subject statistical probability theory summary of course content

nàn ngư, 12 CHAPTER 3: RANDOM QUANTITIES AND THE LAW OF PROBABILITY DISTRIBUTION... -- n n HH HH HH HT rà Hàn rà tr nry 19 I._ PROBABILITY DISTRIBUTION OF DISCRETE RANDOM VECTOR.... PROB

Implementation group:

NO NAME STUDENT

CODE MISSION EVALUATE

H6 Lan Anh 2253401010001

*Requirement |: Prepare + record a video of the statistics section *Requirement 2: Prepare chapters 5+6

100%

Lê Nguyễn Vân Anh 2253401010003

*Requirement |: Prepare + record a video of the statistics section *Requirement 2: Prepare chapters 112

100%

Nguyễn Thị Thụ Hiền 2253401010035

*Requirement |: Prepare + record a video of the probability section *Requirement 2: Prepare chapters 314

100%

Dang Thi Thuy Ngan 2253401010068 *Requirement |:

Prepare + record a video of the probability section *Requirement 2: Prepare chapters 718 100%

INDEX

PROBABILITY CHAPTER 1 SUPPLEMENTARY ANALYSIS OF COMBINATION -c-cccccieree 5 1.1 Mapping LH HH HH HH HH TH Tà Hàn TT TH Tàn TH TT TH 5 1.3 Muldiplication ruÌe - Sàn nh HH TT TT TH TT Hư 5 1.4 Permutafion

1.5 Partial Permutation 1.6 Combimationn oo eee 6 CHAPTER 2: BASIC CONCEPTS AND PROBABILITY FORMULAS 7 PIN C9)).0)n0)02)40).50.950)05A40i055 7 2.1.1 Random phenom€nOH:: 2 2 2 223 E23 HT TH TH TH TT TH TT kệ 7 2.1.2 Randomness (€sf an €V€TIÍ: nền HT HH HH Hà Hà Tư nh nry 7 2.1.3 Relationships between €V€IS: nàn HH HH HH TH 7 2.1.4 ExhaUS(fÏV€ €V€TI(S - TH Hà HH HH Hà Hà Hà To HH TH hà Hà Hà ip 9 V2 (0)79)000À429)0340 10110 9 r N9 na 9 2.2.2 Statistical Ï0orI4Í Hy HH nh HH HH HH Hà Hà Hào TT nà Tờ Hà hp 9 2.2.3 Properties of probabiÌÏlfy cà HH HH HH 9 V00) 3)0000/)9).0 10 0 Tố 10

2.3.1 Probability addition formula

2.3.2 Condifional probabilify .- cành HH HH HH HH Hư 11 2.3.3 Probability multiplicafion ẨormuÌa - - 5 2+ x2 SH HH 11 2.3.4 Full and Bayes probability formulaÏOIS nàn ngư, 12 CHAPTER 3: RANDOM QUANTITIES AND THE LAW OF PROBABILITY DISTRIBUTION 13 A CONCEPT OF RANDOM VARIABL/ES - 2 + nh HH HH ri 13 I0) 0.4008 2.900).0)/A⁄.3.47.3 012701108 13 TI SOME CHARACTTERISTTICS -.- LH HH HH HH HH Hi 13

1I I›f19))0)/89)0/.9 000A 13 1 The expectation (mean) of the random variable X iS: nen 13 2 The variance of the random variable X iS: - HH Hàng, 14 3 The standard deviation of the random variable X ÌS: cà nhe 14 I SOME COMMON PROBABILITY DISTRIBUTION LAWS re 14 1 Hypergeometric distribution: X ~ H((N, Nà, ñ)) vn HH Hee 14 2 Binomial distribution: X ~ B ( nạp ) nà HH HH HH HH HT 15 3 Poisson distribution: X ~ P( Ô ) St the rrrire 15 C CONTINUOUS RANDOM VARIABLES . - S222 ngư 16 X16.) (000) 120 5100091 16 1I I›f19))0)/89)0/.9 000A 16 1 The expectation (mean) of the random variable X iS: nen 16 2 The variance of the random variable X iS: - HH Hàng, 17 3 The standard deviation of the random variable X ÌS: cà nhe 17 Il NORMAL DISTRIBUTION

