Class 12

Balbharti Maharashtra State Board Class 12 Economics Solutions Chapter 3B Elasticity of Demand Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Economics Solutions Chapter 4 Supply Analysis

1. Complete the following statements:

Question 1.
When the supply curve is upward sloping, its slope is ……………….
a) positive
b) negative
c) first positive then negative
d) zero
Answer:
a) positive

Question 2.
An upward movement along the same supply curve shows ………………..
a) contraction of supply
b) decrease in supply
c) expansion of supply
d) increase in supply
Answer:
c) expansion of supply

Question 3.
A rightward shift in supply curve shows ………………..
a) contraction of supply
b) decrease in supply
c) expansion of supply
d) increase in supply
Answer:
d) increase in supply

Question 4.
Other factors remaining constant, when less quantity is supplied only due to a fall in price, it shows ………………..
a) contraction of supply
b) decrease in supply
c) expansion of supply
d) increase in supply
Answer:
a) contraction of supply

Question 5.
Net addition made to the total revenue by selling an extra unit of a commodity is ………………..
a) total Revenue
b) marginal Revenue
c) average Revenue
d) marginal Cost
Answer:
b) marginal Revenue

2. Complete the Correlation:

1) Expansion of supply: Price rises:: Contraction of supply: ………………….
2) Total revenue : …………………. :: Average revenue :TR/TQ
3) Total cost : TFC + TVC :: Average cost : ………………….
4) Demand curve : …………………. :: Supply curve : Upward
5) …………………. : Change in supply :: Other factors constant: Variation of supply
Answers:
(1) Price falls
(2) PxQ
(3) TC ÷ TQ
(4) Downward
(5) Other factor changes

3. Give economic terms:

1) Cost incurred on fIxed factor.
2) Cost incurred per unit of output.
3) Net addition made to total cost of production.
4) Revenue per unit of output sold.
Answers:
(1) Fixed Cost
(2) Average Cost
(3) Marginal Cost
(4) Average Revenue

4. Distinguish between:

Question 1.
Stock and Supply.
Answer:

Stock	Supply
(a) Stock refers to the total quantity of commodity available with producer for sale.	(a)Supply is that part of stock which the seller is willing to offer for sale at a given price.
(b) It is outcome of production. If production increases, stock will also increase.	(b) It is outcome of stock. Stock is the basis of supply.
(c) It is a fund or reservoir and a static concept (inelastic).	(c) It is a flow concept. It changes according to change in price (elastic).
(d) It can exceed supply.	(d) It cannot exceed stock.

Question 2.
Expansion of Supply and Increase in Supply.
Answer:
Expansion / Extension of Supply

1. When the supply of a commodity rises only due to the rise in the price of the commodity, then it is said to be extension in supply.
2. Extension of supply is a case of variation in supply.
3. Rise in price is the only factor due to which supply expands / extends.
4. When there is extension in supply, there is an upward movement on the same supply curve.
   

Increase in Supply :

1. The supply is said to increase if at the same price more is supplied.
2. Increase in supply is a case of changes in supply.
3. Supply increases due to
   (1) fall in cost of production
   (2) improvement in transport facility
   (3) introduction of modern technology
   (4) government subsidies
   (5) more imports etc.
4. When there is an increase in supply, the supply curve shifts to the right of original supply curve.

Question 3.
Contraction of Supply and Decrease in Supply.
Answer:
Contraction of Supply

1. Contraction of supply occurs when quantity supplied of a commodity falls due to a fall in price alone.
2. It is a case of variation in supply.
3. Supply contracts due to fall in price alone.
4. When there is a downward n curve.

Decrease in Supply

1. Decrease in supply occurs when less quantity is supplied at the same price.
2. It is a case of changes in supply.
3. Supply decreases due to –
   (1) increase in cost of production
   (2) transport strike
   (3) outdated technique
   (4) heavy taxes imposed by government.
   (5) more exports etc.
4. When there is curve shifts to curve.
   

Question 4.
Average Revenue and Average Cost.
Answer:

Average Revenue (AR)	Average Cost (AC)
(a) Average revenue refers to average income earned per unit of a sold commodity.	(a)Average cost refers per unit of cost of production of a commodity produced.
(b) It is calculated by dividing total revenue (TR) earned by number of unit sold.	(b) It is calculated by dividing total cost (TC)by number of units of that commodity produced.
(c) Symbolically it in expressed as \( [latex]\frac { Total Revenue }{ Total Quantity sold }\) [/latex]	(c) Symbolically it is expressed as \(\frac { Total Cost }{ Total Quantity produced }\)
E.g. If TR from sale of 10 units of a commodity is Rs. 1000 then, AP =  1000/10 = Rs. 100	E.g. If TC of 100 units a commodity is Rs. 1000 then, AC =  \(\frac { 1000 }{ 100 }\) = Rs. 10

5. Observe the following table and answer the questions.

A) Supply schedule of chocolates

Price in Rs.	Quantity supplied in units
10	200
15	………
20	300
25	350
30	……..
35	……..
40	……..

Question 1.
Complete the above supply schedule.
Answer:

Price in Rs.	Quantity supplied in units
10	200
15	250
20	300
25	350
30	400
35	450
40	500

Question 2.
Draw a diagram for the above supply schedule.
Answer:

Question 3.
State the relationship between price and quantity supplied.
Answer:
This diagram shows the direct relationship / between price and quantity supplied of) chocolates. When its price is ? 10, 200 units ( are supplied and as price rises to 15, 20, 25 ? …. and so on, quantity supplied also rises to )
When the schedule is plotted on the graph we 250, 300, 350 and so on. This is the law of supply of an individual firm.

B) Observe the market supply schedule of potatoes and answer the following questions.

Question 1.
Complete the quantity of potato supplied by the firms to the market in the above table.
Answer:

Question 2.
Draw the market supply curve from the schedule and explain it.
Answer:

When the schedule is plotted on the graph we get a market supply curve ‘SS’ which is upward sloping. This curve shows that as price rises from ₹ 1 to ₹ 2, supply rises from 1oo to 112 kg, but when price rises from ₹ 2 to ₹ 3, supply rises to a greater extent from 112 kg to 155 kg in the market. When price rises to ₹ 4 Supply falls from 155 kg to 154 kg. This may be because of perishable or seasonal good that supply could not Jj be increased and supply falls. This show backward bending supply curve, showing partly an exception to the supply curve.

6. Answer the following questions:

Question 1.
Explain the concept of total cost and total revenue.
Answer:
Total Cost (TC): It is the total expenditure incurred by a fir m on the factors of production required for the production of goods and services. Total cost is the sum of Total Fixed Cost (TFC) and Total Variable Cost (TVC). Total Fixed Cost is the cost incurred on fixed factors of production like land, factory, building, capital, etc. These factors cannot be changed in the short period. They remain constant. Total Variable Cost is the cost incurred on variable factors such as raw – materials, labour, etc. These factors can be varied or changed according to the change in output level. So the variable cost varies. Total Cost = Total Fixed Cost + Total Variable Cost
i.e., TC = TFC + TVC
TC increases as the level of output increases.

Total Revenue :
(Income) refers to total receipts of the firm from its sales of commodity. It is obtained by multiplying the price per unit of the  commodity with the total number of units!; of commodity sold to the consumers. Thus, Total Revenue = Price per unit Total  Number of units of commodity sold.
Example : If the firm sells lo units of a commodity at ₹ 100 per unit then total revenuewifibe TR = 100 x 10. TR= ₹ 1000

Question 2.
Explain determinants of supply.
Answer:

1. Cost of Production : Changes in the price of factors of production like rent, wages, interest affects the cost of production. When cost of production increases, supply decreases.
2. Price of Other Goods : The supply of a given commodity depends on the price of other commodity. E.g. if the price of wheat rises and that of rice remains the constant, then the producer will think of producing J more of wheat. This will affect the supply of rice.
3. price of the Commodity : Price is an important factor influencing the supply. More is supplied at a higher price and less at a lower price. So price and supply are 5 directly related.
4. Climatic Conditions : The supply of j commodity is also influenced by the forces
5. Government Policy : Government policies like taxation, subsidies, industrial policies etc., may encourage or discourage production and supply. A tax on the commodity will raise the cost of production and reduce the supply while a subsidy on the other hand will provide an incentive to increase production and supply.
6. Exports and Imports : When the
   government resort to imports, supply expands, at the same time heavy exports would reduce the supply in the domestic market.
7. Nature of Market : In a competitive market, the supply would be more but in a monopoly market the seller may create artificial scarcity to raise the price.
8. Future Expectation : If future trends indicate a rise in price, the supply decreases at present. On the other hand if the sellers expect the future price to fall, supply would increase in the current period.
9. Technique of Production : Improvement in the technique of production will lead to increase in supply. Application of advanced technology enables the producer to produce goods on large scale at a lower cost and lesser price.
10. Infrastructure Facility : If means of transport and communication are well developed, the extent of market would be wide. i.e. supply will increase.
11. Natural and Man-made Calamities : Natural calamities like earthquake, cyclone, flood etc., will affect the supply in the market. Even man-made calamities like a bomb-blast, affects supply. Even a strike call can affect supply in the market.

7. Answer in detail :

Question 1.
State and explain law of supply with exceptions.
Answer:
Law of Supply :
(A) Introduction : The law of supply was introduced by Dr. Alfred Marshall in his book “Principles of Economics” published in 1890. The law establishes a functional relationship between the price of a commodity and quantity supplied of that commodity. It explains the general tendency of the sellers in offering more goods for sale at a higher price than at a lower price.

(B) Statement of the Law : According to Prof. Alfred Marshall “Other things remaining constant, the higher the price of the commodity, greater is the quantity supplied and lower the price of the commodity, smaller is the quantity supplied.”In other words, quantity supplied of a commodity varies directly with price i.e., with a fall in price supply contract and with a rise in price supply expands.
S = f (P) [S = Supply, P = Price, f = Function of]
The law can be better understood with the help of a market supply schedule and market supply curve.

(C) Market Supply Schedule : Market supply schedule is a tabular representation of various quantities of a commodity offered for sale by all the sellers in the market at different prices during a given period of time. The schedule is a hypothetical one except one price rest are imaginary prices.

The above schedule clearly shows that sellers in general want to sell more at high prices and less at low price. E.g., at a low price of Rs.10 per unit the seller supplies only 100 units per day and at high price of Rs. 50 the supply rises to 500 units of ‘X’ per day.

(D) Market Supply Curve : It is graphical representation of the above market supply schedule. Price is measured on ‘Y’ axis and quantity supplied on ‘X’ axis and above schedule is plotted. We derive a supply curve SS.

Market Supply Schedule

Price of ‘X’ per unit (in ?)	Total Market Supply per day (in units)
10	100
20	200
30	300
40	400
50	500

There are some exceptions to the law of s supply. Following are such cases when supply may fall with the rises in price or rise with the fall in price.

(1) Labour supply : Supply of labour in the ) terms of hours of work is an important exception pointed out by economists. Generally when wages rise, workers work more, but after a certain point if wages continue to rise, supply of labour falls i.e. workers wish to earn more by work in for less hours and supply curve of labour would bend backwards as shown below :

In this figure as wage rate rises up to 0W, ;i supply of labor also rises up to ON, but when wage rate rises to 0W., labour supply falls from ON to 0Nr Hence an exception.

(2) Saving : In case of savings generally it is observed that as the rate of interest rises, savings also rises but some people want to have a fixed regular income by way of interest. They may save less at a higher rate of interest and save more at a lower rate of interest. For example : suppose a person is interested in earning a fixed income of ₹ 800 p.a. then he saves ₹ 10,000/- at 8% rate of interest but when rate of interest increases to 10%, he will save only ₹ 8,000/-.

(3) Future Expectations: If the seller expects a fall in price in future, then he will supply more today even at a low price. But if he expects the prices to rise further in future he will withhold the supply today to supply more in future at a high price.

(4) Need for Cash : When the sellers are in urgent need of liquid cash, then even at a lower price they will offer more goods for sale.

(5) Rare Goods : In case of rare collections such as rare painting, old coins, antique, the law is not applicable as the supply remains fixed. The supply curve is a vertical straight line parallel to Y axis.

(6) Agricultural Goods: Supply of agricultural product is influenced by natural factors like climatic conditions, rainfall etc., which cannot be controlled by man. So in bad weather condition, even at a higher price the supply of agricultural commodities will not increase.

Intext Questions

Question 1.
“Concept of supply is a micro concept but concept of aggregate supply is a macro concept”. Explain. (Textbook Page No. 43)
Answer:
Micro economics studies about economic behavior of units like households, firm market and particular commodities. Whereas macro economics deals with the broad economic concepts like total demand, total supply, national income, etc.
Supply refers to supply of an individual seller and aggregate supply refers to total supply of a commodity.
Hence, supply is a microscopic concept and aggregate supply is macro concept.

Question 2.
What do you mean by aggregate supply? (Textbook Page No. 43)
Answer:
Aggregate supply refers to the minimum amount of sales proceeds which the entrepreneurs expect to receive from the sale of output at a given level of employment.

