Probability & Random Variables (Notes)

HAJVERIANZ PROBABILITY & RANDOM VARIABLES NOTES

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Probability Method in Engineering

Probability:Probability is the likelihood of chance that a particular event will occur. In statistics word probability has two basic meaning.

1) A quantitative measure of uncertainty. 2) Quantitative measure of degree of belief in a situation concept of probability is vital in

statistics. It is used in inferential statistics where we draw conclusion.

Set:A set is a collection of well defined objects. The objects which make a particular set are called element s or members of the set. They are usually denoted by capital letter as (A, B, C). The elements of a set are generally denoted by small letters. V= {a, e, I, o, u} set of voivel B= [all doctors of general hospital]

Sub set:A set B is a sub set of A, if every element of B is also an element of A. A= [1, 2, 3, 4, 5] B= [2, 4, 5] Set B is a sub set of A & A is also called sub set of B. every set is a sub set of itself.

Finite set:If the members of a set are limited it is called finite set. A= {1, 3, 5} B= {days of the week}

Infinite set:If the members of a set are unlimited it is called infinite set. A= {1, 3, 5…} B= {no. of trees in the country}

Null set:A set consisting no elements is called null set or empty set. It is denoted by ᵩ.

Equal set:Two sets A and B are said to be equal set if every element of “A” is an element of “B”. A= {2, 4, 6} B= {2, 4, 6}



Universal set:It is the set consisting all the elements under consideration it is denoted by capital “U”.

Union of two sets:The union of two sets A and B written as AU B is a set that contains the elements either belonging to “A” or “B” or both.

A= {1, 3, 5} B= {2, 4, 6} AUB= {1, 2, 3, 4, 5, 6}

The intersection of two sets:The intersection of two sets A and B written as AB is set of

A= {2, 3, 4} B= {1, 3, 4, 5} A∩B= {3, 4}

Disjoint set:Two sets A and B are said to be disjoint set if they have no element in common. A= {1, 2} B= {3, 4, 5, 6} A∩B= [ ]

A∩B=ᵩor [ ]

Overlapping set:

A BA B A B

A B



Two sets A and B are called overlapping sets if they have some common elements. A= {1, 2, 3, 4} B= {2, 4, 5}

Complement of a set:Complement of a set A w.r.t universal set” S” is a set of elements of S which are denote in A. A` S= {book, pen, notebook, marker} A= {book, notebook} A`= {pen, marker}

Difference of two sets:The difference of sets A and B either A or B is a set of element which belong to A but does not bell on to B. e.g. If A= {1, 2, 3} B= {3, 4, 5} A-B= {1, 2}

A-B

A B

A

A B



What do you mean by the following terms? Product set:

The product set of A and B denoted by A×B is a set consisting of all order pairs it is also called Cartesian after the French mathematician “Rene Disc arts”. A×B= {(x, y), xɛA, yɛB7} (Ɛ is called subset) A= {2, 4, 6} B= {a, b} A×B=(2,a),(2,b),(4,a),(4,b),(6,a),(6,b) B×A=(a,2),(a,4)(a,6),(b,2),(b,4),(b,6)

Factorial :(!)The product of 1st natural no. is called as “factorial” denoted by nil n! =n (n-1) (n-2) (n-3) 6! =6, 5, 4,3,2,1

Per motion:ⁿPr= !

!A permition is an arrangement of all are part of set of objects is and define order. The no. of permutations of n distinct objects taken is at a time. ⁿPr= !

!

⁶p₄= !!=

. . . . .. =360

CombinationA combination is an arrangement of objects without regard to their order. The no. of combinations of n subjects taken all at a time is ⁶c₄= !

! !=15

Define the following term: Experiment:

The term experiment means a process whose results yield a set of data. An experiment may be performing in a laboratory. A field, a class, a hospital, factory etc.

Trial: The single performance of an experiment is called trial.

Outcome:The result obtained from a trail or random experiment is called outcome.

Random experiment:An experiment which produces different results even through it is perform a large no of timer under essentially similar conditions



e.g. The tossing of a coin. Rowing a playing card from the pack of 52 playing cards. A random experiment has the following properties.

i. The experiments can be repeated any no. of times ii. A random trial consists of at least possible two outcomes. iii. Nothing can be set with certainty about the outcome or random experiment. Define simple space and simple point:

Sample space: A set consisting of all possible outcomes that may result from a random experiment is called sample space and denoted by S e.g. Sample space when an electric bulb is tested the result is defected or non defected S= [defective, nondirective] The sample space when a coin is toss is S= [H.T] The sample space when a cubical dice S= [1, 2, 3, 4, 5, 6]

Sample point:The outcome is the elements of the sample space are called the sample point of that case. Define events and types of events:

Event:Any sub set of sample space is called an event. e.g. When a coin is tossed the sample space is S= [H.T] The subset A= [H] B= [T]

Sample event or elementary event:An event that contains only one sample point is called a sample event. e.g. When two coins are tossed the event. A= [H, H] is a sample event when two dice are rolled the event B= [6, 6] that the sum is 12 simple event.

