Probability Theory - I Sem Statistics Complementary Course

PROBABILITY THEORY Study material STATISTICS COMPLEMENTARY COURSE For I SEMESTER B.Sc. MATHEMATICS

(2011 Admission)

UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION CALICUT UNIVERSITY P.O. MALAPPURAM, KERALA, INDIA - 673 635

415

Probability Theory 

 

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UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION Study Material STATISTICS COMPLEMENTARY COURSE I SEMESTER B.Sc. MATHEMATICS

Probability Theory Prepared by : Sri.Z.A.Ashraf Department of Mathematics, Govt. Arts & Science College, Calicut - 18 Edited & scrutinized by : Dr.C.P.Mohammed (Retd.), Poolakkandy House, Nanmanda P.O. Calicut District. © Reserved Probability Theory 

 

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1 PROBABILITY     

1.1 INTRODUCTION    

 

1.3 RANDOM EXPERIMENT  

 

 

 

1.3.1 Sample Space  

   

1.2 EXPERIMENT  

 

 

 

1.3.2 Sample points  

 

 

1.3.3 Event    

 

1.4 CLASSICAL DEFIp[­NITION  

 

 

 

1.4.1 Definition  

 

 

 

1.4.2 Limitations Of Classical Definition   

 

1.5 FREQUENCY DEFINITION  

 

1.6 AXIOMATIC DEFINITION

1.6.1 Borel field or σ-algebra of events:1.6.2 Axioms of probability 1.7 THEOREMS ON PROBABILITY 1.8 CONDITIONAL PROBABILITY 1.9 MULTIPLICATION LAW OF PROBABILITY 1.10 INDEPENDENT EVENTS  

 

1.10.1 Pairwise and Mutual Independents  

1.11 BAYES THEOREM 1.12 MULTIPLE CHOICE QUESTIONS 1.13 EXAMPLES 1.14 PROBLEMS

2 RANDOM VARIABLES  2.1 DEFINITION:2.1.1 Types of Random Variables 2.2 PROBABILITY MASS FUNCTION 2.3 PROBABILITY DENSITY FUNCTION 2.4 DISTRIBUTION FUNCTION OF A R.V

2.4.1 Properties of Distribution function 2.5 MULTIPLE CHOICE QUESTIONS   2.6 EXAMPLES   2.7 PROBLEMS  

3 MATHEMATICAL EXPECTATION    3.1 INTRODUCTION   3.2 DEFINITION   3.2.1 Expectation Of Function Of R.V   3.2.2 Propertie Of Expectation  

3.3 MOMENTS    3.3.1 Moments about A   3.3.2 Moments about origin   3.3.3 Central Moments   3.3.4 Measures Of Central Tendency   3.3.5 Measures Of Variation   3.3.6 Measures Of Skewness    

 

 

3.4 MOMENT GENERATING FUNCTION    

3.3.7 Measures Of Kurtosis  

 

 

3.4.1 Definition  

 

 

 

3.4.2 Properties of mgf  

 

 

 

3.4.3 Characteristic Function  

 

3.5 MULTIPLE CHOICE QUESTIONS  

 

3.6 EXAMPLES  

 

3.7 PROBLEMS  

 

 

 

 

 

 

4 CHANGE OF VARIABLES            

4.1 THEOREM  

        

4.2 MULTIPLE CHOICE QUESTIONS  

        

4.3 EXAMPLES  

        