Find out (Textbook Page No. 43)

If a firm produces 600 units of a commodity in a day and incurs a total cost of ₹ 30,000. Calculate the Average Cost.
Answer:
Average cost refers to the cost of production per unit cost of a commodity. It is calculated by dividing total cost by total quantity of a commodity. Hence,
AC = \(\frac{\mathrm{TC}}{\mathrm{TQ}}=\frac{30,000}{600}\) = ₹ 50 per unit

Find out (Textbook Page No. 43)

If a firm sells 400 units of a commodity at ₹ 10 unit. Calculate the TR and AR.
Answer:
TR = Price X Quantity
= 10 x 400
= 4,000
AR = \(\frac{\mathrm{TR}}{\mathrm{TQ}}\)
= \(\frac{4,000}{400}\)
= ₹ 10

Balbharti Maharashtra State Board Class 12 Economics Solutions Chapter 1 Introduction to Micro and Macro Economics Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Economics Solutions Chapter 1 Introduction to Micro and Macro Economics

1. Choose the correct option:

Question 1.
The branch of economics that deals with the allocation of resources.
a) Microeconomics
b) Macroeconomics
c) Econometrics
d) None of these
Options:
1) a, b and c
2) a and b
3) only a
4) None of these
Answer:
3) only a

Question 2.
Concepts studied under Micro economics.
a) National income
b) General price level
c) Factor pricing
d) Product pricing
Options :
1) b and c
2) b, c and d
3) a, b and c
4) c and d
Answer:
4) c and d

Question 3.
Method adopted in micro economic analysis.
a) Lumping method
b)Aggregative method
c) Slicing method
d) Inclusive method
Options :
1) a, c and d
2) a. b and d
3) only c
4) only a
Answer:
3) only c

Question 4.
Concepts studied under Macro economics.
a) Whole economy
b) Economic development
c) Aggregate supply
d) Product pricing
Options:
1) a, b and c
2) b, c and d
3) only d
4) a, b, c and d
Answer:
1) a, b and c

2. Complete the correlation:

1) Micro economics : Slicing method : : Macro economics: ……………. 1
2) Micro economics: Tree : : Macro economics: …………….. 2
3) Macro economic theory : Income and employment:: Micro economics : ……………. 4
4) Makros : Macro economics:: Mikros : ……………… 3
5) General equilibrium : Macro economics :: …………….. : Micro economics 5
Answers:

1. Lumping method
2. Forest
3. Price theory
4. Micro economics
5. partial equilibrium

3. Identify and explain the concepts from the given illustrations:

Question 1.
Gauri collected the information about the income of a particular firm.
Answer:
Concept: Micro economics / Slicing method.
Explanation : Micro economics refers to the study of small unit from whole economy. Micro economics uses slicing method to split the whole economy into small individual units.
Gauri has used slicing method from micro economics to collect information about the income of a particular firm from various firms.

Question 2.
Ramesh decided to take all decisions related to production, such as what and how to produce?
Answer:
Concept: Free market economy.
Explanation : A free market economy is that economy where the economic decisions regarding production of goods are taken at individual level.

Eg. What to produce? How much to produce? How to produce? etc. decisions are taken by producers.
With the help of free market economy Ramesh has taken decision related to production such as What to produce? and How to produce?

Question 3.
Shabana paid wages to workers in her factory and interest on her bank loan.
Answer:
Concept : Factor Pricing.
Explanation : Theory of factor pricing refers to determining the factor rewards for land, labour, capital and entrepreneur in the form of rent, wages, interest and profit respectively.

Shabana is an entrepreneur who has paid wages to its worker in a factory for production of goods and also paid interest on her bank loan in form of rewards to the factors of production.

4. Answer the following:

Question 1.
Explain the features of Micro economics.
OR
Explain the characteristics or nature of micro economics.
Answer:
Features of Micro Economics :

• Based on certain ssumption : Micro economics is based on ‘ceteris paribus’ assumption i.e., other things remaining constant like full employment, laissez faire policy, perfect competition, pure capitalism, etc.
• Study of Individual units : Micro economics deals with the study of behaviour of small individual units of the economy such as individual units of the economy such as individual consumer, individual firm, individual industries, individual prices, etc.
• Slicing Method : It divides or slices the economy into small units and studies each unit in detail e.g. study of a particular household demand in detail.
• Analysis of Market Structures : Micro economics analyses different market structures such as perfect competition, monopoly, monopolistic competition, oligopoly, etc.
• Use of Marginalism Principle : The term ‘marginal’ means change brought in total by an additional unit. Marginal analysis helps to study a variable through the changes by which producers and consumers take economic decisions using this principle.
• Price Theory : Micro economics is known as price theory because it determines the prices of goods and services as well as prices of factors of production.
• Limited Scope : The study of micro economics is limited to individual economic unit only. It does not deal with macro problems like unemployment, inflation, deflation, poverty, unemployment, population, etc.
• Partial I quilibrium : Micro economics analysis deals with partial equilibrium which analyses equilibrium position of an individual economic unit i.e. individual consumer, individual firm, etc.

Question 2.
Explain the importance of Macro economics.
Answer:
Importance of Macro Economics :

• Functioning of an Economy : It gives an idea of functioning of an economic system and help us to understand the behavioural pattern of aggregate variables.
• Economic fluctuations : It helps to analyse the causes of fluctuation in income, output and employment.
• National Income : It helps to study about National Income and makes possible to formulate correct economic policies.
• Economic Development : It helps us to understand the problems of the developing countries such as poverty, difference in the standards of living, etc., and suggest important steps to achieve economic development.
• Performance of an Economy : It helps us to analyse the performance of an economy where National Income estimates are used to measure the same.
• Study of Macro-economic Variables :
  Study of macro economic variables are important to understand the working of the economy.
• Level of Employment : Macro economics helps to analyse the general level of employment and output in an economy.

Question 3.
Explain the scope of Macro economics.
OR
“Scope of Macro Economics is wide.” Explain.
OR
Macro Economics is comprehensive in nature.
OR
Explain the subject matter of macro economics.
Answer:
Scope of Macro Economics:
The given chart helps us to understand the scope of macro economics.

1. Theory of Income and Employment : It explains which factors determine the level of National Income and employment and what j causes fluctuations in the level of income, output and employment.
To understand how the level of employment is determined, we have to study the consumption function. It includes theory of business cycles.

2. Theory of General Price Level and Inflation: Macro economics analyses shows how the general price level is determined and the causes for fluctuations in it. This study is important for understanding the  problems created by inflation and deflation.

3. Theory of Economic Growth and Development : Macro economics studies the causes of under development and poverty in poor countries and suggests strategies for accelerating growth and development in the country.

4.  Macro theory of Distribution : Macro theory of distribution deals with the relative share of rent, wages, interest and profit in j the total national income of various classes.

5. State with reasons whether you agree or disagree with the following statements:

Question 1.
The scope of micro economics is unlimited.
Answer:
No, I do not agree with this statement.

• Micro economics deals with small or individual units.
• Micro economics is the study of particular firm, particular household, individual prices, wages, incomes, individual industries, particular commodities.
• Micro economics deals with small part of National economy. It does not deal with whole economy like National income, Aggregate demand, Aggregate supply, poverty, inflation, etc.
• Hence, the scope of micro economics is limited.

Question 2.
Macro economics deals with the study of individual behaviour.
OR
Macro economics studies small units.
Answer:
No, I do not agree with this statement.
OR
Macro Economics is the study of I aggregate.
OR
Macro economics is concerned with macro economic variables.
Yes, I agree with this statement.

• Macro Economics studies the behaviour ofthe economy as a whole and not individual behaviour.
• It studies about larger economic units or aggregate economic variables like aggregate demand, aggregate supply, total investment, total savings, total employment, etc.
• It studies the general price level and macro theory of distribution.
• Whereas Micro Economics deals with individual behaviour of the people in the economy. It studies about individual demand, market demand, individual income, price of particular commodity etc.
• According to Prof. Kenneth E. Boulding “Macro Economics deals not with individual; quantities as such, but with aggregates of these quantities, not with individual income but with National Income, not with individual prices but with general price level, not with individual output but with National Output.

Question 3.
Macro economics is different from micro economics.
OR
Macro economics is wider than Micro economics.
OR
There is difference between micro economics and macro economics.
Answer:
Yes, I agree with this statement.

• Micro economics is a study of a particular unit of an economy. On the other hand macro economics is the study of entire economy.
• Micro economics studies individual demand, individual supply, individual income, price determination of particular product, etc. On the other hand macro economics studies aggregate demand, aggregate supply, national income, etc.
• Micro economics follows partial equilibrium analysis and macro economics follows general equilibrium analysis.
• Micro economics uses slicing method for study of small unit and macro economics uses lumping method for study of large unit.
• Therefore, macro economics is different from micro economics.

Question 4.
Micro economics uses slicing method.
Answer:
Yes, I agree with this statement.

• Micro economics deals with small or individual units.
• Micro economics divides or slices the economy into small units and studies each unit in detail.
• It is concerned with microscopic study of these units.
• It is the study of particular firm, particular household, individual prices, wages, incomes, etc.
• Hence, micro economics uses slicing method.

Question 5.
Micro economics is known as Income theory.
Answer:
No, I do not agree with this statement.
OR Micro economics is also known as price theory.
Yes, I agree with this statement.

• Micro Economics is known as ‘Price Theory’.
• The scope of micro economics includes the study of product pricing and factor pricing.
• The theory of product pricing explains how the price of food grains, vegetables, clothes, etc., are determined.
• They are determined by the interaction of market demand and supply forces.
• The theory of factor pricing explains the distribution of factor income such as rent on land, wages to labourers, interest on capital and profit to entrepreneurs.
• The factor prices are also determined by the demand and supply forces.
• Therefore, Micro Economics is also known as ‘Price Theory’.

6. Answer in detail :

Question 1.
Explain the importance of Micro economics.
Answer:
Introduction : Micro economics is the Js branch of economics that studies the behaviour of individuals.
It includes individual prices, wages, income, individual industries, particular commodities, particular household, etc.
(1) Definition :
(a) According to Maurice Dobb – “Micro economics is in fact a microscopic study of l the economy.
(b) According to Prof. A. P. Lerner – “Micro economics consists of looking at the economy ? through a microscope as it were to see how the millions of cell in the body of economy – the individuals or households as consumers and individuals or firms as producers play their part in the working of the whole economics organism.

(2) Meaning:
Micro economics deals with small individual economics units such as an individual ( consumer, individual producer, the price of a particular commodity or factor etc.

(3) Importance :
(a) Price Determination : Micro economics j explains how the prices of different products < and various factprs of production are determined.

(b) Free Market Economy : A free market economy is that economy where the economic decisions are taken at individual levels without intervention by the government. Decisions are regarding production of goods such as What to produce? How much to produce? How to produce? etc.

(c) Foreign Trade : Micro economics also explains gains from foreign trade, effects of tarrifs, factors affecting exchange rate, etc.

(d) Economic Model Building : Micro
economics helps in understanding various complex economic situations with the help of economic models.

(e) Business Decision : Micro economics theories are helpful to businessman for taking important business decision related to determination of cost of production and prices of goods, maximization of output & profit, etc.

(f) Useful to Government : It is useful in formulating and evaluating economic policies including pricing and distribution policies that promote economic welfare. It is useful in determining tax policy, public, expenditure policy, etc.

(g) Basis of Welfare Economics : It explains how optimum use of resources can be made to increase the welfare of the society. It also studies how taxes affect social welfare.

Question 2.
Explain the concept of Macro economics and its features.
Answer:
Introduction : Macro economics is the branch of economics that studies the behaviour and performance of an economy as a whole. It includes inflation, unemployment, working of the monetary system, business cycles, economic policies, etc.

(1) Definition:
(a) J. L. Hansen : “Macro economics is that branch of economics which considers the relationship between large aggregates such as the volume of employment, total amount of savings, investment, national income, etc”.
(b) Prof. Carl Shapiro : “Macro economics deals with the functioning of the economy as a whole. ”

(2) Meaning:
Macro economics is the study of aggregates national income, total employment, total consumption, inflation, total saving, etc.

(3) Features:
(a) Study of Aggregate : Macro economics deals with the study of entire economy. It studies the overall condition in the economy, such as National Income, National Output, Total Employment, General Price levels, etc.

(b) General Price Level : Macro economic studies the determination and changes in general price level which is the average of all prices of goods and services currently being produced in the economy.

(c) policy Oriented : Macro economics is a policy oriented science which is useful in formulating economic policies to promote economic growth, to control inflation and depression, to generate employment, etc.

(d) Lumping Method : Lumping method is the study of the whole economy rather than in part. It considers aggregates like National Income, Total consumption, etc. instead of personal income, PCC, etc.

(e) General Equilibrium Analysis : Macro Economics analysis is based on general equilibrium which deals with the economic system as a whole and studies the inter relationships between the various macro variables in an economy. General equilibrium deals with the behaviour of demand, supply and prices in the whole economy.

(f) Income Theory : Macro economics studies the concept of National Income and its causes of fluctuations that lead to business cycles i.e. inflation and deflation.

(g) Growth Models : Macro economics studies various factors that contribute to economic growth and development. These growth models are used for studying economic development.

(h) Interdependence : There is an element of interdependence among the macro economic variables such as income, output, employment, investment, price level, etc.

Intext Questions

Try this (Textbook Page 6)

Visit the vegetable market in the nearest area and try to get information about income and expenditure items of a particular seller.
Answer:
[Note : Students should do this activity by themselves.]

Balbharti Maharashtra State Board Class 12 History Solutions Chapter 2 European Colonialism Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 History Solutions Chapter 2 European Colonialism

1A. Choose the correct alternative and rewrite the statement.

Question 1.
‘Declaration of Independence was drafted by __________
(a) George Washington
(b) Thomas Jefferson
(c) Lord Amherst
(d) Lord Cornwallis
Answer:
(b) Thomas Jefferson

Question 2.
The second Anglo-Burmese war was fought during the times of __________
(a) Lord Amherst
(b) Lord Dufferin
(c) Lord Dalhousie
(d) Ashley Eden
Answer:
(c) Lord Dalhousie

1B. Find the incorrect pair from group ‘B’ and write the corrected one.

Question 1.