Compound event:An event that contains more than one sample points is called compound event. e.g. When two coins are tossed there are possible results B= [HH, TH, HT]



That at least are head appearing is a compound event.

Certain or soure event:An event consisting of the sample space itself is the soure event and is always accurse. e.g. When we through a dice thronging a no less than 7 is soure or certain event.

Impossible event:An event consisting of the empty (Φ) is called the impossible event because it can never occur. e.g: When a dice is rolled the event.

Exhaustive event:Two events A and B are said to be exhaustive events it they constitute the entire sample space. e.g.: When a fair dice is rolled S= [1, 2, 3, 4, 5, 6] The events A= [1, 3, 5] and B= [2, 4, 6] are exhaustive events because AUB =S

Equally likely events:Two events are said to be equally likely events if they have same chance of occurrence. e.g: When a fair coin is tossed the head is as likely is occur as tail. Similarly all the simple space of a fair dice is equally likely.

Mutually exclusive events:Two events are said to be mutually exclusive events are disjoint if they can’t occur together. A= [1, 3, 5] B= [2, 4, 5] A∩B=Φ e.g: In a toss single coin we get either a head of as tail. But both head and tail cannot occur together. Similarly on tossing a fair dice we get one of six possible outcomes. These outcomes are there for mutually exclusive.

Not mutually exclusive events:Two of more events are said to be not mutually exclusive events if they may occur at the same time. e.g A card is picked at random from a deck of cards. If can be both and ace and a club. Therefore access and clubs are not mutually exclusive events become we can pick the ace of clubs.

Independent events:



Two events A&B are said to be independent if the occurrence of one event does not affect the occurrence of other. e.g In two successive tossing of a fair coin the outcome of a second toss deed not depends of the outcome of the 1st toss. When selection of events is with replacement, then the events are independent.

Dependent event:Two events A&B are said to be dependent event if the occurrence of one event dependent on the occurrence of other... e.g The results of two drawing of a ball from a bag dependent if the ball is not return to the bag after the 1st draw. When the selection of the events is without replacement then event are called dependent event.

DEFINE THE FOLLOWING DEFINATIONS OF PROBABILITY: Classical of periodic:

According to p.s Laplace if an experiment can result in an mutually exclusive, quirkily likely and exhaustive outcomes and M of which are favorable to the occurrence of an event of A, the probability that event A will occur is given by the ratio m/n. P (A) = no. of favorable outcomes/no. of total possible outcomes

Relative frequency or posteriori:If an experiment is repeated a large no. of times say n under uniform conditions and if the event A occur m times, then the probability of occurrence of the event A is defined as the limit of relegating frequency m/n as n times to infinity. P (A) = limit m/n

Axiomatic(mathematical approach)The definition was introduced by Russian mathematician A.N Kolgomorovein 1933. Is based on a set of Axioms where an axioms in a state that is assumed to be true. Let S.P sample space with the sample point A₁,A₂,A₃,……… up to so on each sample point is assigned a real no. P (A₁), p (A₂), P (A₃), .p (An)

Then the probability of any point must obey the following. i. 0< ̲p(A)<̲1 ii. P(A)=1 iii. For two mutually exclusive events A₁,A₂

P (A₁ᴜA₂) = p (A₁) +p (A₂)

Subjective or personality probability:



If refers to the degree of believe of an in individual that the event will occur, based on whatever evidence is available to the individual.

What is mean by conditional probability:If A&B are two events defined on the sample space S, the probability that be occurs given that A has already occurs is called conditional probability of B given A has occurred & denoted by p (B/A). P (B/A) = p (A∩B)/p (A) Where p (A) ≠0 STATE & PROVE ADITIONAL LAW FOR MUTUALLY

EXCLUSIVE EVENTS:STATEMENT:If two events A & B are mutually exclusive then probability that either A or B will occur is the sum of probability of A & B. SYMBOLICALLY:P (AᴜB) = p (A) +p (B) PROOF:Suppose any sample space S containing N sample point, which are equally likely. Let the two events A & B belonging to S & containing n₁ & n₂. Sample point respectively. As events A & B are to be taken as mutually exclusive, so there two events are disjoint & have no point is common. So, events AᴜB will contains (n₁+n₂) sample point. Now using definition probability of P (AᴜB) = . ᴜ

.