4.4 PROBLEMS  

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SYLLUBUS Semester I- PROBABILITY THEORY Module 1. Probability concepts: Random experiment, sample space, event, classical definition, axiomatic definition and relative frequency definition of probability, concept of probability measure. Addition and multiplication theorem (limited to three events). Conditional probability and Bayes Theorem-numerical problems. (25 hours) Module 2. Random variables: Definition- probability distribution of a random variable, probability mass function, probability density function and (cumulative) distribution function and their properties. (15 hours) Module 3. Mathematical Expectations: Expectation of a random variable, moments, relation between raw and central moments, moment generating function (mgf) and its properties. Measures of skewness and kurtosis in terms of moments. Definition of characteristic function and its simple properties. ( 20 hours ) Module 4 Change of variables: Discrete and continuous cases (univariate only), simple problems. (12 hours) Book for Reference 1. V.K. Rohatgi: an Introduction to Probability theory and Mathematic Statististics, Wiley Eastern. 2. S.C. Gupta and V.K.Kapoor: Fundamentals of Mathematical Statistics, Sultan Chand and sons 3. Mood A.M., Graybill. F.A and Boes D.C: Introduction to Theory of Statistics McGraw Hill. 4. John E Freund: Mathematical Statistics (Sixth Edition), Pearson Education (India),New Delhi.

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Chapter 1 PROBABILITY 1.1 INTRODUCTION In our day to day life, we may face many situations where uncertainty plays a vital role. We usually use statements like ”there is a chance for rain today” or ”probably I will get A grade in university examination” etc. In all these contexts the term chance or probably is used to indicate uncertainty. The word Probability is related with the occurrence of uncertainty, and Probability theory is the discipline which tries to quantify the concept of chance or likelihood.

1.2 EXPERIMENT An experiment is an activity which can be repeated in more or less same condition and will have some specific outcome or outcomes. By combining hydrogen with oxygen under certain conditions results in formation of H2O or tossing a coin and checking whether the face value is Head or Tail are examples of experiments. Generally, experiment are of two types:

1. Deterministic experiment:- In deterministic experiment, if we repeat the experiment, outcome or outcomes will not change; provided the initial conditions of the experiment remains same. Experiments like combining hydrogen and oxygen are examples of deterministic experiments. 2. Random experiment or stochastic experiment:- In random experiments, result or outcome may not be same if we repeat the experiment in a similar conditions. But result will be one among a set of possible out-comes. Tossing a coin or throwing a die are typical examples of this type of experiments.

1.3 RANDOM EXPERIMENT A Random experiment may be defined as an experiment with more than one possible outcome, which may be repeated any number of times under more or less similar conditions and the outcomes of which vary irregularly from repetition to repetition. For example, throwing a die and looking on the face value is a random experiment as the face value will be any value from the set {1, 2, 3, 4, 5, 6}, which cannot be predicted before conducting the experiment. Similarly taking a ball from a box containing different coloured balls and looking the colour of the selected ball, inspecting life length of a bulb etc are random experiments. Random experiments are common in all kinds of research activities. Administering a new medicine on a group of people and looking on effectiveness, Taking a product from factory and verifying whether it is defective, Providing training for better academic achievement and measuring its effectiveness etc are some examples of random experiments.

1.3.1 Sample Space The set of all simple outcomes or sample points of a random experiment is called its sample space, which is denoted by the letter S or Ω. Eg: Sample space of the random experiment of throwing a die is S = {1, 2, 3, 4, 5, 6} and sample space of the random experiment of tossing a coin is, S= {H,T }.

1.3.2 Sample points Every elements in the set of all outcomes of the random experiments are called sample points.

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1.3.3 Event Those subsets of the sample space to which probabilities are assigned by probability set function under consideration are called an Event. Eg: Consider the random experiment of throwing a die and the sample space will be, S = {1, 2,3,4,5, 6}. One can define an event E as getting odd number. Then E = {1, 3, 5}. Note that the event E is a subset of sample space S.

Types of Events •

Simple and Compound Event :- If an event has only one outcome, that is, only one sample point it is called simple event. Eg: The event of getting head in the experiment of tossing a coin. If an event has more than one sample points then it is called compound event. Eg: The event of getting odd or even when rolling a die.

• Mutually Exclusive Event :- Two events A and B are said to be mutually exclusive if the occurrence of any one of the events exclude the occurrence of the other.