Group ‘A’	Group ‘B’
1. Togoland	German colony
2. Egypt	British colony
3. Orange Free State	Dutch colony
4. Ivory Coast	Portuguese colony

Answer:
Ivory Coast – French colony

2A. Write the names of historical places/persons/events.

Question 1.
The region from Florida to California on the southern coast of America was under the control of-
Answer:
Spain

Question 2.
The sea voyages for exploration were encouraged in the reign of-
Answer:
Queen Elizabeth I

2B. Choose the correct reason from those given below and complete the sentence.

Question 1.
The British wanted to gain control over Myanmar because __________
(a) they wanted to expand their empire
(b) it was very important to control the natural resources and the market in Myanmar
(c) they wanted to be at the forefront of the colonial competition amongst the European nations
(d) they wanted to teach a lesson to King Thibaw of Myanmar
Answer:
(b) it was very important to control the natural resources and the market in Myanmar

3. Complete the following concept map.

Question 1.

Answer:

4. Write short notes.

Question 1.
Nature of Colonialism.
Answer:

• Colonialism is the act of a developed nation occupying the land of a less developed, distant nation and establishing its rule on the occupied country. Europeans travelled all over the world with various intentions such as the urge for adventures, to earn a name, to discover unknown lands, to search for gold mines, etc.
• The Europeans established colonies wherever they went and in doing so there was a competition to gain economic, social, and political supremacy among the nations. Extreme nationalism, feeling of racial superiority, industrialization, aggressive approach, etc., are the factors that led to the growth of colonialism.
• Thus, Europeans went to America, Australia, and New Zealand. The Europeans also colonised Asia. However, the climate of Asia did not favour them. Favourable or unfavourable, Europeans disposed of the indigenous people in their own lands.
• The Europeans saw these colonies as potential marketplaces to sell their surplus goods which were accumulated as a result of mass production. The Europeans also needed additional sources that would provide ample raw material.
• In the latter half of the 19th century, England built a flourishing trade as a result of the industrial revolution.

Question 2.
American War of Independence.
Answer:
The colonies got into several battles with the British army even after gaining independence. The colonies finally at Saratoga got a decisive victory over the British army. This victory proved to be a turning point for the colonies as the French agreed to support them in their conflict against the British. Eventually, Spain also joined the conflicts in their struggle for independence.

On 7th October 1780, George Washington defeated the British army. On 19th October, British General Lord Cornwallis surrendered and America became independent. The American revolt for independence is also known as the ‘American Revolution’. America proved to the world that the subjects have a right to fight their rulers who deny them their natural rights.

5. Explain the following statements with reasons.

Question 1.
The industrial revolution gave momentum to colonialism.
Answer:
Colonialism was a result of the industrial revolution. Production increased enormously because of the new machines. However, the rate of local consumption was much less compared to the surplus rate of production. Hence, the immediate need of the Europeans was to find new markets for selling their products. It was also necessary that these markets be dependable and easy to dominate.

Question 2.
European nations established colonies in America.
Answer:
The stronger European nations dominated the weaker countries by establishing colonies there and pushed them to subjection. The lands of the original inhabitants in America were seized by the Europeans and were also massacred. The original inhabitants were forced into slavery. The Europeans discovered gold mines and the Spanish colonizers brought African slaves to work in sugarcane and tobacco fields. Farming and mining earned them enormous wealth. Essential raw materials were exported to Spain and the finished goods used to be imported for sale in the local markets for the colonies. The trade of gold and silver also earned huge profits for the king. Witnessing Spain’s prosperity England, Holland and France also began to establish colonies in America.

Class 12 History Chapter 2 European Colonialism Intext Questions and Answers

Try to do this. (Textbook Page No. 11)

Locate the following regions under British domination on the map: Gibraltar, Malta in the Mediterranean Sea; British Guyana, British Honduras, British West Indies, Bermuda and Falkland islands in the Western Hemisphere; Aden, Sri Lanka, Myanmar, Hongkong, India; also, the countries in Africa.
Answer:

Project (Textbook Page No. 18)

Question 1.
Collect information about the biodiversity and sources of minerals in the African continent.
Answer:
1. Meaning of Biodiversity:
Biodiversity consists of two words ‘Biological’ and ‘Diversity’. It refers to all the variety of life that can be found on earth (plants, animals, fungi, and microorganisms) as well as to the communities that they form and the habitats in which they live.

2. Biodiversity in Africa:

• Rich in biodiversity.
• Africa supports the earth’s largest assemblages of large mammals which roam freely in many countries.
• Africa is home to a rich and diverse animal, plant, and marine biodiversity that provide critical ecosystem services.

3. Minerals in Africa:
Africa is rich in mineral reserves and ranks first or second in quantity of world reserves of bauxite, cobalt, industrial diamond, phosphorite, platinum, etc.

Balbharti Maharashtra State Board Class 12 Economics Solutions Chapter 8 Public Finance in India Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Economics Solutions Chapter 8 Public Finance in India

1. Choose the correct option

Question 1.
Optional functions of Government:
a) Protection from external attack
b) Provision of education and health services
c) Provision of social security measures
d) Collection of tax
Options:
1) b and c
2) a, b and c
3) b, c and d
4) All of the above
Answer:
1) b and c

Question 2.
Obligatory functions of the Government:
a) Provision of employment
b) Maintaining internal law and order
c) Welfare measures
d) Exporting goods and services
Options:
1) c and d
2) a and b
3) only b
4) a, c and d
Answer:
3) only b

Question 3.
Public finance is one of those subjects which are on the borderline between economics and politics’ ……………. is the view of
a) Adam Smith
b) Alfred Marshall
e) Prof. Hugh Dalton
d) Prof. Findlay Shirras
Options:
1) only a
2) only b
3) only c
4) only d
Answer:
3) only c

Question 4.
Non-tax sources of revenue:
a) Direct and Indirect Tax
b) Direct Tax and Fees
c) Fees
d) Special Levy
Options:
1) b and c
2) a and c
3) a, b, c andd
4) c and d
Answer:
4) c and d

Question 5.
Trends shown by Public expenditure of any Government shows following trend.
a) Constant
b) Increasing
c) Decreasing
d) Fluctuating
Options:
1) only a
2) only b
3) only c
4) only d
Answer:
2) only b

Question 6.
Identify the right group of pairs from the given options.
I) Direct tax – a) Non-tax revenue
ii) Indirect tax – b) Inflation
iii) Fees and Fines – c) GST
iv) Surplus budget – d) Personal income tax
Options:
a) i-d ii-c iii-b iv-a
b) i-c li-d ill-a iv-b
c) i-d li-c ill-a iv-b
d) i-a li-b iii-c iv-d
Answer:
c) i-d li-c ill-a iv-b

2. Distingwish between following concepts:

Question 1.
Public finance and Private finance.
Answer:

Public Finance	Private Finance
(a) Public finance refers to income and expenditure of public authorities.	(a) Private finance refers to income and expenditure of individual and private sector organisations.
(b) The objective of public finance is to offer maximum social advantage.	(b) The objective of private finance is to fulfil private interest.
(c) Credit availability is more to increase public finance.	(c) Credit availability is limited to increase private finance.
(d) The supply of public finance is more elastic.	(d) The supply of private finance is less elastic.
(e) In case of public finance, government first determines the volume and different ways of it’s expenditure.	(e) In case of private finance, an individual considers income first and then determines the volume of expenditure.

Question 2.
Internal debt and External debt.
Answer:

Internal Debt	External Debt
(a) It refers to borrowings of the government to raise fund within the economy.	(a) It refers to borrowings of the government to raise fund outside the economy.
(b) In case of internal debt, domestic currency is used.	(b) In case of external debt, foreign currency is used.
(c) It is less complex to manage internal debt.	(c) It is more complex to manage external debt.
(d)  E.g. borrowings from RBI, nationalized banks and business organisations within a country.	(d) E.g. borrowings from foreign government and international organisation like IMF, World Bank, etc.

Question 3.
Developmental expenditure and Non developmental expenditure.
Answer:

Developmental Expenditure	Non-developmental Expenditure
(a) The government expenditure which gives productive impact is called developmental expenditure.	(a) The government expenditure which does not yield any direct productive impact, is called non-developmental expenditure.
(b) Developmental expenditure results in the generation of employment, an increase in production, etc.	(b) Non-developmental expenditure does not help to increase employment or production level.
(c) Expenditure on     education, industrial developmental expenditure.	(c) Administration cost, war expenses, etc. are examples of non-developmental expenditure.
(d) They are productive in nature.	(d) They are unproductive in nature.

Question 4.
Special assessment and Special levy.
Answer:

Special Assessment	Special Levy
(a) The charges paid by the citizens for getting certain special facilities by authorities are called special assessment.	(a) Special Levy are the charges levied on those commodities, whose consumption is harmful to human health.
(b) The objective behind taking special assessment is to provide extra special facilities to people.	(b) The objective behind charging special levy is to discourage the consumption of harmful commodities.
(c) Special assessment is taken from residents of a particular area.	(c) Special levy is taken from consumers of particular commodities.
(d) E.g. Special assessment is paid for the provision of special facilities like roads, water supply, etc.	(d) Special levy is paid for using commodities like wine, opium and other intoxicants, etc.

Question 5.
Direct Tax and Indirect tax.
Answer:

Direct Tax	Indirect Tax
(a) It refers to that tax which is paid by a person on whom it is legally imposed.	(a) It refers to that tax which is imposed on one person but paid by the other.
(b) A direct tax is paid by a person on whom it is legally imposed. It cannot be transferred.	(b) Indirect tax is imposed on one person but paid by the other.
(e) In case of public finance, government first determines the volume and different ways of it’s expenditure.	(e) In case of private finance, an individual considers income first and then determines the volume of expenditure.

3. State with reasons whether you agree or disagree with the following statement:

Question 1.
Obligatory function is the only function of the Govenment.
Answer:
No, I do not agree with this statement.
It is essential to perform obligatory functions for the government like protection from external attack, maintaining law and order, defence and civil administration, etc.
But modern government also perform some optional functions also. They are necessary for social and economic development of the country such as provision of education and health services, provision of social security like pensions and other welfare measures.
So, government has to perform obligatory as well as optional functions.

Question 2.
Fines and penalties are a major source of revenue for the Government.
Answer:
No, I do not agree with this statement.
Public revenue is the aggregate income, with the government, comes through various sources.
These sources are classified as tax-revenue and non-tax revenue.
Tax revenue is a compulsory contribution from people to government without getting any special benefits to tax-payers.
So, tax revenue is the major source of revenue for the government.
However, fines and penalties are sources of non-tax revenue.
The government imposes fines and penalties { only on those who violate the laws of a country. So, the income from this source is ) very small.

Question 3.
The goods and services tax (GST) has replaced almost all indirect taxes in India.
Answer:
Yes, I agree with this statement.
The Goods and Service Tax (GST) came into effect in India on 1st July, 2017.
GST is an indirect tax used in India, on the supply of goods and services.
GST simplified the tax system in a country.
GST is different from an excise or sales tax imposed on the manufacture or sale of ; a product. GST is a tax levied on supply of goods and services.
GST replaced almost all indirect taxes like central excise duty, service tax, entry tax, entertainment tax, etc.
Because, GST is a comprehensive tax base with nationwide coverage of goods and ; services.

Question 4.
Democratic Governments do not lead to increase in public expenditure.
Answer:
No, I do not agree with this statement. i Public expenditure is that expenditure which is incurred by the public authority (i.e., central, state and local government). Public expenditure is required for protection of the citizens, for satisfying their collective needs and for promoting economic and social welfare of the people.
In a democratic state, government has to ( perform the obligatory functions like, defence and civil administration, maintaining internal law and order, etc.
Government also performs optional functions like provision of education and health services, provision of social security, etc.
To perform all these functions more efficiently, democratic government leads to increase in public expenditure.

Question 5.
Public finance is more elastic than private finance.
Answer:
Yes, I agree with this statement.
Public finance refers to the income and expenditure of public authorities, whereas, private finance is the income and expenditure of individuals and private sector organisations.

The main objective of public finance is to offer maximum social advantage, while the main motive of private finance is to fulfil private interest.

Public finance is more elastic compare to private finance because credit provision is much more in the market to increase public finance but, credit availability is limited to increase private finance.
It is also possible to the government to adjust revenue and expenditure with one another in case of public finance.

4. Read the given passage and answer the questions:

‘The conventional notion of social security is that the government would make periodic payments to look after people in their old age, ill-health. disability and poverty. This idea should itself change from writing a cheque for the beneficiary to institutional arrangements to care for beneficiaries. including by enabling them to look after themselves. to a large extent.

The write-a-cheque model of social security is a legacy from the rich world at the optimal phase of its demographic transition, when the working population was numerals enough and earning enough to generate the taxes to pay for the care of those not working. This model is ill-suited for less, well- off India with growing life expectancy. increasing urbanization and resultant migration. Social security
under urbanization will be different from social security in a static society.

Question 1.
State the conventional notion of social security.
Answer:
The conventional notion of social security is that the government would make periodic payments to look after people in their old age, ill-health, disability and poverty.

Question 2.
What kind of conceptual change is suggested in the given paragraph.
Answer:
The given paragraph suggests that, the idea should change from writing a cheque for the beneficiary to institutional arrangement to care for beneficiaries.
It will enable them to look after themselves to a large extent.

Question 3.
What is a legacy of social security from the rich world?
Answer:
The write-a-cheque model of social security i is a legacy from the rich world.

Question 4.
Which features of India make the traditional model of social security ill-suited for the
economy?
Answer:
Growing life expectancy, increasing urbanization and resultant migration are the features of India that make the traditional model of social security ill suited for the economy.

