Physical chemistry volume 1

i The characteristic of the logarithm of a number greater than olle is positive and is numerically one less than the number of digits before the decimal point.. the second person can si

P~9a~~ -PHYSICAL CHEMISTRY

loP College, Bulandshahr

IIUlij'

MATHEMATICAL CONCEPTS AND COMPUTER

1 Logarithmic and Antilogarithmic Relations

2 Find out the values of the following

3 Differentiation with Examples

51 Logarithmic, Trigonometric Series

52 Maxima and Minima

53-57 Numerical Problem

58 Functioning, Characteristics, Limitations

Computer Programing Flow Charts

Fortan, Cobol, Basic, Pascal

Operating System

Exercise

o Multiple Choice Questions

o Fill in the Blanks

13

16

22 23

1 Nature of R and its value in different units 56

2 Short account of kinetic theory of gases and derivation of 57 kinetic equation

3 (a) Short account of kinetic theory of gases Derivation of 59

PV = RT and show how the various gas laws are consistent with it?

(b) Expression for kinetic energy of one mole of gas

4 Values of C v and Cp from kinetic equation and variation of 62

CplC v with molecular complexity of the gas

5 Distribution of molecular velocities of Maxwell's law 65

6 (a) Average velocity, root me~n square velocity and most 67

probable velocity and relation among them

(b) Calculation of RMS velocity from kinetic theory of gases 67

8 Ideal gas and its difference with a real gas 70

9 (a) Limitations of PV = RT and improvements suggested by 70

vander Waals Derivation of vander Waals equation

(b) Units of vander Waals constants 70 (c) Show that effective volume of gas molecules is four 70 times greater than actual volume of molecules

10 (a) Critical phenomenon, calculation and determination of 75

critical constants, short note on continuity of state

11 Some short questions on vander Waals equation 79

(a) Various equation of state

(b) Law of corresponding states

(c) Mean free path

(d) Critical phenomenon and its utility

(e) Collision frequency

(0 Law of equipartition of energy

(g) Specific heat ratio

(h) Boyle temperature

(i) Continuity of state

13 Methods for producing cold and liquefaction of gases, 86 inversion temperature

IIn"j'"

CHEMICAL AND PHASE EQUILmRIUM

I Chemical Equilibrium I

1 Law of mass action and eqUilibrium constant

2 Short notes on the following :

(i) Work function (ii) Free energy

3 Thermodynamic derivation of law of mass action

103-171

103

104

108

5 Thermodynamlc··-derivation of van't Hoff isochore or van't 111 Hoff equation

6 Thermodynamic derivation of Clapeyron equation and 114 Clausius-Clapeyron equation

7 Le-Chatelier-Braun principle and applications to different 118 equilibria

I Phase Equilibrium I

1 Explanation and illustration of phase, component and degree 126

of freedom

2 Phase rule and its thermodynamic derivation -129

3 Explain: Can all four phases in a one component system 131 co-exist in equilibrium?

4 Application of phase rule to water system

5 Application of phase rule to sulphur system

8 Application of phase rule to lead-silver system 140

9 Application of phase rul~ to potassium iodide and water 142 system

10 Determination of number of phases and components of 143 different systems

12 Determination of number of phases, components and degree 144

of freedom of different systems

13 Ideal solutions, vapour pressure of such solutions 144

14 Non-ideal or real solutions, vapour pressure curves of 147 completely miscible binary solutions

15 Theory of fractional distillation of binary solutions

16 (a) Theory of partially miscible liquid pairs, e.g.,

(i) Phenol-water system

Oi) Triethyl-amine water system

150

154

(iii) Nicotine-water system

(b) Influence of impurities on critical solution temperature 154

I Distribution Law I

1 Nemst's distribution law limitations applications 157

2 Nemst's distribution law modification when the solute 162 undergoes dissociation or association

o Numerical Problems

o Multiple Choice Questions

o Fill in the Blanks

3 Preparation of colloidal solutions of AS2S3 Fe(OH)3 gold 178 sulphur silicic acid carbon mastic iodine

(xii) Gold number

(xiii) Stability of lyophilic colloids

(xiv) Iso-electric point

(xv) Emulsion

(xvi) Gel

(xvii)Electrical double layer or Zera potential

5 (a)

(b)

Origin and significance of charge on a colloidal particle 198

Gold, Fe(OH)3, gelatin, blood, sulphur, AS2S3

(a) A sulphur sol is coagulated by adding a little electrolyte, whereas a gelatin sol is apparently unaffected

(b) What happens when a colloidal solution of gold is brought under the influence of electric field?

(c) What happens when an electrolyte is added to colloidal solution of gold?

(d) What happens when a beam of light is passed through a colloidal solution of gold?

(e) A colloidal solution is stabilised by addition of gelatin

(f) Presence of H2S is essential in AS2S3 sol though H2S ionises and should precipitate the sol

(g) Why ferric chloride or alum is used for stoppage of bleeding?

7 Applications of colloids in chemistry

8 Sol-gel transformation

9 Note on thixotropy

o Multiple Choice Questions

o Fill in the Blanks

o True or False

III~I'M

200 2C4

1 (a) Explain the terms: rate of chemical reaction, velocity 211

coefficient, molecularity and order of reaction

(b) Difference between molecularity and order of reaction (c) Why reactions of higher orders are rare?

(d) Factors which affect reaction rates?