• Exhaustive Event:- Let E1, E2, E3. . . En are n events of sample space then E0 is are said to be exhaustive if [n i=1Ei = S.

1.4 CLASSICAL DEFINITION There are different approaches in defining probability. The oldest and simplest one is classical definition or mathematical definition.

1.4.1 Definition If a random experiment results in n exhaustive, mutually exclusive and equally likely cases, m of them are favourable (m ≤ n) to the occurrence of an event A, then the probability of A, denoted as P(A), is defined as P(A) =

 

Note:1. We can also say the above statement as the ’odds in favour of A are m : (n - m) or odds against A are (n - m) : n.

2. For any event A, 0 ≤P(A) ≤1 3. If A is a sure event, then P(A) = 1

4. If A is an impossible event, then P(A) = 0

1.4.2 Limitations Of Classical Definition Classical definition of probability is very easy to understand. But the definition may not be applicable in all situations. Following are some of the limitations of classical definition of probability.

1. If the events cannot be considered as equally likely, classical definition fails. 2. When the total number of possible outcomes n become infinite this definition cannot be applied. Probability Theory 

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1.5 FREQUENCY DEFINITION Let the trials be repeated over a large number of times under essentially homogeneous condition. The limit of the ratio of the number of times of an event A happens (say, m) to the total number of trials (say n), as the number of trials tends to infinity is called the probability P(A) = lim

of the event A .

 

1.6 AXIOMATIC DEFINITION Axiomatic definition of probability was introduced by Russian mathematician A.N. Kolmogrov and it approaches probability as a measure. To explain axioms of probability we have to define Borel field.

1.6.1 Borel field or σ-algebra of events:Let S be a non-empty set and F be a collection of subsets of S. Then F is called borel field (σ- field) if 1.

is non-empty

2. The elements of

are subsets of S

thenAc

3. IfA

4. The union of any countable collection of elements of for

= 1, 2,... then A1

A2

is an element of

. i.e. if Ai

... 

1.6.2 Axioms of probability A probability function P[.] is a set function with domain A (σ- field) and counter domain the interval[0,1] which satisfies the following axioms:

1. P(A) 0 for every A

 

2. P(S)=1 3. If A1, A2,... is a sequence of mutually exclusive events in A thenP[

  

A ] = Σ    P(A ).

1.7 THEOREMS ON PROBABILITY Theorem-1 If φ denotes impossible event then P(φ) = 0 Proof:- Let S be the sample space and φ be null (impossible) event. We have S = S. Therefore, P(S [ φ) = P (S). i.e P (S) + P (φ) = P (S) i.e 1 + P (φ) = 1. Hence P (φ) = 0. proved.

Theorem-2 For any event A, P (Ac) = 1 - P (A) Proof:- If Ac is the complement of A, we have A Ac = S. Then P (A P (Ac) = 1. i.e P (Ac) = 1 - P (A). proved Probability Theory 

Ac) = P (S).i.e P (A) + Page 7 

Theorem -3 (Addition Theorem) If A and B are two events, then P (A Proof:- We have A

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

(Ac B)

B =A

B) = P (A) + P (Ac

therefore P (A

B)

B) ................. (1)

Now, B =(A B)  (Ac B) and hence P (B) = P (A B)+P (Ac (B) - P (A B) ........................................................ (2)

B) ie; P (Ac

B) = P

Substituting (2) in (1), P (A

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

B). proved

Note 1:- If the two events A and B are disjoints P(A

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

Note 2:- For three events A,B,C P(A

B

C)=P(A)+P(B)+P(C)-P(A B)-P(B C)-P(A C)+P(A B C).

Note 3:-For n events A1,A2,...An P(A1 A2 . . .  . . An).

An) = ∑

P(A )- ∑∑P(A

A ) + ∑∑∑P(Ai Aj Ak)-. . . + (-1)(n+1)P(A1 A2

.

Theorem 4 For every sequence of events A1, A2,..., P[

  

A ]) P(A1)+P(A2)+...