5. Answer the following:

Question 1.
State the types and importance of Government budget.
Answer:
(B) Importance of Budget:
Budget is important in number of ways.

(1) Tax rates presented in the budget indicates disposable income of the tax payer. It also determines the development of business and individuals.
(2) Government expenditure is also a part of budget. This public expenditure on defence, administration, infrastructure, education, health care, etc. affects the lives of the citizens and overall economy.
(3) Government uses budget as a medium for implementing economy policies in the country.
(4) Budgetary actions of the government affect production size and distribution of income, utilization of human and material resources of the country.
Thus, implementing suitable budgetary policy is very important for overall development of the economy.

Question 2.
Explain the principles of taxation.
Answer:
Principles of taxation are also called canons of taxation. There are four principles (canons) of taxation, propounded by Adam Smith
(1) Canon of Equity or Equality
(2) Canon of Certainty
(3) Canon of Convenience
(4) Canon of Economy

They are explained as follows :

1. Canon of Equity or Equality : According to Adam Smith, every person should pay taxes to the government in proportion to his ability to pay.
Canon of equity or equality means rich people should pay more tax as compared to poor.

2. Canon of Certainty : Adam Smith suggested that the tax payer should know in advance that, how much tax he has to pay, at what time and in what form he has to pay tax to the government.

3. Canon of Convenience : According to this principle, every tax should be levied in such a manner and at such a time that, it becomes convenient to the tax payer to make payment.

4. Canon of Economy : This principle suggests that the cost of tax collection should be the minimum. If tax is collected economically, then such a tax is considered to be a good tax.
Every citizen of a country has to pay tax, imposed upon him as it is compulsory contribution to the government.
Tax is a major source of revenue to the government.
Therefore, public authority (Government) must consider all the principles (canons) of taxation in the preparation and implementation of tax system.

Question 3.
Explain non-tax sources of revenue of the Government.
Answer:
Non-tax revenue refers to the revenue received by the government from various ? sources other than taxes.
The sources of non-tax revenue are as follows:

Fees : It refers to charges paid, in return for certain specific services rendered by the government. E.g. fees paid for registration of house, car, education fees, etc.

Prices of Public Goods and Services : Various types of goods and services are produced, supplied and sold by modern government to the citizens. It; is added to public revenue when people s purchase them and pay their prices.

Special Assessment : It is special kind of tax, which is levied by local government on the residents of a particular area. In exchange of it, government provides some special facilities to them.

Fines and Penalties : It is imposed by government on those who violate the laws of the country.
E.g. a traffic police charges fine and collects money if someone violates traffic rule. The objective behind collection of fines and penalties is not to earn money but to discourage the people from violating the laws framed by the government It is small source of income.

Gifts, Grants and Donations : The government receives gifts from its citizens and others. It is included in public revenue. The government may also get grants from foreign government and institutions for general and specific purposes.
Foreign aid is also an important form of public revenue for developing country like India. However, this source of revenue is uncertain in nature.

Special Levies : It refers to the charges levied by government on those commodities, whose consumption is harmful to human health.
Special levies are paid for using commodities like wine, opium and other intoxicants. Special levy is imposed, not to earn income, but to discourage the people from using harmful products.

Borrowings: Government borrows to raise fund because government expenditure generally exceeds government revenue, in a welfare state.

When government borrows from foreign government or international organisations, it is known as external debt. It is more popular source of public revenue for investment in development of projects. Thus, public revenue in form of non-tax sources play very important role in socio¬economic development of a country. Explain the classification of public expenditure.

6. Answer in detail :

Question 1.
Explain various reasons for the growth of public
expenditure.
Answer:
Public expense is the expense incurred by the government (central, state and) local government). It is necessary for the protection of the citizens, for satisfying collective needs of the society and for ? promoting economy.and social welfare of a S country.

Public expense is necessary to perform various functions of the government Public ; expense consists of revenue and capital expenses as well as developmental and non-developmental expenses.

1. High Growth of Population : In a developing country like India, population is rising rapidly. Therefore, government has to incur greater expense to fulfil the needs of growing population.

(2) Growing Urbanisation : Due to expansion of urban sector, government expense increases. Government has to make the provision of water supply, roads, energy, schools and colleges, public transport, hospitals, welfare centres, sanitation, drainage system, etc.
It leads to growth in public expense.

(3) Public Health Care : Public health is a top most priority of modern welfare state. Government undertakes public vaccination programme, maintenance of dispensaries, maternity care and child welfare centres, etc
.
(4) Democracy : There is a democracy in India. A democratic form of government is very expensive due to regular elections and other public works.
It leads to growth in public expense.

(5) An increase in Defence Expense : Government has to incur defence expense to protect the country from external attacks as well as to maintain law and order in a country. Hence, an increase in defence expense leads to growth of public expense.

(6) Disaster Management : Many natural calamities like earthquakes, flood, cyclones, Covid-19 and man-made problems like social unrest, economic instability, etc. occur frequently. In such cases, government has to spend for disaster management which increases public expense.

(7) Infrastructure Development : It is necessary to make provision of economic infrastructure like energy, transport, communication and social infrastructure like education, health, etc. for rapid economic development of a country.
Thus, development of infrastructure facilities results into growth of public expense.

(8) Inflation : Due to inflation, prices of goods and services tend to rise. When government buys goods and services from the market for development of a country, government has to pay higher cost which raises public expense.

(9) Industrial Development : An increase in production depends upon industrial development. It leads to an increase in level of employment and overall economy growth. So, government implements various schemes and programmes for industrial development.
It results into growth of public expense.

(10) Increase in Government Activities : The modern government performs various obligatory and optional functions for social and economic development of a country. It requires huge fund to spend on education, public health, public recreation, social welfare schemes, etc.

Many other functions like maintenance of roads, lighting, public streets, construction of public houses, protecting life and property, public vaccination, garbage collection and ; disposal, prevention and control of epidemics etc. lead to growth of public expenses.

Government also spends on provision of pure water supply, removal of slums, checking ) food adulteration, etc.
All these factors are responsible for the j growth of public expense.

Intext Questions

Find out (Textbook Page 70) :

More examples of obligatory and optional functions of the government.
Answer:

Obligatory Functions of Govt.	Optional Functions of Govt.
(a) Supply and maintenance of water works.	(a) Construction of public parks and gardens.
(b) Extinguishing fires and protecting life and property when fire occurs.	(b) Town planning
(c) Prevention and control of epidemics (e.g. Corona)	(c) Housing for low income group
(d) Garbage collection and disposal	(d) Construction and maintenance of rest- houses
(e) Public vaccination	(e) Organising cultural events, sports etc.

Find out (Textbook Page 73) :

Reasons for growth in public expenditure other than given in the text-book.
Answer:

1. Rise in per capita income.
2. Rural development.
3. Provision of transport and communication.
4. Reducing inequalities.
5. Expenditure on social services (like food, housing, education etc.)
6. Effects of war.
7. Pressure of social progress.

Find out (Textbook Page 73) :

Important Social Welfare Schemes in India.
Answer:

Name of the Scheme	Date of Launch	Main Objective
1. Atal Pension Yojana	May 2015	A contribution based program for poor people to receive pension.
2. Deen Dayal Upadhyaya Grameen Kaushalya Yojana	Sept. 2014	Providing gainful employment to rural Youth, through training.
3. Pradhan Mantri Gramin Awaas Yojana	June 2015	Providing financial assistance to rural poor for their houses.
4. Integrated Child Development Services	Oct. 1975	To tackle malnutrition and health problems in children below 6 years.
5. Midday Meal Scheme	Aug. 1995	Lunch (free of cost) to school children on all working days.
6. Suraksha Bima Yojana	May 2015	Accidental insurance with a premium of Rs. 12/- per year.
7.Rashtriya Krishi Vikas Yojana	Aug. 2007	Provision for development of agriculture and its allied sector

 

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

(I) Choose the correct alternative.

Question 1.
F(x) is c.d.f. of discreter r.v. X whose p.m.f. is given by P(x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\), for x = 0, 1, 2, 3, 4 & P(x) = 0 otherwise then F(5) = __________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{4}\)
(d) 1
Answer:
(d) 1

Question 2.
F(x) is c.d.f. of discrete r.v. X whose distribution is

then F(-3) = __________
(a) 0
(b) 1
(c) 0.2
(d) 0.15
Answer:
(a) 0

Question 3.
X : number obtained on uppermost face when a fair die is thrown then E(X) = __________
(a) 3.0
(b) 3.5
(c) 4.0
(d) 4.5
Answer:
(b) 3.5

Question 4.
If p.m.f. of r.v. X is given below.

then Var(X) = __________
(a) p²
(b) q²
(c) pq
(d) 2pq
Answer:
(d) 2pq

Question 5.
The expected value of the sum of two numbers obtained when two fair dice are rolled is __________
(a) 5
(b) 6
(c) 7
(d) 8
Answer:
(c) 7

Question 6.
Given p.d.f. of a continuous r.v. X as
f(x) = \(\frac{x^{2}}{3}\) for -1 < x < 2
= 0 otherwise then F(1) =
(a) \(\frac{1}{9}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{3}{9}\)
(d) \(\frac{4}{9}\)
Answer:
(b) \(\frac{2}{9}\)

Question 7.
X is r.v. with p.d.f.
f(x) = \(\frac{k}{\sqrt{x}}\), 0 < x < 4
= 0 otherwise then E(X) = __________
(a) \(\frac{1}{3}\)
(b) \(\frac{4}{3}\)
(c) \(\frac{2}{3}\)
(d) 1
Answer:
(b) \(\frac{4}{3}\)

Question 8.
If X follows B(20, \(\frac{1}{10}\)) then E(X) = __________
(a) 2
(b) 5
(c) 4
(d) 3
Answer:
(a) 2

Question 9.
If E(X) = m and Var(X) = m then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
Answer:
(b) Possion distribution

Question 10.
If E(X) > Var(X) then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
Answer:
(a) Binomial distribution

(II) Fill in the blanks.

Question 1.
The values of discrete r.v. are generally obtained by __________
Answer:
counting

Question 2.
The values of continuous r.v. are generally obtained by __________
Answer:
measurement

Question 3.
If X is dicrete random variable takes the values x₁, x₂, x₃, …… x_n then \(\sum_{i=1}^{n} p\left(x_{i}\right)\) = __________
Answer:
1

Question 4.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = \(\frac{x-1}{3}\) for x = 1, 2, 3, and p(x) = 0 otherwise then F(4) = __________
Answer:
1

Question 5.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\) for x = 0, 1, 2, 3, 4, and p(x) = 0 otherwise then F(-1) = __________
Answer:
0

Question 6.
E(X) is considered to be __________ of the probability distribution of X.
Answer:
centre of gravity

Question 7.
If X is continuous r.v. and f(x_i) = P(X ≤ x_i) = \(\int_{-\infty}^{x_{i}} f(x) d x\) then f(x) is called __________
Answer:
Cumulative Distribution Function

Question 8.
In Binomial distribution probability of success ________ from trial to trial.
Answer:
remains constant/independent

Question 9.
In Binomial distribution, if n is very large and probability success of p is very small such that np = m (constant) then ________ distribution is applied.
Answer:
Possion

(III) State whether each of the following is True or False.

Question 1.
If P(X = x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\) for x = 0, 1, 2, 3, 4, then F(5) = \(\frac{1}{4}\) when f(x) is c.d.f.
Answer:
False

Question 2.

If F(x) is c.d.f. of discrete r.v. X then F(-3) = 0.
Answer:
True

Question 3.
X is the number obtained on the uppermost face when a die is thrown the E(X) = 3.5.
Answer:
True

Question 4.
If p.m.f. of discrete r.v.X is

then E(X) = 2p.
Answer:
True

Question 5.
The p.m.f. of a r.v. X is p(x) = \(\frac{2 x}{n(n+1)}\), x = 1, 2,……n
= 0 otherwise,
Then E(X) = \(\frac{2 n+1}{3}\)
Answer:
True

Question 6.
If f(x) = kx (1 – x) for 0 < x < 1
= 0 otherwise then k = 12
Answer:
False

Question 7.
If X ~ B(n, p) and n = 6 and P(X = 4) = P(X = 2) then p = \(\frac{1}{2}\).
Answer:
True

Question 8.
If r.v. X assumes values 1, 2, 3,………, n with equal probabilities then E(X) = \(\frac{(n+1)}{2}\)
Answer:
True

Question 9.
If r.v. X assumes the values 1, 2, 3,………, 9 with equal probabilities, E(X) = 5.
Answer:
True

(IV) Solve the following problems.

Part – I

Question 1.
Identify the random variable as discrete or continuous in each of the following. Identify its range if it is discrete.
(i) An economist is interested in knowing the number of unemployed graduates in the town with a population of 1 lakh.
Solution:
X = No. of unemployed graduates in a town.
∵ The population of the town is 1 lakh
∴ X takes finite values
∴ X is a Discrete Random Variable
∴ Range of = {0, 1, 2, 4, …. 1,00,000}

(ii) Amount of syrup prescribed by a physician.
Solution:
X : Amount of syrup prescribed.
∴ X Takes infinite values
∴ X is a Continuous Random Variable.

(iii) A person on a high protein diet is interested in the weight gained in a week.
Solution:
X : Gain in weight in a week.
X takes infinite values
∴ X is a Continuous Random Variable.

(iv) Twelve of 20 white rats available for an experiment are male. A scientist randomly selects 5 rats and counts the number of female rats among them.
Solution:
X : No. of female rats selected
X takes finite values.
∴ X is a Discrete Random Variable.
Range of X = {0, 1, 2, 3, 4, 5}

(v) A highway safety group is interested in the speed (km/hrs) of a car at a checkpoint.
Solution:
X : Speed of car in km/hr
X takes infinite values
∴ X is a Continuous Random Variable.