= ₁ ₂

= ₁+ ₂

= p (A₁) +p (A₂) The law can be generalizing for more than two events.

S

STATE & PROOF ADITIONAL LAW FOR NOT MUTUALLY EXCLUSIVE EVENT:STATEMENT:

A B

n₁ n₂



IF A & B are not mutually exclusive events then probability that A or B or both occurs. P (AᴜB) = p (A) +p (B)-p (A∩B) PROOF:Suppose that sample space S containing N sample point which is equally likely. The two events A & B which are not mutually exclusive, belong to S. we can write events B in the form of two mutually exclusive events. B= (A∩B) ᴜ (B-A∩B) as events A∩B & (B-A∩B) are mutually exclusive.

S

So, By applying probability P (B) = P (A∩B) +P (B-A∩B) (1) P (AᴜB) = P (A)-P (B-A∩B) (2) By adding eq. (1) & (2) P (AᴜB)-P (B) =P (A)-P (A∩B) & P (AᴜB) =P (A) +P (B)-P (A∩B) This law is called general law of addition of probability.

STATE & PROOF MULTIPLICATION LAW OF INDEPENDENT EVENTS:STATEMENT:If A & B are independent events having none zero probability. P (A∩B) = P (A).P (B) PROOF:Suppose a sample space

S₁ S₂

A∩B

A B

A

(m)

B

(n)



S₁ has in sample point. The event a belonging to s₁ has m₁ sample point. Similarly the sample space s₂ has n sample point out of which n₁ are favorable to events B belonging to s₂. So probability of P (A) = m₁/m P (B) = n₁/n Since A & B are independent events the total sample points for combined events for A&B will be m*n & favorable sample point for joint events A∩B will be m₁*n₁P (A∩B) = m₁*n₁/m*n = m₁/m * n₁/n

P (A∩B) = p (A).p (B) The probability p (A∩B) is called joint probability of A & B. And p (A) & p (B) Are called majonal probability of A & B respectively.

STATE & PROFF MULTIPLACTION LAW FOR DEPDENDENT:STATEMENT:If A & B are two dependent events then P (A∩B) = p (A). P (B/A) = p (B). P (A/B) PROOF:Suppose there is a sample space S containing n sample points which are equally likely. Let events A has m₁ favorable sample points & event B has m₂ sample points & m₃favorable sample points thus, P (A) = m₁/n P (B) = m₂/n P (A∩B) = m₃/n

A∩B P (A∩B) =m₃/n* m₁/n

= m₁/n * m₃/n

= m₁/n * mob/n/m₁/n =P (A)*p (A∩B)/p (A)

A

m₁

B

m₂



= p (A) * p (B/A) Similarly: P (A∩B) = m₃/n = m₃/n * m₂/m₂ = m₂/n * m₃/m₂ = m₂/n * m₃/n/m₂/n = p (B)* p (A∩B)/p (B) = p (B). P (A/B) = p (B). P (A/B) This is called general rule for multiplication of probability.

IF S= [1,2,3,4,5,6,7,8,9,10]A = [1, 2, 3, 4] B = [2, 4, 6, 8] C = [1, 3, 5, 7] D = [2, 4]

i. AᴜB ii. AᴜC iii. A∩C iv. C∩B v. C′ vi. A′

Answer: i. AᴜB= [1,2,3,4,6,8] ii. BᴜC= [1,2,3,4,5,7] iii. A∩C =[1,3] iv. C∩B=[] v. C′= [2,4,6,8,9,10] vi. A′=[5,6,7,8,9,10]

HOW MANY DISTINCT FOUR DIGIT NO CAN BE PERFORM FOR THE FOLLOWING INTIGER 1,2,3,4,5,6 IF EACH INTIGER USE ONLY ONE:

i. ⁶P₄= 6! / (6-4)! = 360 ii. ⁵P₃= 5! / (5-3)! = 60 iii. ⁸P₃= 8! / (8-3)! = 312

ⁿ C ᵣ= !! !

i. ⁸ C ₃= 8! / (8-3)! 3! = 52.4 ii. ⁵ C ₂ = 5! / (5-2)! 2! = 10 iii. ⁹ C ₄ = 9! / (9-4)! 4! = 126



A FAIR COIN IS TOSSED MAKE A SAMPLE SPACE AND FIND THE PROBABILITY:S= [H, T]

i. A head appear ii. A tail app ear iii. No head appear

N (S) = 2 Answer:

i. P (H)= n (A) / n (S) = 1 / 2 ii. P (T)= 1 / 2 iii. N (no head or tail)= 1

n (c) = 1 P (c) = 1 /2

TWO COINS ARE TOSSED MAKE A SAMPLE SPACE & THE FIND THE PROBABILITY:S= [HH, HT, TH, TT]

n (S) = 4 i. No head appear : n(A)= [TT]