Proof:- By considering the infinite operations into union of disjoint events,   

i.e.   

P

A

Ai

A1(Ac1 A2) 

A ) ≤P(A1) + P(Ac1 A2) + P(Ac 1

P(A1) + P(A2) + P(A3) +……… i.e. P

(Ac 1 Ac 2 A3)…………

  

A)≤∑

P(A

.

Ac 2 A3)+....

Since P(Ac1 A2)≤P(A2) etc.,

)

1.8 CONDITIONAL PROBABILITY Let A and B be any two events. The probability of the event A given that the event B has already occured is known as conditional probability of A given B, denoted by P(A|B), and is defined as P(A|B) = P(A \ B)/P(B), provided P(B)6= 0 Similarly the conditional probability of B given A is defined as, P(B|A) = P(A \ B)/P(A), Provided P(A) 6= 0 Remarks:1. For P(B) > 0P(A|B) ~ P(A) 2. P(A|B) is not defined if P(B) = 0 3. P(B|B)=1 Probability Theory 

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1.9 MULTIPLICATION LAW OF PROBABILITY For any two events A and B P(A B) = P(A)P(B|A) provided P(A) > 0 or P(A B) = P(B)P(A|B) provided P(B) > 0 , where P(A|B) and P(B|A) are the conditional probabilities of A and B respectively Proof Suppose the sample space consists of N sample points of which NA are favourable for the occurrence of A and NB are favourable for the occurrence of B. Let NAB be the number of occurrence favourable to the compound event A B, then the conditional probabilities are given by P(A) =

,P(B) =

and P(A

B) = =

 〰

 Now the conditional probability P(A|B) refers to the

sample space of NB occurrences. Out of which NAB occurrences pertaining to the occurrence of A Therefore P(A|B) = Now P(A B) = P(A

B) =

N

AB

N

AB

N

/N = N

AB

AB

/NB and P(B|A) = NAB/NA

/NA.NA/N = P(A|B)P(A) ........................ (1)

/N = NAB/NB.NB/N = P(A|B)P(B) ...................(2)

Thus we get P(A

B) = P(A)P(B|A), P(A) >0 =P(B)P(A|B),P(B) >0

1.10 INDEPENDENT EVENTS Two or more events are said to be independent if the probability of any one them is not affected by the supplementary knowledge concerning the materialisation of any number of the remaining events.

1.10.1 Pairwise and Mutual Independents A set of events A1, A2, . .. ,An are said to be pairwise independent if every pair of different events are independent. i.e ,P(A Aj) = P(A )P(Aj) for all and j,  =6 j. A set of events A1, A2 . . . An are said to be mutually independent if P(Ai Aj ……….  Ar) Ar) of A1,A2,…………An = P(Ai)P(Aj)………….P(Ar) for every subset (Ai, Aj,

Note:- If A and B are two independent events,then P(A B) = P(A).P(B).

1.11 BAYES THEOREM Let S be a sample space partitioned into n mutually exclusive event B1, B2, . . . Bn such that P(Bi) > 0, = 1, 2, 3……….. n. Let A be any event of S for which P(A) ≠ 0, Then the probability for the event B ( 牵 = 1, 2, . . . n) given the event A is P(B /A) = (P(B )P(A/B ) i

∑

Probability Theory 

i

P(Bj )P(A/Bj)

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Proof:From the definition of conditional probability, P(B /A) =

P(A B ) P(A)

=

P(Bi)P(A/Bi) P(A)

But A=A Therefore P(A) = P[U

(A

…………… . .