1 Standard normal distribution: T ~ Ì (Ô;Ï) sánh HH HH Hee 17 2 Normal distribution: X ~ N (u, ơ2) —— 18 CHAPTER 3.2 RANDOM VEC TOR n n HH HH HH HT rà Hàn rà tr nry 19 I._ PROBABILITY DISTRIBUTION OF DISCRETE RANDOM VECTOR 19 1 Component probabilify distribufion (margin distribution) . - cà sec 19 2 Condidional probability distribufion nành HH HH HH HH 20 1I PROBABILITY DISTRIBUTION OF CONTINUOUS RANDOM VECTOR

1 Simultaneous density function of (X,Y) cà nghe 2 Component densify funcfion nh nàng TH ki 22 3 Condifional densify ŸunCfÏOH nàng HH HH TH TT TH 23

CHAPTER 1 SUPPLEMENTARY ANALYSIS OF COMBINATION 1.1 Mapping

1.2 Addition rule Identifying signs: number of ways, cases, options that when removing one option, method, case, a certain task can still be completed

mi+imt Mn EX: There are 5 types of flowers in the carton: 2 red flowers, 2 yellow flowers, | blue flower, | white flower and | rose Randomly take | type of flower from the box to arrange, how many ways are there to choose?

- Solution:

How to choose red flowers: 2 How to choose yellow flowers: 2 How to choose blue flower: | How to choose white flower: |

How to choose pink flower: | LK) 2+2+1+1+1 =7 how to choose flowers for arrangement 1.3 Multiplication rule

Identifying signs: number of stages, all stages must be completed to get results

mim2 ™n

EX: A store sells shirts with 2 types of sizes including size 39, with 5 different colors and size 40 with 4 different colors How to choose 2 shirts of 2 sizes?

Solution: » = Shirt size 39 has 5 options

8 Shirt size 40 has 4 options O 5.4 =20 how to choose 2 shirts of 2 sizes

1.4 Permutation

Identification signs: Get all, sort all elements

Note: The permutations are all the same in composition, only different by the arrangement order of the elements in the group

members participating We know that class 12C5 has 25 boys How many ways are there to choose 7 boys in class 12C5 to compete in basketball?

Solution:

25) 480700 how to choose

7 B— C25 = 71(25-7)!

CHAPTER 2: BASIC CONCEPTS AND PROBABILITY FORMULAS 2.1, RANDOMNESS TEST AND EVENT

2.1.1, Random phenomenon:

2.1.2 Randomness test and event:

Sign sample space: ©) EX: Roll | dice of the same suit O ={1;2;3;4;5;6} Solution:

Let A be the event of “an even number appearing” A={2;4:6} Let B be the event of “an odd number appearing” B={1;3;5} EX: In a flower basket there are 2 types of flowers available for sale Know that these two types of flowers have type | products, type 2 products and waste products What is the simple event?

Solution: Let A be the simple event, A={2 type | flowers; 2 type 2 flowers; 2 waste products }

EX: Need to get 5 cans of water in a box containing 5 cans of beer and 3 cans of soft drinks Then which event is the sure event and which event is the empty event?