P (A) = n (A) / n(S) = 1 / 4 ii. One head appear:

B= [HT, TH] P (one head) = 2 / 4 = 1 / 2

iii. One tail & one head appears: n (c) = 2 P (one tail, one head) = 2 / 4 = 1 / 2

iv. One of more than one head appear: D= [HH, HT, TH] n (D) = 3 P (one of more than one head appear) = 3 / 4

WHEN THREE COINS ARE TOSSED FIND THE PROBABILITY:a. No head appears:

S= [HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, TTH] n (s) = 8 A (TTT) n (A) = 1 P (A) = 1 / 8

b. One head appears: B= [HTT, TTH, THT] n (B) = 3 P (B) = 3 / 8

c. Two head appears: C= [HHT, THH, HTH] N(c) = 3



P(c) = 3 / 8 d. Three head appears:

D= [HHH] n (D) = 1 P (D) = 1 / 8

e. All tails appears: E= [TTT] n (E) = 1 P (E) = 1 / 8

f. More than one head: F= [HHH, HHT, THH, HTH] N (F) = 4 P (F) = 4 / 8

g. More than three head: G= [0] n (G) = 0 P (G) = o / 8 = 0

A FAIR DICE IS ROLLED MAKE SAMPLE SPACE & FIND THE PROBABILITY:S= [1, 2, 3, 4, 5, 6]

I. The odd no’s: A= [1, 3, 5] N (A) = 3 P (A) = 3 / 6 = 1 / 2

II. The number is multiple of three: B= [3, 6] N (B) = 2 / 6= 1 / 3

III. The number is even of odd: C= [1, 2, 3, 4, 5, 6] N (C) = 6 P (C) = 6 / 6 = 1

IV. The number if one or more than one: D= [1, 2, 3, 4, 5, 6] n (D) = 6 P (D) = 6 / 6= 1

V. The prime no’s appears: E= [2, 3, 5] N (E) = 3 P (E) = 3 / 6= 1/2

TWO FAIR DICE ARE ROLLED FIND THE PROBABILITY:a) The sum is even.



b) 1st dice have a six. c) The sum is twelve. d) Both dice have same outcome. e) The product of both dice is 12. f) The sum id less than 2.

DICE-1DICE-2

1 2 3 4 5

1 1,1 1,2 1,3 1,4 1,52 2,1 2,2 2,3 2,4 2,53 3,1 3,2 3,3 3,4 3,54 4,1 4,2 4,3 4,4 4,55 5,1 5,2 5,3 5,4 5,56 6,1 6,2 6,3 6,4 6,5

a) A=[(1,1)(3,1)(5.1)(2,2)(4,2)(6,2)(1,3)(3,3)(5,3)(2,4)(4,4)(6,4)(1,5)(3,5)(5,5)(2,6)(4,6)(6,6)] N (A) = 18 P (A) = 18 / 36= 1 / 2

b) B= [(6,1)(6,3)(6,4)(6,5)(6,6)(6,2)] N (B) = 6 P (B) = 6 / 36= 1 / 6

c) C= [(6,6)] N (C) = 1 P (C) = 1 / 36

d) D= [(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)] n (D) = 6 P (D) = 6 / 36= 1 / 6

e) E= [(2,6)(2,4)(4,3)(6,2)] n (E) = 4 P (E) = 4 / 36= 1 / 9

f) F= 0 N (F) = 0 P (F) = 0 / 36= 0

A BAG CONTAINS FOUR RED & SIX GREEN BALLS, OUT OF WHICH THREE BALLS ARE DRAWN. FIND THE PROBABILITY:

1) 2 red , 1 green 2) All red balls 3) 1 green 4) No red ball

Answer: 1) A=



⁴C₂= 4! / (4-2)! 2! = 6 ⁶C₁= 6! / (6-1)! 1! = 6 N (S) = ¹⁰C₃= 10! / (10-3)! 3! = 120 P (A) = 36 / 120

2) B(all red)= ⁴C₃= 4! / (4-3)! 3! = 4 ⁶Cₒ= 6! / (6-0)! 0! = 1 P (B) = 4 / 120= 1 / 30