)=

S =A (U B )] =∑

P(Bi)P(A/Bi

Substituting (2) in (1) we get P(B /A) =

(1)

(A

B)

…… (2)

P(B )P(A/B )

∑

P(B )P(A/B ). proved

1.12 MULTIPLE CHOICE QUESTIONS 1. Two coins are tossed simultaneously. If one of them turned head, what is the probability that the other one also turn head? (a) 0.01 (b) 0.05 (c) 0.25 (d) 0.50 2. In the following functions, which constitute S=a,b,c as a probability space: (a) p(a) = 0.5p(b) = 0.3p(c) = 0.3 (b) p(a) = 0.5p(b) = 0.2p(c) = 0.3 (c) p(a) = 0.5p(b) = -0.3p(c) = 0.3 (d) p(a) = 0.5p(b) = 0.7p(c) = - 0.2 3. If an unbaised coin is tossed once, then the two events head and tail are (a) Mutually exclusive (b) Equally likely (c) Exhaustive (d) All the above 4. If the letters of teh word ’REGULATIONS’ be arranged at rando what is the chance that there will be exactly 4 letters between R and E (a)

10/55

(b)

6/55

(c)

10/25

(d)

None of thsese

5. An event that can be split into further events is known as Probability Theory 

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(a) Simple event (b) Complex event (c) Composite event (d) None of these 6. If A and B are two events such that P(A) = 1/4 , P(B) = 1/3 and P(A be P(A/B)

B) = 1/2 what will

(a) 1/4 (b) 1/2 (c) 1/3 (d) None of these 7. Sangeetha speaks truth in 20% cases and Jaseena speaks truth in 30% cases, what is the probability that they will contradict each other in a particular issue. (a) 1/4 (b) 1/2 (c) 1/3 (d) None of these 8. If A and B are independent events, what is P(A/B) (a) P(B/A) (b) P(A) (c) P(B) (d) None of these 9. Which one of the following statement(s) are true (a) If A and B are mutually exclusive P(B [ B) = P(A) + P(B) (b) If A and B are mutually independent P(B \ B) = P(A)P(B) (c) Both of these are true (d) None of these 10. If A and B are independent events then which of the following statements are true (a) A and Bc are independent (b) Ac and Bc are independent (c) Ac and Bc are independent (d) All the above

1.13 EXAMPLES Probability Theory 

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Example 1 Three coins are tossed. What is the probability of getting 1. All Heads 2. Exactly one Head 3. Exactly two Heads 4. Atleast one head Solution Here the sample space is: S= (HHH,HHT,HTH,THH,TTH,THT,HTT,TTT) Therefore total number of cases is 8.

1. There is only one case with all heads as (HHH), therefore, number of favourable cases is1. Hence P (All heads) = .

2. There are three cases with exactly one head as (HTT ,THT ,TTH) Hence P (exactly one head) = .

3. There are three cases with exactly two heads, as (HHT, HTH, THH) Hence P (exactly two heads) =

.

4. In all 7 cases except (TTT) there is atleast one Head, Hence P(atleast one head)=

Example 2 A box contains 5 red,3 white 6 blue balls.If 3 balls are drawn at random, determine the probability that 1. All 3 are blue 2. 2 are red 1 is white 3. One of each colour is drawn Solution Given that the box contains 5 red, 3 white 6 blue balls. If three balls are drawn at random, there will be 14C3 possible outcomes.

1.

Here, since there are 6 blue balls the number of favourable cases will be 6C3 .

Therefore the P(getting 3 blue balls) =

6C3 14C3 

= 91 5

5C2 X 3C1  15 = 182  14C3  5C2 X 3C1 X 6C1  And P( One red, One blue and oOne white balls) = = 14C3 

2. Similarly, P(getting 2 Red and 1 white ball)= 3.

45 182 

Example 3 A card is drawn from a well shuffled pack of playing cards.Find the probability that 1. A club 2. A king 3. The ace of spade

Solution Probability Theory 

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There are 52 different cards in a pack of playing cards. Total number of cases= 52C1 = 52

1. There will be 13 club cards in a packet. Therefore favourable cases= 13C1 = 13. Hence probability of getting a club will be

