribution if ∑ 𝑓(𝑥) = 1. The function can serve as the probability distribution function of a random variable (𝑋). When we continue our calculation it would results as adding all the results:- 𝑃 (𝑋 = 𝑥) = 1.

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CHAPTER 4: CONCLUSION 4.1 Summary From the result obtained, it is able to classify the probabilities according to the specific distribution between discrete and continuous. On that note, it can be learn on how to solve discrete distribution amd cumulative distribution. The expected value and variance properties of discrete distribution can be conclude.

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