3) C= ⁴C₂= 4! / (4-2)! 2! = 6 ⁶C₁= 6! / (6-1)! 1! = 6 P (C) = 36 / 120

4) D= ⁴Cₒ= 4! / (4-0)! 0! = 1 ⁶C₃= 6! / (6-3)! 3! = 20 P (D) = 20 / 120= 1 / 6

A COIN IS TOSSED TWICE IF A & B ARE 1ST POINT & TAIL ON 2ND

POINT RESPECTIVELY. FIND THE PROBABILITY:S= [HH, HT, TH, TT] N (S) = 4 A= [HH, HT] n (A) = 2 B= [TH, TT] n (B) = 2 P (A) = 2 / 4 = 1 / 2 P (B) = 2 / 4= 1 / 2 AᴜB= [HT] N (A∩B) = 1 P (A∩B) = 1 / 4 P (AᴜB) = P (A) + P (B) - P (A∩B) = 1 /2 + 1/ 2 – 1 / 4 = 3 / 4

TWO DICE ARE ROLLED IF A & B ARE RESPICTIVELY EVENTS THAT THE SUM OF POINT IS A & BOTH DICE SHOULD GIVEN ODD NO FIND PROBABILITY:S= 36 A= [(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)]



N (A) = 5 P (A) = 5 / 36 B= [(1,1)(1,3)(1,5)(3,1)(3,3)(3,5)(5,1)(5,3)(5,5)] n (B) = 9 P (B) = 9 / 36= 1 / 4 N (A∩B) = [(3, 5) (5, 3)] n (A∩B) = 2 P (A∩B) = 2 / 36= 1 / 18 P (A∩B) = P (A) + P (B) - P (A∩B) = 5 / 36 + 9/ 36 – 2 / 36 = 1 / 3

A PAIR OF DICE IS ROLLED FIND THE PROBABILITY THAT THE SUM OF THE UPPER MOST DOTS ID EITHER 6 OR 9.n (S) = 36 Let A be the event the sum of upper most dot is 6. A= [(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)] N (A) = 5 P (A) = 5 / 36 Let B the event that the sum of upper most is 9. B= [(3, 6) (4, 5) (5, 4) (6, 3)] N (B) = 4 P (B) = 4 / 36 Since A&B are mutually exclusive events. P (AᴜB) = p (A) + p (B) = 5/ 36 + 4 / 36 = 1 / 4

TO TOSSING TWO COINS FIND PROBABILITY:(I) PROBABILITY OF TWO HEADS GIVEN THAT A HEAD ON THE 1ST. (II) THE PROBABILITY OF TWO HEADS GIVEN THAT AT LEAST ONE HEADS.

S= [HH, HT, TH, TT] N (S) = 4 Let A be the event that the head appear on the 1st coin. A= [HH, HT]

n (A) = 2

P (A) = 2 / 4= ½ Let B the event that two head appears. B= [HH]



N (B) = 1 P (B) = 1 / 4 A∩B = [HH] n (A∩B) = 1 P (A∩B) = 1 / 4

A & B ARE TWO INDEPENDENT EVENTS IF PROBABILITY OF:P () = 0.40, p (B) = 0.30

FIND THE PROBABILITY:(I) P (A∩B) (II) P (AᴜB)

SINCE A&B ARE INDEPENDENT EVENTS THEREFORE (I) P (A∩B) = P (A). P (B)

= (0.40). (0.30)= 0.12 (II) P (AᴜB)= P (A) + P (B) – P(A∩B) = 0.40 + 0.30 – 0.12 = 0.58

DEFINE PERMUTATION , COMBINATION & DISCRIMINATE BETWEEN TWO:LET A= [1, 3, 5, 7] B= [2, 4, 6, 8] C= [1, 2, 3, 4, 5] S= [1, 2, 3, 4, 5, 6, 7, 8,]

1) A∩B= [1,3,5,7]∩[2,4,6,8] = []

2) C∩A= [1,2,3,4,5]∩[1,3,5,7] = [1, 3, and 5]

3) A∩C= [1,3,5,7]∩[1,2,3,4,5] = [1, 3, and 5]

4) B∩C= [2,4,6,8]∩[1,2,3,4,5] = [2, 4] PERMOTATION:Is an arrangement of all or part of a set of object in a definite order? The no of permutation n distinct object taken r at a time. ⁿ P ᵣ= n! / (n-r)!