Question 2.
The probability distribution of a discrete r.v. X is as follows.

(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:
(i) Assuming that the given distribution is a p.m.f. of X
∴ Each P(X = x) ≥ 0 for x = 1, 2, 3, 4, 5, 6
k ≥ 0
ΣP(X = x) = 1 and
k + 2k + 3k + 4k + 5k + 6k = 1
∴ 21k = 1 ∴ k = \(\frac{1}{21}\)

(ii) P(X ≤ 4) = 1 – P(X > 4)
= 1 – [P(X = 5) + P(X = 6)]
= 1 – [latex]\frac{5}{21}+\frac{6}{21}[/latex]
= 1 – \(\frac{11}{21}\)
= \(\frac{10}{21}\)
P(2 < X < 6) = p(3) + p(4) + p(5)
= 3k + 4k + 5k
= \(\frac{3}{21}+\frac{4}{21}+\frac{5}{21}\)
= \(\frac{12}{21}\)
= \(\frac{4}{7}\)

(iii) P(X ≥ 3) = p(3) + p(4) + p(5) + p(6)
= 3k + 4k + 5k + 6k

Question 3.
Following is the probability distribution of an r.v. X.

Find the probability that
(i) X is positive.
(ii) X is non-negative.
(iii) X is odd.
(iv) X is even.
Solution:
(i) P(X is positive)
P(X = 0) = p(1) + p(2) + p(3)
= 0.25 + 0.15 + 0.10
= 0.50

(ii) P(X is non-negative)
P(X ≥ 0) = p(0) + p(1) + p(2) + p(3)
= 0.20 + 0.25 + 0.15 + 0.10
= 0.70

(iii) P(X is odd)
P(X = -3, -1, 1, 3)
= p(- 3) +p(-1) + p(1) + p(3)
= 0.05 + 0.15 + 0.25 + 0.10
= 0.55

(iv) P(X is even)
= 1 – P(X is odd)
= 1 – 0.55
= 0.45

Question 4.
The p.m.f of a r.v. X is given by
\(P(X=x)= \begin{cases}\left(\begin{array}{l}
5 \\
x
\end{array}\right) \frac{1}{2^{5}}, & x=0,1,2,3,4,5 . \\
0 & \text { otherwise }\end{cases}\)
Show that P(X ≤ 2) = P(X ≥ 3).
Solution:
For x = 0, 1, 2, 3, 4, 5

Question 5.
In the following probability distribution of an r.v. X

Find a and obtain the c.d.f. of X.
Solution:
Given distribution is p.m.f. of r.v. X
ΣP(X = x) = 1
∴ p(1) + p(2) + p(3) + p(4) + p(5) = 1

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p.m.f. of X.
Solution:
A fair coin is tossed 4 times
∴ Sample space contains 16 outcomes
Let X = Number of heads obtained
∴ X takes the values x = 0, 1, 2, 3, 4.
∴ The number of heads obtained in a toss is an even

Question 7.
Find the probability of the number of successes in two tosses of a die, where success is defined as (i) number greater than 4 (ii) six appearing in at least one toss.
Solution:
S : A die is tossed two times
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)}
n(S) = 36
(i) X : No. is greater than 4
Range of X = {0, 1, 2}

(ii) X : Six appears on aleast one die.
Range of X = {0, 1, 2}

Question 8.
A random variable X has the following probability distribution.

Determine (i) k, (ii) P(X < 3), (iii) P(X > 6), (iv) P(0 < X < 3).
Solution:
(i) It is a p.m.f. of r.v. X
Σp(x) = 1
p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
9k + 10k² = 1
10k² + 9k – 1 = 0
10k² +10k – k – 1 = 0
∴ 10k(k + 1) – 1(k + 1) = 0
∴ (10k – 1) (k + 1) = 0
∴ 10k – 1 = 0r k + 1 = 0
∴ k = \(\frac{1}{10}\) or k = -1
k = -1 is not accepted, p(x) ≥ 0, ∀ x ∈ R
∴ k = \(\frac{1}{10}\)

(ii) P(X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

(iii) P(X > 6) = p(7)
= 7k² + k
= \(7\left(\frac{1}{10}\right)^{2}+\frac{1}{10}\)
= \(\frac{7}{100}+\frac{1}{10}\)
= \(\frac{17}{100}\)

(iv) P(0 < X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

Question 9.
The following is the c.d.f. of a r.v. X.

Find the probability distribution of X and P(-1 ≤ X ≤ 2).
Solution:

P(-1 ≤ X ≤ 2) = p(-1) + p(0) + p(1) + p(2)
= 0.2 + 0.15 + 0.10 + 0.10
= 0.55

Question 10.
Find the expected value and variance of the r.v. X if its probability distribution is as follows.
(i)

Solution:

(ii)

Solution:
E(X) = Σx . p(x)

(iii)

Solution:

(iv)

Solution:

= 1.25
S.D. of X = σ_x = √Var(X)
= √1.25
= 1.118

Question 11.
A player tosses two coins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears, and ₹ 2 if no head appears. Find the expected value and variance of the winning amount.
Solution:
S : Two fair coin are tossed
S = {HH, HT, TT, TH}
n(S) = 4
∴ Range of X = {0, 1, 2}
∴ Let Y = amount received corresponds to values of X

Expected winning amount
E(Y) = Σpy = \(\frac{22}{4}\) = ₹ 5.5
V(Y) = Σpy² – (Σpy)²
= \(\frac{154}{4}\) – (5.5)²
= 38.5 – 30.25
= ₹ 8.25

Question 12.
Let the p.m.f. of the r.v. X be
\(p(x)= \begin{cases}\frac{3-x}{10} & \text { for } x=-1,0,1,2 \\ 0 & \text { otherwise }\end{cases}\)
Calculate E(X) and Var(X).
Solution:

Question 13.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
\(f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
We know that

Question 14.
The p.d.f. of the r.v. X is given by
\(f(x)= \begin{cases}\frac{1}{2 a} & \text { for } 0<x<2 a \\ 0 & \text { otherwise }\end{cases}\)
Show that P(X < \(\frac{a}{2}\)) = P(X > \(\frac{3a}{2}\))
Solution:

Question 15.
Determine k if
\(f(x)= \begin{cases}k e^{-\theta x} & \text { for } 0 \leq x<\infty, \theta>0 \\ 0 & \text { otherwise }\end{cases}\)
is the p.d.f. of the r.v. X. Also find P(X > \(\frac{1}{\theta}\)). Find M if P(0 < X < M) = \(\frac{1}{2}\)
Solution:
We know that

Question 16.
The p.d.f. of the r.v. X is given by
\(f_{x}(x)=\left\{\begin{array}{l}
\frac{k}{\sqrt{x}}, 0<x<4 \\
0, \text { otherwise }
\end{array}\right.\)
Determine k, c.d.f. of X and hence find P(X ≤ 2) and P(X ≥ 1).
Solution:
We know that

Question 17.
Let X denote the reaction temperature (in °C) of a certain chemical process. Let X be a continuous r.v. with p.d.f.
\(f(x)= \begin{cases}\frac{1}{10}, & -5 \leq x \leq 5 \\ 0, & \text { otherwise }\end{cases}\)
Compute P(X < 0).
Solution:
Given p.d.f. is f(x) = \(\frac{1}{10}\), for -5 ≤ x ≤ 5
Let its c.d.f. F(x) be given by

Part – II

Question 1.
Let X ~ B(10, 0.2). Find (i) P(X = 1) (ii) P(X ≥ 1) (iii) P(X ≤ 8)
Solution:
X ~ B(10, 0.2)
n = 10, p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
(i) P(X = 1) = ¹⁰C₁ (0.2)¹ (0.8)⁹ = 0.2684

(ii) P(X ≥ 1) = 1 – P(X < 1)
= 1 – P(X = 0)
= 1 – ¹⁰C₀ (0.2)⁰ (0.8)¹⁰
= 1 – 0.1074
= 0.8926

(iii) P(X ≤ 8) = 1 – P(x > 1)
= 1 – [p(9) + p(10)]
= 1 – [¹⁰C₉ (0.2)⁹ (0.8)¹ + ¹⁰C₁₀ (0.2)¹⁰]
= 1 – 0.00000041984
= 0.9999

Question 2.
Let X ~ B(n, p) (i) If n = 10 and E(X) = 5, find p and Var(X), (ii) If E(X) = 5 and Var(X) = 2.5, find n and p.
Solution:
X ~ B(n, p)
(i) n = 10, E(X) = 5
∴ np = 5
∴ 10p = 5
∴ p = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
V(X) = npq
= 10 × \(\frac{1}{2}\) × \(\frac{1}{2}\)
= 2.5

(ii) E(X) = 5, V(X) = 2.5
∴ np = 5, ∴ npq = 2.5
∴ 5q = 2.5
∴ q = \(\frac{2.5}{5}\) = 0.5, p = 1 – 0.5 = 0.5
But np = 5
∴ n(0.5) = 5
∴ n = 10

Question 3.
If a fair coin is tossed 4 times, find the probability that it shows (i) 3 heads, (ii) head in the first 2 tosses, and tail in the last 2 tosses.
Solution:
n : No. of times a coin is tossed
∴ n = 4
X : No. of heads
P : Probability of getting heads

Question 4.
The probability that a bomb will hit the target is 0.8. Find the probability that, out of 5 bombs, exactly 2 will miss the target.
Solution:
X : No. of bombs miss the target
p : Probability that bomb miss the target
∴ q = 0.8
∴ p = 1 – q = 1 – 0.8 = 0.2
n = No. of bombs = 5
∴ X ~ B(5, 0.2)
∴ p(x) = ⁿC_x p^x q^n-x
P(X = 2) = ⁵C₂ (0.2)² (0.8)^5-2
= 10 × 0.04 × (0.8)³
= 10 × 0.04 × 0.512
= 0.4 × 0.512
= 0.2048

Question 5.
The probability that a lamp in the classroom will burn is 0.3. 3 lamps are fitted in the classroom. The classroom is unusable if the number of lamps burning in it is less than 2. Find the probability that the classroom can not be used on a random occasion.
Solution:
X : No. of lamps not burning
p : Probability that the lamp is not burning
∴ q = 0.3
∴ p = 1 – q = 1 – 0.3 = 0.7
n = No. of lamps fitted = 3
∴ X ~ B(3, 0.7)
∴ p(x) = ⁿC_x p^x q^n-x
P(classroom cannot be used)
P(X < 2) = p(0) + p(1)
= ³C₀ (0.7)⁰ (0.3)^3-0 + ³C₁ (0.7)¹ (0.3)^3-1
= 1 × 1 × (0.3)³ + 3 × 0.7 × (0.3)²
= (0.3)2 [0.3 + 3 × 0.7]
= 0.09 [0.3 + 2.1]
= 0.09 [2.4]
= 0.216

Question 6.
A large chain retailer purchases an electric device from the manufacturer. The manufacturer indicates that the defective rate of the device is 10%. The inspector of the retailer randomly selects 4 items from a shipment. Find the probability that the inspector finds at most one defective item in the 4 selected items.
Solution:
X : No. of defective items
n : No. of items selected = 4
p : Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
P(At most one defective item)
P(X ≤ 1) = p(0) + p(1)
= ⁴C₀ (0.1)⁰ (0.9)^4-0 + ⁴C₁ (0.1)¹ (0.9)^4-1
= 1 × 1 × (0.9)⁴ + 4 × 0.1 × (0.9)³
= (0.9)3 [0.9 + 4 × 0.1]
= (0.9)3 × [0.9 + 0.4]
= 0.729 × 1.3
= 0.9477

Question 7.
The probability that a component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 components tested survive.
Solution:
p = 0.6, q = 1 – 0.6 = 0.4, n = 4
x = 2
∴ p(x) = ⁿC_x p^x q^n-x
P(X = 2) = ⁴C₂ (0.6)² (0.4)² = 0.3456

Question 8.
An examination consists of 5 multiple choice questions, in each of which the candidate has to decide which one of 4 suggested answers is correct. A completely unprepared student guesses each answer randomly. Find the probability that this student gets 4 or more correct answers.
Solution:
n : No. of multiple-choice questions
∴ n = 5
X : No. of correct answers
p : Probability of getting correct answer
∵ There are 4 options out of which one is correct
∴ p = \(\frac{1}{4}\)
∴ q = 1 – p = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
∵ X ~ B(5, \(\frac{1}{4}\))
∴ p(x) = ⁿC_x p^x q^n-x
P(Four or more correct answers)
P(X ≥ 4) = p(4) + p(5)

Question 9.
The probability that a machine will produce all bolts in a production run with in the specification is 0.9. A sample of 3 machines is taken at random. Calculate the probability that all machines will produce all bolts in a production run within the specification.
Solution:
n : No. of samples selected
∴ n = 3
X : No. of bolts produce by machines
p : Probability of getting bolts
∴ p = 0.9
∴ q = 1 – p = 1 – 0.9 = 0.1
∴ X ~ B(3, 0.9)
∴ p(x) = ⁿC_x p^x q^n-x
P(Machine will produce all bolts)
P(X = 3) = ³C₃ (0.9)³ (0.1)^3-3
= 1 × (0.9)³ × (0.1)⁰
= 1 × (0.9)³ × 1
= (0.9)³
= 0.729