Let A be the sure event A ={ get at least 2 cans of beer } Let B be the empty event B={get 5 cans of soft drinks} 2.1.3 Relationships between events:

a.) Equivalent relationship - Pull along when ACB - Equivalent to when ACB and BCA Sign: A=B EX: Toss a dice, let A be the event "the die comes up 5" and B is the event "the dice comes up odd", then ACB

EX: Check 2 t-shirts Let A be the event "there is at least 1 defective shirt" and B the event "there is | defective shirt or 2 defective shirts", then A=B

b.) Sum and product of 2 events Sign: S U T or S+T

Q

EX: Consider the test of incubating 2 chicken eggs Let Ni: “The ith fruit blooms” (i=1;2)

Ki: “The ith fruit does not bloom” (i=1;2) A: “There is | fruit blooming”

Then, the sample space of the test is: O={K1K2;N1K2:K1N2;NiN2} The following events are simple events: @1= Kiko; @2=NiK2; @3=KiN2 ; @4=NiN2

Event A is not elementary because A=NiK2 U KiN2

A=Q\A EX: From a batch containing 10 genuine products and 3 waste products, 11 products are randomly selected

=> A and B are two mutually exclusive events

Chart VEN AUB ANB A and B conflict A.A Opposition

2.1.4 Exhaustive events Identification signs: Conflict with each other, and the total sum is equal Q EX: There are 4 coat racks Choose one shirt from each shelf Let Ai be the event “the coat is taken from the ith shelf”, i = 1,4

When {A1;A2;A3;A4} is the exhaustive events 2.2, PROBABILITY OF EVENT

1 Both admitted candidates are female 2 There is at least | female student admitted

2.2.3 Properties of probability 1.) IfA is an arbitrary event 0 < p(A) <1

2.) p(Ø) =0; p(©) = I

3.) IfA CB then p(A) < p(B)

2.3 PROBABILITY FORMULA 2.3.1 Probability addition formula

e IfA and B are two arbitrary events

p(ATB) = p(A) + p(B) - p(A.B)

e IfA and B are two mutually exclusive events

p(AtB) = p(A) + p(B)

e If {Ai} G=l, ,n) is mutually incompatible then

p(Ai + Az + + An) = p(At) + p(A2) + + p(An)

EX: A group has 30 investors of all types, including 13 securities investors, 17 equipment investors and 10 investors in both securities and equipment Find the probability that that person will meet the equipment investor

Solution 1 Let Ais “a partner who meets with a securities or equipment investor”

B is “a partner who meets stock investors" C is “a partner who meets equipment investors"

Solution 1; Let A be the event "take at least 1 red candy"

Aj 1s the event “getting ith red candy”, G=0,1,2,3) {Au; Az; A3} pairwise conflict

chce cậc]} , Cả C?j - 17

=>P(A) = P(Au) + P(A¿) +P(A3) = S34 + Sh Sh

Cio Cio Cio CŸCỷ _ 17

Solution 2: P(A) = 1 - P(Ao) = 1-374 =

EX: A group of 10 employees includes 3 men and 7 women, including 2 30-year-old men and 3 30-year-old women Randomly select | employee from that group Let A be "the selected employee who is female", B is "the selected employee who is 30 years old” Calculate P(A|B), P(BJA)?

1.) 0< p(A|B) s 1, VA CQ 2.) If Ac C then p(A|B) < p(C|B)

3.) P(A|B) = 1-p(A|B) 2.3.3 Probability multiplication formula

e If Aand B are two tensely independent events

p(An 8) = p(B)p(A|B) = p(A)p(B|A)|

e IfA and B are two independent events

e [fn events Aj, i= l, n are not independent then

| p(A1Ao An) = p(A1).p(Ag|A1) p(AnlAt An-t) |

EX: On Christmas, Mr A sold | large pine tree and | small pine tree The probability of selling a large pine tree is 0.9 If a large pine tree is sold, the probability of selling a small pine tree is 0.7 If a large pine tree cannot be sold, the probability of selling a small pine tree is 0.2 Knowing that Mr A sold at least 1 pine tree What is the probability that Mr A can sell both trees?