⁶P₄= 6! / (6-4)! = 360



COMBINATION:Is an arrangement of object without regard to their order? The no of combination objects taken all at a time is ⁿ C ᵣ= n! / (n-r)! r! ⁶ C ₄= 6! / (6-4)! 4! = 15 EVELUATE THE FOLLOWING:(I) 7! = 7.6.5.4.3.2.1 = 5040 (II) 16! / 8! = 51891840 (III) 8! =8.7.6.5.4.3.2.1 = 40320 (IV) 4! = 4.3.2.1 = 24 (V) ⁸ P ₆= 8! / (8-6)! = 20160 (VI) ⁵ P ₅= 5! / (5-5)! = 120 (VII) ⁵ C ₃= 5! / (5-3)! 3! = 10 (VIII) ⁶ C ₂= 6! / (6-2)! 2! = 180

LETA= [1, 4] B= [2, 3] C= [3] B is the sub set of universal set S= [1, 2, 3, 4] A*B= [1,4]*[2,4]= (1,2)(1,3)(4,2)(4,3) B*A= [2,3]*[1,4]= (2,1)(2,4)(3,1)(3,4) A*A= [1,4]*[1,4]= (1,1)(1,4)(4,1)(4,4) B*B= [2,3]*[2,3]= (2,2)(2,3)(3,2)(3,3)

FIND THE PROBABILITY OF EACH OF THE FOLLOWING:A) A head appears in tossing a fair coin. B) A 5 appears in rolling a 6 faced cubical dice C) An event no appears when a perfect cubical dice is rolled.

Sol: (I) When a coin is tossed the sample is S= [HT] it means

n (S) = 2 Let A be the event that given head A= [H] n (A) = 1 / 2

(II) When a 6 pack dice is rolled the sample space. S= [1, 2, 3, 4, 5, 6] n (S)= 6 let



B be the event that 5 appears on the dice B= [5] n (B)= 1 p (B) = 1 / 6

(III) When a perfect cubical dice is rolled the sample space is S= [1, 2, 3, 4, 5, 6] n (S)= 6 let C be the event that even no appears C= [2, 4, 6] n (C) = 3 P (C) = 3 / 6 = 1 / 2

WHAT IS THE PROBABILITY OF A CARD OD DIAMOND FROM A PACK OF PLAYING CARD CONSISTING OF THE USUALLY 52 CARDS?Sol: The sample space consisting of 52 possible outcomes it means n (S) = 52 Let A be the event that the card is diamond n (A) = 13 / 52 = 1 / 4

SHOW THAT IN A SINGLE THROUGH WITH TWO DICE THE CHANCE OF THROWING TWO DICES THE CHANCE OF THROWING MORE THAN 7 IS EQUAL TO THAT OF THROWING LESS THEN 7.n (S) = 36 Let A be the event that sum is less than 7. A= [(1,1)(1,2)(1,3)(1,4)(1,5)(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(4,1)(4,2)(5,1)] n (A) = 15 P (A) = 15 / 36 Let B the event that the sum is more then 7, B= [(2,6)(3,5)(3,6)(4,4)(4,5)(4,6)(5,3)(5,4)(5,5)(5,6)(6,2)(6,3)(6,4)(6,5)(6,6)] n (B) = 15 P (B) = 15 / 36 Hence P (A) = p (B) = 15 / 36



WHAT IS THE PROBABILITY OF THROWING EITHER 7 OR MORE THEN 10 WITH TWO DICE?Sol: When two fair dice are rolled the sample space is S= 36 n (S) = 36 Let A be the event sum of dots is 7. A= [(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)] n (A) = 6 P (A) = 6 / 36 = 1 / 6 Let B is the event that sum of dots more then 10. B= [(5, 6) (6, 5) (6, 6)] n (B) = 3 P (B) = 3 / 36 = 1 P (AᴜB) = p (A) + p (B) = 6 / 36 + 3 / 36 = 1 / 4 PROBABILITY DISTRIBUTATION:

DEFINE RANDOM VARIABLE:A variable whose values are determined by the outcomes of a random experiment is called a random variable. Thus the no [H] or [T] in tossing of two coins & sum of dots obtained with a pair of dice are then example of random variable. A random variable is also called a chance variable. The random variables are usually denoted by capital letter such as X, Y, Z & their values by small letters x, y, z respectively. There are two types of random variable. (1) Discrete variable (2) Continues variable

WHAT IS DISCRETE RANDOM VARIABLE EXPLAIN ITS PROBABILITY DISTRIBUTATION:(A) DISCRETE RANDOM VARIABLE:

A random variable X is defined to be discrete random variable if it assumed finite or countable infinite no of values. e.g. The no of defective blabs in a lot. The no of road accident on motorway per day etc.

(B)DISCRETE PROBABILITY DISTRIBUTATION:



Let X is a discrete random variable assumed the values. X₁, X₂, X₃, X₄, X₅, .Xn P (X₁), p (X₂), p (X₃)…………….. P (An) Then the ordure pair [X, p (X₁)], [X₂, p (X₃] ________ [Xn, p (Xn)] Where i= 1, 2, 3, 4, 5, .n is called discrete probability distributation of X.