 =

2. There are 4 kings in a packet. Therefore number of favourable cases = 4C1 and probability of getting a king will be

=

3. In a pack of cards there will be only one ace of spade and hence probability of getting an ace of spade will be =

Example 4 Given P(A) = 0.30, P(B) = 0.78, P(A 1. P(Ac

B) = 0.16.. Evaluate

Bc)

2. P(Ac[ Bc) 3. P(A

Bc)

Solution Given P(A)=0.30, P(B)=0.78, P(A B)=0.16 1. P(Ac Bc) = P(A B)c = 1 - P(A B) =1-P(A)+P(B)P(A B) =1-0.30+0.78-0.16 =0.08 2. P(Ac Bc) = P(A B)c = 1 - P(A B) =1-0.16=0.84 3. P(A

Bc) = P[A - (A B)] =P(A)-P(A B)

=0.30-0.16=0.14.

Example 5 The probability that a student passes Statistics course is passes both Statistics and mathematics course is one course is

.

and the probability that he

.The probability that he passes atleast

what is the probability that he passes mathematics course?

Solution Let A be the event of a student passes Statistics course and B be the same student passes Mathematics course.

Given P(A) =

 

, P(A

Probability Theory 

B) =

,

P(A B) =45 . We have to find P(B). Page 13 

By addition theorem, P(A B) =P(A)+P(B)-P(A B) i.e, = + P(B) Therefore P(B) =

-  +

=

Example 6 A problem in statistics is given to 3 students A,B and C whose chances of solving it are  ,  and respectively. What is the probability that the problem will be solved? Solution Let us define the events as A

-The problem is solved by the student A

B

-The problem is solved by the student B

C

-The problem is solved by the student C

Then P(A) = ,P(B) =  andP(C) = The problem will be solved if atleast one of them solves the problem. That means we have to find P(A B C). Now P(A B C)

= 1 - P(A B C)c = 1 - P(Ac Bc

Cc)

=1 - ((1 - )(1 - )(1 - )) =

Example 7 If two dice are thrown, what is the probability that the sum is equal to 9? Solution When two dice are thrown, there will be 36 mutually exclusive and equally likely cases like (1, 1), (1,2), (1,3) .. . (6,6). Among these cases, exactly 4 cases viz. (3, 6), (6, 3), (4, 5), (5,4) are favourable to the event sum equal to 9. Therefore P(sum equal to9) =

Example 8 What is the probability of getting 53 Sundays in a leap year Solution We have in a leap year there will be 366 days. So there will be 52 complete weeks and two days remaining. These two days may be any one from following 7 pairs: (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday) and (Saturday, Sunday). Here two pairs are favourable towards sunday. Hence probability of a leap year containing 53 sudnay will be

Example 9 Let A and B be two events associated with an experiment and suppose P(A) = 0.5 while P(AorB) = Probability Theory 

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0.8 .let P(B) = p. For what values of p are A and B independent? Solution If A and B are independent,we know that P(A ⇒ 0.8 =0.5 + p - 0.5 p

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

⇒ 0.5p= 0.3 ⇒ p=

.

Example 10 Suppose there are two bags with first bag contains 3 white and 2 black balls, second bag contains 2 white and 4 black balls. One ball is transferred from bag I to bag II and then a ball is drawn from the latter. It happens to be white. What is the probability that the transferred ball is white? Solution Let A be the event f drawing a white ball; B1 is the event of transferring a white ball from bag I and B2 is the event of transferring a black ball from bag I to bag II. Then P(B1) =

.

and P(B2) =

Also P(A/B1) =

and P(A/B2) =

P[transefered ball was white given that the ball drawn is white] = P(B1/A) From Bayes theorem, P(B1/A) =

  /

/

.

   .

   /

 .

  =

  ,    ,         ,

=

1.14 PROBLEMS 1.

For two events A and B, if P(A) = 0.3 ,P(B) = 0.2 and P(AB) = 0.1 , find the probability that exactly one of the events will happen.

2.

The chance of two students to win a competition are

and

respectively. If they participate in the same

competition what is the probability that at least one will win? 3.