Question 10.
A computer installation has 3 terminals. The probability that anyone terminal requires attention during a week is 0.1, independent of other terminals. Find the probabilities that (i) 0 (ii) 1 terminal requires attention during a week.
Solution:
n : No. of terminals
∴ n = 3
X : No. of terminals need attention
p : Probability of getting terminals need attention
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
∵ X ~ B(3, 0.1)
∴ p(x) = ⁿC_x p^x q^n-x
(i) P(No attention)
∴ P(X = 0) = ³C₀ × (0.1)⁰ (0.9)^3-1
= 1 × 1 × (0.9)³
= 0.729

(ii) P(One terminal need attention)
∴ P(X = 1) = ³C₁ (0.1)¹ (0.9)^3-1
= 3 × 0.1 × (0.9)²
= 0.3 × 0.81
= 0.243

Question 11.
In a large school, 80% of the students like mathematics. A visitor asks each of 4 students, selected at random, whether they like mathematics, (i) Calculate the probabilities of obtaining an answer yes from all of the selected students, (ii) Find the probability that the visitor obtains the answer yes from at least 3 students.
Solution:
X : No. of students like mathematics
p: Probability that students like mathematics
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
n : No. of students selected
∴ n = 4
∵ X ~ B(4, 0.8)
∴ p(x) = ⁿC_x p^x q^n-x
(i) P(All students like mathematics)
∴ P(X = 4) = ⁴C₄ (0.8)⁴ (0.2)^4-4
= 1 × (0.8)⁴ × (0.2)⁰
= 1 × (0.8)⁴ × 1
= 0.4096

(ii) P(Atleast 3 students like mathematics)
∴ P(X ≥ 3) = p(3) + p(4)
= ⁴C₃ (0.8)³ (0.2)^4-3 + 0.4096
= 4 × (0.8)³ (0.2)¹ + 0.4096
= 0.8 × (0.8)³ + 0.4096
= (0.8)⁴ × 0.4096
= 0.4096 + 0.4096
= 0.8192

Question 12.
It is observed that it rains on 10 days out of 30 days. Find the probability that
(i) it rains on exactly 3 days of a week.
(ii) it rains at most 2 days a week.
Solution:
X : No. of days it rains in a week
p : Probability that it rains
∴ p = \(\frac{10}{30}=\frac{1}{3}\)
∴ q = 1 – p = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
n : No. of days in a week
∴ n = 7
∴ X ~ B(7, \(\frac{1}{3}\))
(i) P(Rains on Exactly 3 days of a week)

(ii) P(Rains on at most 2 days of a week)
∴ P(X ≤ 2) = p(0) + p(1) + p(2)

Question 13.
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, find variance of X.
Solution:
X : Follows Possion Distribution

∴ m = 1
∴ Mean = m = Variance of X = 1

Question 14.
If X has Poisson distribution with parameter m, such that
\(\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}\)
find probabilities P(X = 1) and P(X = 2), when X follows Poisson distribution with m = 2 and P(X = 0) = 0.1353.
Solution:
Given that the random variable X follows the Poisson distribution with parameter m = 2
i.e. X ~ P(2)
Its p.m.f. is satisfying the given equation.
\(\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}\)
When x = 0,
\(\frac{\mathrm{P}(\mathrm{X}=1)}{\mathrm{P}(\mathrm{X}=0)}=\frac{2}{0+1}\)
P(X = 1) = 2P(X = 0)
= 2(0.1353)
= 0.2706
When x = 1,
\(\frac{\mathrm{P}(\mathrm{X}=2)}{\mathrm{P}(\mathrm{X}=1)}=\frac{2}{1+1}\)
P(X = 2) = P(X = 1) = 0.2706

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 1.
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e^-1 = 0.3678.
Solution:
∵ m = 1
∵ X follows Poisson Distribution

= e^-m × 1 + e^-m × 1
= e^-1 + e^-1
= 2 × e^-1
= 2 × 0.3678
= 0.7356

Question 2.
If X ~ P(\(\frac{1}{2}\)), then find P(X = 3) given e^-0.5 = 0.6065.
Solution:

Question 3.
If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e^-3 = 0.0497
Solution:
∵ X follows Poisson Distribution

Question 4.
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives (i) only two complaints on a given day, (ii) at most two complaints on a given day. Use e^-4 = 0.0183.
Solution:
∵ m = 1
∵ X ~ P(m = 4)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
X = No. of complaints recieved
(i) P(Only two complaints on a given day)

(ii) P(Atmost two complaints on a given day)
P(X ≤ 2) = p(0) + p(1) + p(2)
= \(\frac{e^{-4} \times 4^{0}}{0 !}+\frac{e^{-4} \times 4^{1}}{1 !}\) + 0.1464
= e^-4 + e^-4 × 4 + 0.1464
= e^-4 [1 + 4] + 0.1464
= 0.0183 × 5 + 0.1464
= 0.0915 + 0.1464
= 0.2379

Question 5.
A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows a Poisson distribution with a mean of 1.5. Find the probability that
(i) no car is used on a given day.
(ii) some demand is refused on a given day, given e^-1.5 = 0.2231.
Solution:
Let X = No. of demands for a car on any day
∴ No. of cars hired
n = 2
m = 1.5
∵ X ~ P(m = 1.5)

Question 6.
Defects on plywood sheets occur at random with an average of one defect per 50 sq. ft. Find the probability that such a sheet has (i) no defect, (ii) at least one defect. Use e^-1 = 0.3678.
Solution:
∵ X = No. of defects on a plywood sheet
∵ m = -1
∵ X ~ P(m = -1)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(No defect)
P(X = 0) = \(\frac{e^{-1} \times 1^{0}}{0 !}\)
= e^-1
= 0.3678

(ii) P(At least one defect)
P(X ≥ 1) = 1 – P(X < 1)
= 1 – p(0)
= 1 – 0.3678
= 0.6322

Question 7.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has
(i) exactly 5 rats
(ii) more than 5 rats
(iii) between 5 and 7 rats, inclusive. Given e^-5 = 0.0067.
Solution:
X = No. of rats
∵ m = 5
∴ X ~ P(m = 5)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(Exactly five rats)

(ii) P(More than five rats)
P(X > 5) = 1 – P(X ≤ 5)

(iii) P(between 5 and 7 rats, inclusive)
P(5 ≤ x ≤ 7) = p(5) + p(6) + p(7)

= 0.0067 × 3125 × 0.02
= 0.0067 × 62.5
= 0.42

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 1.
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of (i) 2 successes (ii) at least 3 successes (iii) at most 2 successes.
Solution:
X: Getting an odd no.
p: Probability of getting an odd no.
A die is thrown 4 times
∴ n = 4
∵ p = \(\frac{3}{6}=\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∵ X ~ B(3, \(\frac{1}{2}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(Two Successes)

(ii) P(Atleast 3 Successes)

(iii) P(Atmost 2 Successes)

Question 2.
A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
n: No. of times die is thrown = 3
X: No. of doublets
p: Probability of getting doublets
Getting a doublet means, same no. is obtained on 2 throws of a die
There are 36 outcomes
No. of doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Question 3.
There are 10% defective items in a large bulk of items. What is the probability that a sample of 4 items will include not more than one defective item?
Solution:
n: No of sample items = 4
X: No of defective items
p: Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
X ~ B(4, 0.1)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} \mathrm{q}^{n-x}\)
P(Not include more than 1 defective)
P(X ≤ 1) = p(0) + p(1)
= ⁴C₀ (0.1)⁰ (0.9)⁴ + ⁴C₁ (0.1)¹ (0.9)^4-1
= 1 × 1 × (0.9)⁴ + 4 × 0.1 × (0.9)³
= (0.9)³ [0.9 + 0.4]
= (0.9)³ × 1.3
= 0.977

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that (i) all the five cards are spades, (ii) only 3 cards are spades, (iii) none is a spade.
Solution:
X: No. of spade cards
Number of cards drawn
∴ n = 5
p: Probability of getting spade card

(i) P(All five cards are spades)

(ii) P(Only 3 cards are spades)

(iii) P(None is a spade)

Question 5.
The probability that a bulb produced by a factory will use fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of (i) X = 0, (ii) X ≤ 1, (iii) X > 1, (iv) X ≥ 1.
Solution:
X : No. of bulbs fuse after 200 days of use
p : Probability of getting fuse bulbs
No. of bulbs in a sample
∴ n = 5
∴ p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
∵ X ~ B(5, 0.2)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(X = 0) = ⁵C₀ (0.2)⁰ (0.8)^5-0
= 1 × 1 × (0.8)⁵
= (0.8)⁵

(ii) P(X ≤ 1) = p(0) + p(1)
= ⁵C₀ (0.2)⁰ (0.8)^5-0 + ⁵C₁ (0.2)¹ (0.8)^5-1
= 1 × 1 × (0.8)⁵ + 5 × 0.2 × (0.8)⁴
= (0.8)⁴ [0.8 + 1]
= 1.8 × (0.8)⁴

(iii) P(X > 1) = 1 – [p(0) + p(1)]
= 1 – 1.8 × (0.8)⁴

(iv) P(X ≥ 1) = 1 – p(0)
= 1 – (0.8)⁵

Question 6.
10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
Solution:
X : No. of balls drawn marked with the digit 0
n : No. of balls drawn
∴ n = 4
p : Probability of balls marked with 0.
∴ p = \(\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
p(x) = \({ }^{n} C_{x} p^{x} q^{n-x}\)
P(None of the ball is marked with digit 0)

Question 7.
In a multiple-choice test with three possible answers for each of the five questions, what is the probability of a candidate getting four or more correct answers by random choice?
Solution:
n: No. of Questions
∴ n = 5
X: No. of correct answers by guessing
p: Probability of getting correct answers

Question 8.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
X : No. of sixes in 6 throws
n : No. of times dice thrown
∴ n = 6
p : Probability of getting six
∴ p = \(\frac{1}{6}\)
∴ q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
∵ X ~ B(6, \(\frac{1}{6}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
P(At most 2 sixes)
P(X ≤ 2) = p(0) + p(1) + p(2)

Question 9.
Given that X ~ B(n, p),
(i) if n = 10 and p = 0.4, find E(X) and Var(X).
(ii) if p = 0.6 and E(X) = 6, find n and Var(X).
(iii) if n = 25, E(X) = 10, find p and Var(X).
(iv) if n = 10, E(X) = 8, find Var(X).
Solution:
∵ X ~ B (n, p), E(X) = np, V(X) = npq, q = 1 – p
(i) E(X) = np = 10 × 0.4 = 4
∵ q = 1 – p = 1 – 0.4 = 0.6
V(X) = npq = 10 × 0.4 × 0.6 = 2.4

(ii) ∵ p = 0.6
∴ q = 1 – p = 1 – 0.6 = 0.4
E(X) = np
∴ 6 = n × 0.6
∴ n = 10
∴ V(X) = npq = 10 × 0.6 × 0.4 = 2.4

(iii) E(X) = np
∴ 10 = 25 × p
∴ p = 0.4
∴ q = 1, p = 1 – 0.4 = 0.6
∴ S.D.(X) = √V(X)
= \(\sqrt{n p q}\)
= \(\sqrt{25 \times 0.4 \times 0.6}\)
= √6
= 2.4494

(iv) ∵ E(X) = np
∴ 8 = 10p
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
∵ V(X) = npq = 10 × 0.8 × 0.2 = 1.6

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2

Question 1.
Check whether each of the following is p.d.f.
(i) \(f(x)= \begin{cases}x & \text { for } 0 \leq x \leq 1 \\ 2-x & \text { for } 1<x \leq 2\end{cases}\)
Solution:
Given function is
f(x) = x, 0 ≤ x ≤ 1
Each f(x) ≥ 0, as x ≥ 0.


∴ The given function is a p.d.f. of x.

(ii) f(x) = 2 for 0 < x < 1
Solution:
Given function is
f(x) = 2 for 0 < x < 1 Each f(x) > 0,

∴ The given function is not a p.d.f.