(C) CONDITIONED OF PROBABILITY DISTRIBUTATION:(i) There must be a finite [0-1] against every possible value of the random variable 0≤ p

(X)≤ 1 (ii) The sum of all possibility must be equal to 1.

P (x) = 1 Discrete probability distributation X may be expressing any of the following 3 form: (1) Tabular form (2) Equation form (3) Graphical form

TABULAR FORM:For expressing discrete probability in tabular form we take all possible value of random variable in a Colum of correspondence probabilities in another Colum. Probability distributation of X X p(x) X₁ p x₁) X₂ p (x₂) X₃ p(x₃)

Xn p (Xn) ∑ P (x) = 1

EQUATION FORM:For expressing probability distributation in equation form we find an equation in involving the random variable which gives the probability for every possible value of the random variable. The particular equation which generates probabilities of the random variable at its possible value is known as probability function & abbreviated.

GRAPHIC FORM:The discrete probability distributation is usually displayed by a bar diagram. Probability histogram;

BAR DIAGRAM:



In bar diagram we take values of the random variable X₁, X₂, X₃, X₄, .Xn along x-axis & probability along y-axis

PROBABILITY HISTOOGRAM:To display probability histogram we take values of random variable along x-axis & probability along y-axis. Adjacent rectangle are drawn against each rectangle are drawn against each values such that the height of each rectangle is equal to the probability at that point & width of each rectangle is 1.

DEFINE CONTINUS RANDOM VARIABLE, WHAT IS PROBABILITY DENSITY FUNCTION, WRITE DOWN ITS PROPERTIES?CONTINOUS RANDOM VARIABLE:A random variable is defined to be a continues random to be a continues random variable. If it can assume unlimited values within a given range of possible values. e.g. Weight of student of 1st year, temperature of a room, rain fall amount recorded a hill station.

PROBABILITY DENSITY FUNCTION:Let x be a continues random variable which can take the value on the interval [a,b] Then the probability density function (p.d.f) of X has following two properties. (1) The value of the function must be greater than or equal to zero. For every value of x.

f (x) ≥0 For all value of X.

(2) The area under the curve & above x-axis is equal to one. The probability density function of a random variable cannot be expressed in tabular form, but it can be expressed in graph of in equation form. The mathematical equation is known as P.D.F.

0%

50%

100%

x1 x2x3

Xn

2 3 4 7

Series 1

Series 1



WHAT IS MATHEMATICAL EXPECTION OF DISCRETE RANDOM VARIABLE , WRITE DOWN ITS PROPERTIES:MATHEMATICAL EXPECTATION:If a discrete random variable X assumed the values X₁, X₂, X₃, .Xn with respect probabilities P (X₁) p(X₂) p (X₃)…… p (Xn) Such that the sum of probabilities is equal to one. ∑ P (Xi) = 1, then the mathematical expectation of exceptive value E (X) = X₁. P (X₁) + X. p (₂) + ……+ Xn. P (Xn) = ∑ Xi. P (Xi) μ = E (X) = ∑ X. p (X) E (X) is also called mean of x. which is denoted by (μ) Mean = μ= E (X) = ∑ X. p (X) If random variables is denoted by y then exceptive of y. E (y) = E (y) = ∑ y. p (y)

PROPERTIES OF EXPECTIVE OR LAW OF EXPECTATIVE:1) IF a is the constant then expected value of constant is constant itself.

E (a) = a 2) If y is a random variable & a,b are constant

E (ax) = a.E(X)+b 3) The expected value of the product of two independent random variable is equal to the product

of their individual expected values. If x,y are independent random variables then E (xy) = E (X). E(y)

4) The expected value of the sum or difference of two random variables is equal to the sum or difference to the expected of the individual random variables E (x + y) = E (x) + E (y) E (x-y) = E (x) - E (y)

5) The expected value of the deviation of random variable from its own expected value is equal to zero. E [X – E (x)] = 0 E [x – μ] = o

DEFINE VARIANS OF DISCRETE RANDOM VARIABLR , ALSO EXPLAIN ITS PROPERTIES:VARIANS:The variants of a probability distributation is defined as the expected value of the squares of the daviations of a random variable from its mean. If x is random variable with mean expected of E (x) = μ The variants of x is denoted by var(x) = σ² = E [x – E (x)] ² (a-b) ² = a² - 2ab +b² = E [x² + E (x) ² - 2xE (x)]



= E (x²) + [E(x)] ² - 2E(x). E (x) = E (x²) + [E(x) ²] - 2[E(x)] ² = E (x²) - μ² Var(x) = σ²= ∑ x². P (x) - μ² S.D = σ = ∑ . − ²S.D = standard deviation(x)