A box contains 2 silver and 4 gold coins and a second box contains 4 silver and 3 gold coins. If a coin is selected at random from one one box, what is the probability that it is a gold coin?

4.

An integer is chosen at random from the first hundred integers. What is the probability that the integer chosen is divisible by 3 or 5?

5.

A, B and C tosses a coin in order. The first one to toss head will win. What are their respective probabilites.

6.

A box contains 5 red, 6 white and 3 blue balls. If three balls are drawn at random what is the probability that all are blue?

7.

Sunitha and Saleena stand in a circle with 10 other girls. If the arrangement of 12 girls are at random, find the chance that there are exactly three girls between Sunitha and Saleena?

8.

Two dice are thrown at random and face value is noted, If A is the event that first die shows 2 or 5 or 6 and B is the event of sum of the face values is 9, check whether A and B are independent. ?

9.

Three firms A, B, and C supply 25%, 35% and 40% of chairs needed to a college. Past experience shows that 5%, 4%, and 2% of the chairs produced by these companies are defective. If a chair was found to be defective, what is the probability that chair was supplied by firm A?

10. Three urns are given each containing red and white balls. Urn I contains 6 red and 4 white balls, Urn II contains 2 red and 6 white balls and Urn III contains 1 red and 2 white balls. An urn is selected at random and a ball is drawn. If the ball is red what is the chance that it is from first Urn?

Probability Theory 

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Chapter 2 RANDOM VARIABLES By the term random variable we mean a variable which takes real values in accordance with the change in the outcome of the random experiment. Or in other words, it is a real valued function defined over the sample space Ω. Hence its domain will be Ω and its range will be real line R. Random variable is also known as stochastic variable and it is denoted as X(ω) or simply X.

2.1 Definition:Let (Ω,S) be a sample space. A finite single-valued function that maps Ω into is called a random variable, denoted by , if the inverse images under of all Borel sets in are events. i. e. X-1(B) = [ space

:

(

)

B] S for all B .

For example, let us consider a coin tossing experiment. Here the sample = will be Ω = (H, T). Now define a variable X(ω) such that X(ω) 8

      0 ifω  Head Here X takes real values either 0 or 1, in accordance with the outcome Head or   1 ifω  Tail Tail. Thus X is a random variable.

2.1.1 Types of Random Variables:Random variable (r.v ) can be either discrete or continuous. 1. Discrete :A r.v is said to be discrete if its range includes finite number of values or countable infinite number of values. For example, Number of road accidents occurs in a day in a city. 2. Continuous :A r.v which can assume any value from a specified interval of the form [a,b] is known as continous r.v. For example, Lifetime of mobile phone or height of a randomly selected student from your college etc can be treated as continuous r.vs. Note:If X is a rv on (Ω, S) then aX + b will also be a r.v on (Ω, S) where a, b are constants. Also noted that If X is r.v, then X2 and

 

will also be rv’s.

2.2 Probability mass function If X is a discrete r.v, then the probability mass function (pmf) or probability distribution of X is defined as: P(X=  ) Properties:1. As

is a probability function,

2. The total probability Σ

 

0.

= 1.

2.3 Probability Density Function Probability Theory 

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If X is a continuous random variable, then the probability density function or probability distribution of X is defined as

≤ X ≤  + d  ) Properties:-

= P(

 

is a probability function,

1. As

∞ ∞

2. The total probability

0

= 1.

2.4 Distribution Function of a r.v Let X be a r.v, either discrete or continuous, its cumulative distribution function or simply distribution function is defined as : F( ) = P(X ≤ ) ∑

∞ = if X is discrete and ∞

∞

if X is continuous.

2.4.1 Properties of Distribution function if F(x) is the distribution function of a r.v. X, then 1. F(-∞) =0 2. F(∞)=1 3. If a < b, then F(a) ≤ F(b) (i.e F( ) is non decreasing)

, lim

4. F(x) is continuous from the right ie

) = F(

5.