Question 2.
The following is the p.d.f. of a r.v. X.
\(f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}\)
Find (i) P(X < 1.5), (ii) P(1 < X < 2), (iii) P(X > 2)
Solution:

Question 3.
It is felt that error in measurement of reaction temperature (in Celsius) in an experiment is a continuous r.v. with p.d.f.
\(f(x)= \begin{cases}\frac{x^{3}}{64} & \text { for } 0 \leq x \leq 4 \\ 0 & \text { otherwise }\end{cases}\)
(i) Verify whether f(x) is a p.d.f.
(ii) Find P(0 < X ≤ 1).
(iii) Find the probability that X is between 1 and 3.
Solution:
(i) f(x) is p.d.f. of r.v. X if
(a) f(x) ≥ 0, ∀ x ∈ R
(b) \(\int_{0}^{4} f(x) d x\) = 1

Question 4.
Find k, if the following function represents the p.d.f. of a r.v. X.
(i) \(f(x)= \begin{cases}k x & \text { for } 0<x<2 \\ 0 & \text { otherwise }\end{cases}\)
Also find P[\(\frac{1}{4}\) < X < \(\frac{1}{2}\)]
Solution:

(ii) \(f(x)= \begin{cases}k x(1-x) & \text { for } 0<x<1 \\ 0 & \text { otherwise }\end{cases}\)
Also find (a) P[\(\frac{1}{4}\) < X < \(\frac{1}{2}\)], (b) P[X < \(\frac{1}{2}\)]
Solution:
We know that

Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected student and suppose that X has p.d.f.
\(f(x)= \begin{cases}0.5 x & \text { for } 0 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Calculate (i) P(X ≤ 1), (ii) P(0.5 ≤ X ≤ 1.5), (iii) P(X ≥ 1.5).
Solution:
Given p.d.f. of X is f(x) = 0.5x for 0 ≤ x ≤ 2
∴ Its c.d.f. F(x) is given by

(i) P(X < 1) = F(1)
= 0.25(1)²
= 0.25

(ii) P(0.5 < X < 1.5) = F(1.5) – F(0.5)
= 0.25(1.5)² – 0.25(0.5)²
= 0.25[2.25 – 0.25]
= 0.25(2)
= 0.5

(iii) P(X ≥ 1.5) = 1 – P(X ≤ 1.5)
= 1 – F(1.5)
= 1 – 0.25(1.5)²
= 1 – 0.25(2.25)
= 1 – 0.5625
= 0.4375

Question 6.
Suppose X is the waiting time (in minutes) for a bus and its p.d.f. is given by
\(f(x)=\left\{\begin{array}{cl}
\frac{1}{5} & \text { for } 0 \leq x \leq 5 \\
0 & \text { otherwise }
\end{array}\right.\)
Find the probability that (i) waiting time is between 1 and 3 minutes, (ii) waiting time is more than 4 minutes.
Solution:
p.d.f. of r.v. X is given by
f(x) = \(\frac{1}{5}\) for 0 ≤ x ≤ 5
This is a constant function.
(i) Probability that waiting time X is between 1 and 3 minutes

(ii) Probability that waiting time X is more than 4 minutes

Question 7.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
\(f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
Since given f(x) is a p.d.f. of r.v. X
Since -2 ≤ x ≤ 2
∴ x² ≤ 4
∴ 4 – x² ≥ 0
∴ k(4 – x²) ≥ 0
∴ k ≥ 0 [∵ f(x) ≥ 0]

Question 8.
Following is the p.d.f. of a continuous r.v. X.
\(f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}\)
(i) Find an expression for the c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7, and 5.
Solution:
The p.d.f. of a continuous r.v. X is
\(f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}\)
(i) c.d.f. of continuous r.v. X is given by

(ii) F(0.5) = \(\frac{(0.5)^{2}}{16}=\frac{0.25}{16}=\frac{1}{64}\) = 0.015
F(1.7) = \(\frac{(1.7)^{2}}{16}=\frac{2.89}{16}\) = 0.18
For any of x greater than or equal to 4, F(x) = 1
∴ F(5) = 1

Question 9.
The p.d.f. of a continuous r.v. X is
\(f(x)=\left\{\begin{array}{cl}
\frac{3 x^{2}}{8} & \text { for } 0<x<2 \\
0 & \text { otherwise }
\end{array}\right.\)
Determine the c.d.f. of X and hence find (i) P(X < 1), (ii) P(X < -2), (iii) P(X > 0), (iv) P(1 < X < 2).
Solution:
The p.d.f. of a continuous r.v. X is

Question 10.
If a r.v. X has p.d.f.
\(f(x)= \begin{cases}\frac{c}{x} & \text { for } 1<x<3, c>0 \\ 0 & \text { otherwise }\end{cases}\)
Find c, E(X) and V(X). Also find f(x).
Solution:
The p.d.f. of r.v. X is
f(x) = \(\frac{c}{x}\), 1 < x < 3, c > 0
For p.d.f. of X, we have

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Question 1.
Let X represent the difference between a number of heads and the number of tails obtained when a coin is tossed 6 times. What are the possible values of X?
Solution:
∵ A coin is tossed 6 times
S = {6H and 0T, 5H and 1T, 4H and 2T, 3H and 3T, 2H and 4T, 1H and 5T, 0H and 6T}
X: Difference between no. of heads and no. of tails.
X = 6 – 0 = 6
X = 5 – 1 = 4
X = 4 – 2 = 2
X = 3 – 3 = 0
X = 2 – 4 = -2
X = 1 – 5 = -4
X = 0 – 6 = -6
X = {-6, -4, -2, 0, 2, 4, 6}

Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:
S : Two bolts are drawn from the Urn
S = {RR, RB, BR, BB}
X : No. of black balls
∴ X = {0, 1, 2}

Question 3.
Determine whether each of the following is a probability distribution. Give reasons for your answer.
(i)

Solution:
Here, P(X = x) ≥ 0, ∀ x and
\(\sum_{x=0}^{2} \mathrm{P}(\mathrm{X}=x)\) = p(0) + p(1) + p(2)
= 0.4 + 0.4 + 0.2
= 1
∴ The function is a p.m.f.

(ii)

Solution:
Here, p(3) = -0.1 < 0
∴ P(X = x) ≯ 0, ∀ x
∴ The function is not a p.m.f.

(iii)

Solution:
Here, P(X = x) ≥ 0, ∀ x and
\(\sum_{x=0}^{2} \mathrm{P}(\mathrm{X}=x)\) = p(0) + p(1) + p(2)
= 0.1 + 0.6 + 0.3
= 1
∴ The function is a p.m.f.

(iv)

Solution:
Here, P(Z = z) ≥ 0, ∀ z and
\(\sum_{x=-1}^{3} \mathrm{P}(\mathrm{Z}=z)\) = p(-1) + p(0) + p(1) + p(2) + p(3)
= 0.05 + 0 + 0.4 + 0.2 + 0.3
= 0.95
≠ 1
∴ The function is not a p.m.f.

(v)

Solution:
Here, P(Y = y) ≥ 0, ∀ y and
\(\sum_{x=-1}^{2} \mathrm{P}(\mathrm{Y}=y)\) = p(-1) + p(0) + p(1)
= 0.1 + 0.6 + 0.2
= 0.9
≠ 1
∴ The function is not a p.m.f.

(vi)

Solution:
Here, P(X = x) ≥ 0, ∀ x and
\(\sum_{x=0}^{0} \mathrm{P}(X=x)\) = p(-2) + p(-1) + p(0)
= 0.3 + 0.4 + 0.2
= 0.9
≠ 1
∴ The function is not a p.m.f.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin,
(ii) number of trails in three tosses of a coin,
(iii) number of heads in four tosses of a coin.
Solution:
(i) S: Coin is tossed two times
S = {HH, HT, TH, TT}
n(S) = 4
X: No. of heads
Range of X = {0, 1, 2}
p.m.f. Table

(ii) S: 3 coin are tossed
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
X: No. of heads
Range of X = {0, 1, 2, 3}
p.m.f. Table

(iii) S: Four coin are tossed
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
n(S) = 16
X: No. of heads
Range of X = {0, 1, 2, 3, 4}
p.m.f. Table

Question 5.
Find the probability distribution of the number of successes in two tosses of a die if successes are defined as getting a number greater than 4.
Solution:
S = A die is tossed 2 times
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)}
n(S) = 36
X = No. getting greater than 4
Range of X = {0, 1, 2}
p(0) = \(\frac{16}{36}=\frac{4}{9}\)
p(1) = \(\frac{16}{36}=\frac{4}{9}\)
p(2) = \(\frac{4}{36}=\frac{1}{9}\)

Question 6.
A sample of 4 bulbs is drawn at random with replacement from a lot of 30 bulbs which includes 6 defective bulbs. Find the probability distribution of the number of defective bulbs.
Solution:
Total no. of bulbs = 30
No. of defective bulbs = 6
A sample of 4 bulbs are drawn from 30 bulbs.
∴ n(S) = \({ }^{30} \mathrm{C}_{4}\)
∴ No. of non-defective bulbs = 24
Let X = No. of defective bulbs drawn in sample of 4 bulbs.

Question 7.
A coin is biased so that the head is 3 times as likely to occur as the tail. Find the probability distribution of a number of tails in two tosses.
Solution:
Here, the head is 3 times as likely to occur as the tail.
i.e., If 4 times coin is tossed, 3 times there will be a head and 1 time there will be the tail.
∴ p(H) = \(\frac{3}{4}\) and p(T) = \(\frac{1}{4}\)
Let X : No. of tails in two tosses.
And coin is tossed twice.
∴ X = {0, 1, 2}
For X = 0,
p(0) = p(both heads)
= p(H) × p(H)
= \(\frac{3}{4} \times \frac{3}{4}\)
= \(\frac{9}{16}\)
For X = 1,
p(1) = p(HT or TH)
= p(HT) + p(TH)
= p(H) × p(T) + p(T) × p(H)
= \(\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}\)
= \(\frac{6}{16}\)
For X = 2,
p(2) = p(both tails)
= p(T) × p(T)
= \(\frac{1}{4} \times \frac{1}{4}\)
= \(\frac{1}{16}\)
The probability distribution of the number of tails in two tosses is

Question 8.
A random variable X has the following probability distribution:

Determine (i) k, (ii) P(X < 3), (iii) P(0 < X < 3), (iv) P(X > 4).
Solution:
(i) It is a p.m.f. of r.v. X
∴ Σp(x) = 1
∴ p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
∴ k + 2k + 2k + 3k + k² + 2k² + (7k² + k) = 1
∴ 10k² + 9k = 1
∴ 10k² + 9k – 1 = 0
∴ 10k² + 10k – k – 1 = 0
∴ 10k(k + 1) – (k + 1) = 0
∴ (10k – 1)(k + 1) = 0
∴ 10k – 1 = 0 or k + 1 = 0
∴ k = \(\frac{1}{10}\) or k = -1
but k = -1 is not accepted
∴ k = \(\frac{1}{2}\) is accepted

(ii) P(X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

(iii) P(0 < X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

(iv) P(X > 4) = p(5) + p(6) + p(7)
= k² + 2k² + (7k² + k)
= 10k² + k
= \(10\left(\frac{1}{10}\right)^{2}+\frac{1}{10}\)
= \(\frac{2}{10}\)
= \(\frac{1}{5}\)

Question 9.
Find expected value and variance of X using the following p.m.f.

Solution:

E(X) = Σxp = -0.05
V(X) = Σx²p – (Σxp)²
= 2.25 – (-0.05)²
= 2.25 – 0.0025
= 2.2475

Question 10.
Find expected value and variance of X, the number on the uppermost face of a fair die.
Solution:
S : A fair die is thrown
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
X: No obtained on uppermost face of die
Range of X = {1, 2, 3, 4, 5, 6}

E(X) = Σxp = \(\frac{21}{6}=\frac{7}{2}\) = 3.5
V(X) = Σx²p – (Σxp)²
= \(\frac{91}{6}\) – (3.5)²
= 15.17 – 12.25
= 2.92

Question 11.
Find the mean of the number of heads in three tosses of a fair coin.
Solution:
S : A coin is tossed 3 times
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Range of X = {0, 1, 2, 3}

∴ Mean = E(X) = Σxp = \(\frac{12}{8}=\frac{3}{2}\) = 1.5

Question 12.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
S : Two dice are thrown
S = {(1, 1), (1, 2), (1, 3), ……, (6, 6)}
n(S) = 36
Range of X = {0, 1, 2}
First 6 positive integers are 1, 2, 3, 4, 5, 6
X = Larger two numbers selected
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)}
n(S) = 36

∴ E(X) = Σxp = \(\frac{12}{36}=\frac{1}{3}\)

Question 13.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Solution:
First 6 positive integers are 1, 2, 3, 4, 5, 6
X : The larger of the selected two numbers
S = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
n(S) = 30

E(X) = Σxp = \(\frac{140}{30}=\frac{14}{3}\) = 4.67

Question 14.
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance of X.
Solution:
S : Two fair dice are rolled
S = {(1, 1), (1, 2), (1, 4), ……, (6, 6)}
n(S) = 36
X : Sum of the two numbers.
Range of X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}


V(X) = Σx²p – (Σxp)²
= \(\frac{1952}{36}-\left(\frac{252}{36}\right)^{2}\)
= 54.22 – (7)²
= 5.22
SD(X) = √V(X) = √5.22 = 2.28

Question 15.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. If X denotes the age of a randomly selected student, find the probability distribution of X. Find the mean and variance of X.
Solution:

Question 16.
70% of the member’s favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and V(X).
Solution:

E(X) = Σxp = 0.7
V(X) = Σx²p – (Σxp)²
= 0.7 – (0.7)²
= 0.7 – 0.49
= 0.21

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

(I) Choose the correct alternative.