PROPERTIES OF VARIANS:i) The variants of a constant are zero. If a is constant then var (a)= 0 ii) The variants is independent of origin

Var (x+a) = var (x) Var (x-a) = var (x)

iii) When all the values are multiplied with a constant , the variants of the values is multiple with square of the constant Var (ax) = a². (Var (x)) Var (1/a. x) = 1/a² var (x)

iv) The variants of the sum or difference of two independent variables is the sum of their respective variances. Var (x+y) = var (x) + var (y) Var (x-y) = var (x) - var (y)

WHAT IS DISTRIBUTATION FUNCTION? WRITE DOWN ITS PROPERITIES:DISTRIBUTATION FUNCTION:A function showing probability that a random variable X has a value less than or equal to x is called distributation function or cumulative distributatuion function. The distributation function is denoted by F (x) = p (X≤x)

PROPERTIES OF DISTRIBUTION FUNCTION:The distribution has two following properties: (1) F (-∞)= 0

E (∞) = 1 Which means that F(x) is an increasing functioning having range from 0 t0 1.

(2) If x₁<x₂ then F(x₁)< F (x₂) TWO ELECTRIC LIGHT BULB ARE TESTED LET Y DENOTED THE

NO OF NON DEFECTIVE BULBS. THE POSSIBLE OUTCOMES WHEN TWO BULBS ARE TESTED ARE GIVEN BELOW.D= DEFECTIVE BULBN = NON DEFECTIVE BULB



OUTCOMES Y THE PROBABILITY FUNCTION OF Y

D 0 0/4N D 1 ¼D N 1 ¼N 2 2/4

4/4= 1

MEAN &VARIANS OR AVERAGE VARIANS:y P(y) Y(p(y)) y².p(y)0 0.3 0 01 0.3 0.3 0.32 0.2 0.4 0.83 0.1 0.3 0.94 0.1 0.4 1.6

Answer = 3. Average / mean∑ y= y.p(y) = 1.4 Variants σ²= ∑ y². P (y) - [∑ y. p(y ²)] = 3.6 – (1.4)² = 1.64

FROM THE FOLLOWING PROBABILITY DISTRIBUTATION FIND THE AVERAGE (MEAN)

y P(y) y [p(y)] y².p(y)0 0.3 0 01 0.3 0.3 0.32 0.2 0,4 0.83 0.1 0.3 0.94 0.1 0.4 1.6

∑ Y. p(y) = 1.4 ∑ y². P(y) = 3.6 Average / mean = ∑ y. p(y) = 1.4 Varians = ∑ y². P(y) – [∑y. p(y) ²]

= 3.6 – (1.4)²

= 1.64



WHAT ARE RANDOM NUMBERS? HOW CAN THEY BE GENERAITED EXPLAIN THE APPLICATIONS OF RANDOM NO’S?RANDOM NUMBER:Random number is the numbers obtained by some random process. These numbers are assumed to be randomly and uniformly distributed and are prepared by combining by the numbers 0,1,2,3,4,5,6,7,8,9. Each of these has equal chance of 1/10 of being selected. These numbers are combined into two-digit, three digits… numbers according to use. Two digit random numbers (00, 01, 02, and 99) are 100 in number and each of them has an equal probability of 1/100 of being selected. Three digit random numbers (000, 001, 002, 044, 194, and .999) are 1000 in number and each of them has an equal probability of 1/1000 of being selected. GENERATION OF RANDOM NUMBERS:Random numbers can be generated manually as well as mechanically. A table of random numbers can be generated manually by drawing cards from numbered card. The numbers 0, 1, 2, 9 are written on slips of paper and by shuffling they are drawn recorded and then are replaced before nest draw. Similarly, random numbers can be generated by rotating a numbered wheel. The mechanical method has then very much simplified the generation of random numbers. Calculators and computers are used in mechanical method. APPLICATION OF RANDOM NUMBERS:

1. One off the most important use of random numbers is the selection of simple random sample from finite population. Let consider there are 100 students in a class room and we want to select 10 students and random. The students can be numbered from 00 to 99. Then we consult the random number table and note the ten two- digit random numbers. We can read the random numbers table from any place we read the first two columns of random number table, the random numbers are 10, 37, 08,99,12,66,31,35,63 and 73. These ten random numbers represents those ten students who have been selected in our sample.

2. Random numbers are used in simulation techniques which are useful in the situations where actual experiment cannot be performed or cost of conducting. What are random numbers? How can they be generated explain the application of random no’s. Experiment is high for example we can obtain the outcomes of tossing a coin a large no. of time by the use of random no’s.

MALIK ALI RAZA

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Probability & Random Variables (Notes)