Let P(A) 1s the probability of “selling both trees” and P(C) is the probability of “selling at least | tree”

P(Anc) 0,9.0,7 P(Cc) 1—0,1.0,8

EX: Aquarium I has 3 goldfish and 4 brown fish, aquarium II has 5 goldfish and 3 brown fish Observed a fish jumping from lake I to lake I Calculate the probability that the fish jumping into lake IT is a goldfish?

lution: chart (PAM t 99%4 and fish 2 gold) = 2.2 —p-$ LÊ 4 5 38

p(fish 1 brown va ca 2 gold) = = 79.79

5 p(truck passes X to the oil pump) = 50° 0,1 = 0,025

2 p(car passes X to the oil pump) = 59° 0,2 = 0,02

13 ( p(motorbike passes X to the oil pump) = 30° 0,15 = 0,0975

A CONCEPT OF RANDOM VARIABLES

The random variable (BNN) X of the experiment with sample space © is the mapping:

Xx: OQ 9R

wor X(@)=x The value x is called the value of the random variable X Random variables include 2 types:

- Discrete random variable - Continuous random variable

B DISCRETE RANDOM VARIABLE I SOME CHARACTERISTICS

Let X be a discrete random variable with a probability distribution table:

The expectation of the random variable X?is: | E(X)? = Đ*x‡p,

2 The variance of the random variable X is:

| Varx = E(X)* — (EX)? |

3 The standard deviation of the random variable X is:

o = ~vVarx

EX: Given arandom variable X with a probability distribution table:

X 1 2 3 P 0,4 0,3 0,3 Calculate the expectation, variance, and standard deviation of X

In there: p =" 9 =1-p EX: A shipment has N = 40 light bulbs, including 10 broken light bulbs, randomly take 5 light bulbs to check Let X be the random variable indicating the number of broken light bulbs among the 5 light bulbs taken out

a) Make a table of the probability distribution of X

b) Calculate the average number of failed bulbs among the bulbs removed and find the variance of X

Solution: a) We have: X = lo, 1;2; 3; 4; 5} và N = 40,N¿ = 10,n = 5P X € H(40,10,5)

So we have the probability distribution table of X:

EX = np; VarX = npq; ModX = xo:np-qsxo<np-qt1

np —q < ModX <np—q+1>79,8 < ModX < 80,8> ModX = 80 The highest number of living trees after a period of planting is 80 trees 3 Poisson distribution: X ~ P( 2 )

EX: At gas station H, on average every 10 minutes, there are 15 motorbikes coming to fill up gas Knowing that the number of motorbikes coming to refuel at this gas station in a period of t minutes is a random variable with a Poisson distribution

a) Find the probability that in a period of 7 minutes, at least 4 motorbikes will come to fill up gas at this gas station

b) Find the probability that in a period of 15 minutes, from 20 to 25 motorbikes will come to refuel at gas station H

Let X and Y be the number of motorbikes that come to refuel at gas station H in 7 minutes and in 15 minutes, respectively

Under the assumption: X~P(A); Y~P(A); 4 = “= 10,5; A = 22,5 a) The probability that in a 7-minute period at least 4 motorbikes will come to fill up

with gas is:

P(X <4) =1-P(X <4) =1- [P(X = 0) + P(X =1) + P(X = 2) + P(X =3)

40 _ 1 _ L2 _ 3 _ =1- (re Ay Tre “tae 3+ 31° 4) = 0,99

b) The probability that in 15 minutes there will be from 20 to 25 motorbikes coming to fill up with gas is:

IL RANDOM QUANTITY 1 The expectation (mean) of the random variable X is:

2 The variance of the random variable X is:

VarX = E(X)? — (EX)?

3 The standard deviation of the random variable X is:

2 Normal distribution: X ~ N (u,Ø?)

a Define:

b Characteristics: | ModX = EX =p; VarX = 0? c Probability:

pasxsd)=0C 35) 0 34)

EX: The lifespan of a type of equipment provided by company A has a (approximately) normal distribution with a mean lifespan of 1500 hours and a standard deviation of 150 hours The product will be warranted by company A if its lifespan is less than 1200 hours

a) Calculate the percentage of equipment provided by company A that is warranted b) If company A wants the warranty rate to be 1%, how many hours must the

warranty period be? Solution:

a) Let X be the lifespan of that type of equipment (hours)

We have: X~ N(1500; 1507)