Question 1.
In sequencing, an optimal path is one that minimizes ___________
(a) Elapsed time
(b) Idle time
(c) Both (a) and (b)
(d) Ready time
Answer:
(c) Both (a) and (b)

Question 2.
If job A to D have processing times as 5, 6, 8, 4 on first machine and 4, 7, 9, 10 on second machine then the optimal sequence is:
(a) CDAB
(b) DBCA
(c) BCDA
(d) ABCD
Answer:
(b) DBCA

Question 3.
The objective of sequence problem is
(a) to find the order in which jobs are to be made
(b) to find the time required for the completing all the job on hand
(c) to find the sequence in which jobs on hand are to be processed to minimize the total time required for processing the jobs
(d) to maximize the cost
Answer:
(c) to find the sequence in which jobs on hand are to be processed to minimize the total time required for processing the jobs

Question 4.
If there are n jobs and m machines, then there will be ___________ sequences of doing the jobs.
(a) mn
(b) m(n!)
(c) n^m
(d) (n!)^m
Answer:
(d) (n!)^m

Question 5.
The Assignment Problem is solved by
(a) Simple method
(b) Hungarian method
(c) Vector method
(d) Graphical method
Answer:
(b) Hungarian method

Question 6.
In solving 2 machine and n jobs sequencing problem, the following assumption is wrong
(a) No passing is allowed
(b) Processing times are known
(c) Handling times is negligible
(d) The time of passing depends on the order of machining
Answer:
(d) The time of passing depends on the order of machining

Question 7.
To use the Hungarian method, a profit maximization assignments problem requires
(a) Converting all profit to opportunity losses
(b) A dummy person or job
(c) Matrix expansion
(d) Finding the maximum number of lines to cover all the zeros in the reduced matrix
Answer:
(a) Converting all profits to opportunity losses

Question 8.
Using the Hungarian method the optimal assignment obtained for the following assignment problem to minimize the total cost is:

(a) 1 – C, 2 – B, 3 – D, 4 – A
(b) 1 – B, 2 – C, 3 – A, 4 – D
(c) 1 – A, 2 – B, 3 – C, 4 – D
(d) 1 – D, 2 – A, 3 – B, 4 – C
Answer:
(a) 1 – C, 2 – B, 3 – D, 4 – A

Question 9.
The assignment problem is said to be unbalanced if
(a) Number of rows is greater than the number of columns
(b) Number of rows is lesser than number of columns
(c) Number of rows is equal to the number of columns
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 10.
The assignment problem is said to be balanced if
(a) Number of rows is greater than the number of columns
(b) Number of rows is lesser than number of columns
(c) Number of rows is equal to the number of columns
(d) If the entry of rows is zero
Answer:
(c) Number of rows is equal to number of columns

Question 11.
The assignment problem is said to be balanced if it is a
(a) Square matrix
(b) Rectangular matrix
(c) Unit matrix
(d) Triangular matrix
Answer:
(a) Square matrix

Question 12.
In an assignment problem if the number of rows is greater than the number of columns then
(a) Dummy column is added
(b) Dummy row is added
(c) Row with cost 1 is added
(d) Column with cost 1 is added
Answer:
(a) Dummy column is added

Question 13.
In a 3 machine and 5 jobs problem, the least of processing times on machines A, B, and C are 5, 1 and 3 hours and the highest processing times are 9, 5 and 7 respectively, then it can be converted to a 2 machine problem if the order of the machines is:
(a) B – A – C
(b) A – B – C
(c) C – B – A
(d) Any order
Answer:
(b) A – B – C

Question 14.
The objective of an assignment problem is to assign
(a) Number of jobs to equal number of persons at maximum cost
(b) Number of jobs to equal number of persons at minimum cost
(c) Only the maximize cost
(d) Only to minimize cost
Answer:
(b) Number of jobs to equal number of persons at minimum cost

(II) Fill in the blanks.

Question 1.
An assignment problem is said to be unbalanced when ___________
Answer:
the number of rows is not equal to the number of columns

Question 2.
When the number of rows is equal to the Number of columns then the problem is said to be ___________ assignment problem.
Answer:
balanced

Question 3.
For solving assignment problem the matrix should be a ___________
Answer:
square matrix

Question 4.
If the given matrix is not a ___________ matrix, the assignment problem is called an unbalanced problem.
Answer:
square

Question 5.
A dummy row(s) or column(s) with the cost elements as ___________ the matrix of an unbalanced assignment problem as a square matrix.
Answer:
zero

Question 6.
The time interval between starting the first job and completing the last, job including the idle time (if any) in a particular order by the given set of machines is called ___________
Answer:
Total elapsed time

Question 7.
The time for which a machine j does not have a job to process to the start of job i is called ___________
Answer:
Idle time

Question 8.
The maximization assignment problem is transformed to minimization problem by subtracting each entry in the table from the ___________ value in the table.
Answer:
maximum

Question 9.
When the assignment problem has more than one solution, then it is ___________ optimal solution.
Answer:
multiple

Question 10.
The time required for printing four books A, B, C, and D is 5, 8, 10, and 7 hours. While its data entry requires 7, 4, 3, and 6 hrs respectively. The sequence that minimizes total elapsed time is ___________
Answer:
A – D – B – C

(III) State whether each of the following is True or False.

Question 1.
One machine – one job is not an assumption in solving sequencing problems.
Answer:
False

Question 2.
If there are two least processing times for machine A and machine B, priority is given for the processing time which has the lowest time of the adjacent machine.
Answer:
True

Question 3.
To convert the assignment problem into a maximization problem, the smallest element in the matrix is deducted from all other elements.
Answer:
False

Question 4.
The Hungarian method operates on the principle of matrix reduction, whereby the cost table is reduced to a set of opportunity costs.
Answer:
True

Question 5.
In a sequencing problem, the processing times are dependent on the order of processing the jobs on machines.
Answer:
False

Question 6.
The optimal assignment is made in the Hungarian method to cells in the reduced matrix that contain a Zero.
Answer:
True

Question 7.
Using the Hungarian method, the optimal solution to an assignment problem is fund when the minimum number of lines required to cover the zero cells in the reduced matrix equals the number of people.
Answer:
True

Question 8.
In an assignment problem, if a number of columns are greater than the number of rows, then a dummy column is added.
Answer:
False

Question 9.
The purpose of a dummy row or column in an assignment problem is to obtain a balance between a total number of activities and a total number of resources.
Answer:
True

Question 10.
One of the assumptions made while sequencing n jobs on 2 machines is: two jobs must be loaded at a time on any machine.
Answer:
False

(IV) Solve the following problems.

Part – I

Question 1.
A plant manager has four subordinates, and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. This estimate of the times each man would take to perform each task is given in the effectiveness matrix below.

How should the tasks be allocated, one to a man, as to minimize the total man-hours?
Solution:
The hr matrix is given by

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get

The minimum no. of lines covering ail the zeros (4) is equal to the order of the matrix (4)
∴ The assignment is possible.

The assignment is
A → I, B → III, C → II, D → IV
For the minimum hrs. take the corresponding value from the hr matrix.
Minimum hrs = 7 + 3 + 18 + 9 = 37 hrs

Question 2.
A dairy plant has five milk tankers, I, II, III, IV & V. These milk tankers are to be used on five delivery routes A, B, C, D & E. The distances (in kms) between the dairy plant and the delivery routes are given in the following distance matrix.

How should the milk tankers be assigned to the chilling centre so as to minimize the distance travelled?
Solution:
The distance matrix is given by

Subtracting row minimum from all values in that row we get

Subtracting column minimum from each value in that column we get

The number of lines covering all the zeros (3) is less than the order of the matrix (5) so the assignment is not possible. The modification is required.
The minimum uncovered value (15) is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same. We get

The minimum lines covering all the zeros (4) are less than the order of the matrix (5) so the assignment is not possible. The modification is required the minimum uncovered value (5) is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same we get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) So assignment is possible.
The assignment is
A → II, B → III, C → V, D → I, E → IV
Total minimum distance is = 120 + 120 + 175 + 40 + 70 = 525 kms.

Question 3.
Solve the following assignment problem to maximize sales:

Solution:
As it is a maximization problem so we need to convert it into a minimization problem.
Subtracting all the values from the maximum value (19) we get

Also, it is an unbalanced problem so we need to add a dummy row (E) with all values zero, we get

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get the same matrix

The minimum number of lines covering all the zero (4) is less than the order of the matrix (5) So assignment is not possible. The modification is required. The minimum uncovered value (2) is subtracted from the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get

The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so the assignment is not possible. The modification is required. The minimum uncovered value (1) is subtracted from the uncovered value and added to the values at the intersection. The values on the lines remain the same. We get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.

The assignment is
A → V, B → II, C → IV, D → III, E → I
No salesman goes to I as E is a dummy row.
For the maximum value take the corresponding values from the original matrix.
We get Maximum value = 15 + 19 + 14 + 17 + 0 = 65 units

Question 4.
The estimated sales (tons) per month in four different cities by five different managers are given below:

Find out the assignment of managers to cities in order to maximize sales.
Solution:
This is a maximizing problem. To convert it into minimizing problem subtract all the values of the matrix from the maximum (largest) value (39) we get

Also as it is an unbalanced problem so we have to add a dummy column (T) with all the values as zero. We get

Subtracting row minimum from all values in that row we get the same matrix
Subtracting column minimum from all values in that column we get

The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so assignments are not possible. The modification is required. The minimum uncovered value (1) is subtracted from the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.

So I → S, II → T, III → Q, IV → P, V → R.
As T is dummy manager II is not given any city.
To find the maximum sales we take the corresponding value from the original matrix
Total maximum sales = 35 + 39 + 36 + 35 = 145 tons

Question 5.
Consider the problem of assigning five operators to five machines. The assignment costs are given in the following table.

Operator A cannot be assigned to machine 3 and operator C cannot be assigned to machine 4. Find the optimal assignment schedule.
Solution:
This is a restricted assignment problem, so we assign a very high cost (oo) to the prohibited cells we get

Subtracting row minimum from all values in that row we get.

Subtracting column minimum from all values in that column we get

As the minimum number of lines covering all the zeros (4) is equal to the order of the matrix (5) so the assignment is not possible. The modification is required. The minimum uncovered value (2) is subtracted from all the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get

As the minimum number of lines covering all the zeros (5) is equal to the order of the matrix, assignment is the possible

So A → 4, B → 3, C → 2, D → 1, E → 5
For the minimum cost take the corresponding values from the cost matrix we get
Total minimum cost = 3 + 3 + 4 + 3 + 7 = 20 units

Question 6.
A chartered accountant’s firm has accepted five new cases. The estimated number of days required by each of their five employees for each case are given below, where-means that the particular employee can not be assigned the particular case. Determine the optimal assignment of cases of the employees so that the total number of days required to complete these five cases will be minimum. Also, find the minimum number of days.

Solution:
This is a restricted assignment problem so we assign a very high cost (∞) to all the prohibited cells. The day matrix becomes

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get

The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so the assignment is not possible, The modification is required. The minimum uncovered value (1) is subtracted from all the uncovered values and added to the values at the intersection. The values on the lines remain the same, we get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.

So E₁ → I, E₂ → IV, E₃ → II, E₄ → V, E₅ → III
To find the minimum number of days we take the corresponding values from the day matrix.
Total minimum number of days = 6 + 6 + 6 + 6 + 3 = 27 days

Part – II

Question 1.
A readymade garments manufacture has to process 7 items through two stages of production, namely cutting and sewing. The time taken in hours for each of these items in different stages are given below:

Find the sequence in which these items are to be processed through these stages so as to minimize the total processing time. Also, find the idle time of each machine.
Solution:
Let A = cutting and B = sewing. So we have

Observe min {A, B} = 2 for item 1 for B.

The problem reduces to

Now min {A, B} = 3 for item 3 for A

The problem reduces to

New min {A , B} = 4 for item 4 for A.

The problem reduces to

Now min(A, B} = 5 for item 6 for B

The problem reduces to

Now min {A, B} = 6 for item 5 for A and item 2 for B

Now only 7 is left
∴ The optimal sequence is

Worktable

Total elapsed time = 46 hrs
Idle time for A (cutting) = 46 – 44 = 2 hrs
Idle time for B (Sewing) = 4 hrs

Question 2.
Five jobs must pass through a lathe and a surface grinder, in that order. The processing times in hours are shown below. Determine the optimal sequence of the jobs. Also, find the idle time of each machine.

Solution:
Let A = lathe and B = surface grinder. We have

Observe min {A, B} = 1 for job II for A

The problem reduces to

Now min {A, B} = 2 for job IV for A

The problem reduces to

Now min {A, B} = 3 for job I for B

The problem reduces to

Now min {A, B} = 5 for jobs III and V for A
∴ We have two options

or

We take the first one.
Worktable

Total elapsed time = 21 hrs
Idle time for A (lathe) = 21 – 17 = 4 hrs
Idle time for B (surface grinder) = 3 hrs

Question 3.
Find the sequence that minimizes the total elapsed time to complete the following jobs. Each job is processed in order AB.

Determine the sequence for the jobs so as to minimize the processing time. Find the total elapsed time and the idle time for both machines.
Solution:
Observe min {A, B} = 3 for job VII on B.

The problem reduces to

Now min {A, B} = 4 for job IV on B.

The problem reduces to

Now min {A, B} = 5 for job III & V on A. we have two options

or

We take the first one
The problem reduces to

Now min {A, B} = 5 for job II on A

The problem reduces to

Now min {A, B} = 7 for a job I on B and for job VI on A
∴ The optional sequence is

Worktable

Total elapsed time = 55 units
Idle time for A = 55 – 52 = 3 units
Idle time for B = 9 units.

Question 4.
A toy manufacturing company has five types of toys. Each toy has to go through three machines A, B, C in the order ABC. The time required in hours for each process is given in the following table.

Solve the problem for minimizing the total elapsed time.
Solution:
Min A = 12, Max B = 12
As min A ≥ max B.
The problem can be converted into two machine problems.
Let G and H be two fictitious machines such that G = A + B and H = B + C, We get

Now min {G, H} = 16 for type 3 on G

The problem reduces to

Min (G, H} = 18 for type 1, 4 & 5 on H
We have more than one option, we take

Now only type 2 is left.
∴ The optional sequence is

Worktable

Total elapsed time = 102 hours
Idle time for A = 102 – 84 = 18 hours
Idle time for B = 54 + (102 – 94) = 62 hours
Idle time for C = 38 hours

Question 5.
A foreman wants to process 4 different jobs on three machines: a shaping machine, a drilling machine, and a tapping, the sequence of operations being shaping-drilling-tapping. Decide the optimal sequence for the four jobs to minimize the total elapsed time. Also, find the total elapsed time and the idle time for every machine.

Solution:
The time matrix is

Min A = 8, Max B = 8, as min A ≥ max B.
The problem can be converted into a two-machine problem.
Let G and H be two fictitious machines such that
G = A + B and H = B + C we get

Observe min (G, H} = 12 for job 2 on H

The problem reduces to

Now min {G, H} = 14 for job 3 on G and job 4 on H

Now only job 1 is left.
∴ The optimal sequence is

Worktable

Total elapsed time = 74 min
Idle time for A (shapping) = 74 – 62 = 12 min
Idle time for B (Drilling) = 47 + (74 – 70) = 51 min
Idle time for C (trapping) = 31 min