) - F(

 =

-1)

2.5 MULTIPLE CHOICE QUESTIONS 1. Values taken by a r.v will be always a (a)

Positive integer

(b)

Positive real number

(c)

Real Number

(d)

None of these

2. If the pdf of a r.v is given as

=    

kx if 0  2  0 otherwise

Find the value of k (a)

1

(b)

1/2

(c)

1/4

(d)

None of thsese

3. If X is a r.v, P(X ~ x) is known as

4.

(a)

Probability mass function

(b)

Probabiliyt density function

(c)

Probability distribution function

(d)

None of thsese

If f(x) is the pdf of X then P f(x) is equal to

Probability Theory 

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(a) 0 (b) -1 (c) ∞ (d) 1 5.

The distribution function of a rv will be (a) Increasing function (b) Decreasing function (c) Constant function (d) Non-decreasing function

6.

If F(x) is the distribution function of r.v X, P(a ≤ X ≤ b) will be (a) (b) (c)

7.

(d)  鑪(b) - (a) Value of k in the following pdf is (a)

 

(b)

 

(c)

 

(d) 8.

F(a) - F(b) F(b) - F(a) a-b x:

0

1

2

(x) : k

k

2k

None of these 

If the pdf of a continuous rv is given as

( ) = 2  if 0≤ ≤1 0 otherwise

What is the value of P(X = (a) (b) (c) (d) 9.

0.5  1  0  None of these 

If X1 and X2 are two r.vs which one of the following statements are true (a) X1 + X2 will be a rv. (b) cX1 & cX2 will be a rv where c is a cont. (c) X1 - X2 will be a rv. (d) All the above

10. Which one of the following statements are ture (a) The maximum value of f(x) is 1 (b) The maximum value of F(x) is 1 (c) Both (a) and (b) (d) None of these

Probability Theory 

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2.6 EXAMPLES EXAMPLE 1 Check whether the following function is a pdf or not ½ 0  2  0 otherwise

=     Solution If

( ) is a pdf, we have

_

d 

Here ∫  ( ) d  =

( )d  = 1   |

|

2 0

1

EXAMPLE 2 Check whether following is a pdf.   for

( )=

= 1,2, 3, ………………

0 elsewhere

Solution If

( ) is a pdf, we have ∑

Here   ( ) =

 

( )=1  … … … … … … …

  ≠1

so, given f(x) is not pdf EXAMPLE 3 If the pdf of a r.v is given as f(x) =

  if

= 1,2, 3,4, 5

Find the distribution function of X ?

0 elsewhere

Solution The Distribution function is defined as F(x) = P(X ≤ x) Therefore the Distribution function of given pdf is

0

F(x) =

1

if < 1 if 1 ≤ < 2 if 2 ≤

b) = 0.05

3x2dx

=

3x2dx

⇒

1

3x2 = 0.05 ⇒ 1-b3 = 0.05 ⇒ b = 0.98

EXAMPLE 7 Let the distribution function of X be 0 if < 0 F (x) 0≤ ≤1 1 >1 Find P(2X+3 ≤ 3.6) Solution P(2X + 3 ≤ 3.6) = P(X ≤ =

.

P(X ≤ 0.3)

= F(0.3) = 0.3

EXAMPLE 8 If a continous r.v X has the pdf 2x if 0 < x ≤ 1 ( )

0

elsewhere

Find the distribution function F(X) Solution

F(x) =

∞

(t) dt =

2tdt = [t2] = x2

Therefore the distribution function is 0 if < 0 2 F (x) 0≤ 1 EXAMPLE 9 Suppose the life in weeks of a certain kind of computers has the pdf: when x ≥ 100 ( )

0

when x < 100

What is the probability that none of three such computers will have to be replaced during the first Probability Theory  

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150 weeks of operation?

Solution Probability that a computer will last for first 150 weeks is given by P(X ≤150) = P(0