2 Zero order reaction, rate expression, characteristics 215

3 Half order reaction, rate expression, characteristic 216

4 First order reaction, rate expression, characteristics, examples 217

(b) Study of kinetics of hydrolysis of methyl acetate

6 Half life period for a first order reaction 220

examples and study of kinetics

8 Half life period for a second order reaction 224

9 Third order reaction, rate expressions, characteristics and 225 examples

10 nth order reactions, rate equation and characteristic 227

11 Methods employed in determining the order of reaction 228

12 Energy of activation and temperature coefficient 230

13 Activation energy, potential energy barrier and Arrhenius law 233

14 Collision theory for unimolecular reactions 235

15 Mathematical treatment of transition state theory, comparison 236 with collision theory

I Catalysis I

1 Catalyst, catalysis, types and classification of catalysis 251 characteristics of catalytic reactions

(a) Catalytic promoters

(b) Catalytic poisons

3 Theories of catalysis, industrial applications of catalysts 261

4 Enzyme catalysis, characteristics and examples of enzyme 265 catalysis, kinetics of enzyme catalysis

Definition: If a b = c; then exponent 'b' is called the logarithm of number

'c' to the base 'a' and is written as log c = b, e.g., J~ = 81 ~ logarithm of

81 to the base 3 is 4, i.e., log3 81 = 4

Note: a b = c, is called the exponential form and loga c = b is called the

logarithmic form, i.e.,

(i)

(ii)

Laws of Logarithms

T3 = 0.125 log2 0.125 =-3

The logarithm of a product is equal to the slim of logarithms of its factors

logo (m X 11) = log" III + log" /I

loga (m X n xp) = loga m + logan + lo~p

Remember:

10& (m + n);/: logam + logan

[II] Second Law (Quotient law) :

The logarithm of a fractl~n is equal to the difference between the logarithm of numerator and (he logarithm of denominator

Remember:

loga m

-1 ;/: loga m - loga n oga Il

[III] Third Law (Power law) :

The logarithm of a power of a number is equal to the logarithm of the number multiplied by the power

(a) Logarithms to the base 10 are known as common logarithms

(b) If no base is given, the base is always taken as 10

(c) Logarithm of a number to the same base is always one, i.e.,

loga a = 1; 10glO 10 = 1 and so on

(d) The logarithm of 1 to any base is zero, i.e.,

loga 1 = 0; logs 1 = 0; 10glO 1 = 0 and so on

(e) 10gJO 1 = 0; 10gJO 10 = 1;

10gIO 100 = 2 [.: 10gIO 100 = 10gJO 102 = 2 log 10 10 = 2 xl = 2] Similarly, 10g]O 1000 = 3; 10gIO 10000 = 4 and so on

Example: If log 2 = 0.3010 and log 3 = 0.4771; find the value of:

(i) log 6 (ii) log 5

(': log 10= 1)

(iii) log J24 = log (24)1/2 = ~ log (23 X 3)

[I] Common Logarithms : Logarithms to the base 10 are known as

common Logarithms If no base is given, the base is always taken as 10 [II] Characteristics and Mantissa: The logarithm of anum ber con-sists of two parts :

(i) Characteristic-It is the integral part of the logarithm

(ii) Mantissa-It is the fractional or decimal part of the logarithm

For exampLe, in log 273 = 2.4362, the integral part is 2 and the decimal part is 4362

Therefore, characteristic = 2 and mantissa = 4362

[III] How to Find Characteristic?

(i) The characteristic of the logarithm of a number greater than olle

is positive and is numerically one less than the number of digits before the decimal point

In number 475.8~ the number of digits before the decimal point is three Characteristic of log 475.8 = 2, i.e., (3 - 1 = 2)

Similarly, Characteristic oflog 4758 = 3, i.e., (4 - 1 = 3)

Characteristic of log 47.58 = 1, i.e., (2 - 1 = 1)

Characteristic of log 4.758 = 0, i.e., (l - 1 = 0)

(il) The characteristic of the logarithm of a number less than one in negative and is numerically one more than the number of zeros immediately after decimal point

The number 0.004758 is less t;lan one and the number of zeros mediately after decimal point in it are two

im-~ : Characteristic of log 0.004758 = - (2 + 1) = -3, which is also written

as 3

Note: To find the characteristic of the logarithm of a number less than one, count the number of zeros immediately after the decimal point and add one to it The number so obtained with negati~ sign gives the characteristic Characteristic of log 0.3257 = - 1 = 1

[Since, the number of ~eros after decimal point = 0] Characteristic of log 0.03257 = - 2 = 2

[Since, the number of zeros aftet; decimal point = 1] Characteristic of log 0.0003257 = -4 = 4 and so on

[IV] How to Find Mantissa?

The mantissa of the logarithm of a number can be obtained from the logarithmic table

30

31

32

A logarithmic table consists of three parts:

(1) A column at the extreme left contains two digit numbers starting from 10 to 99

(2) Ten columns headed by the digits O 1,2, 3,4,5,6, 7, 8,9 (3) Nine more columns headed by digits 1,2,3,4,5,6.7,8,9

A part of the logarithmic table is given below: (Difference to be added)

4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 134 678 JO 11 13

4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 1 34 678 JO 11 12

5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 134 678

(a) To find the mantissa of the logarithm of one digit number

Let the number be 3

: Mantissa of log 3 = Value of the number 30 under O

= 0.477l

(b) To find the mantissa of the logarithm of two digit number

Let the number be 32

: Mantissa of log 32 = Value of the number 32 under O

= 0.5051

9 11 12

(c) To find the mantissa of the logarithm of three digit number

Let the number be 325

: Mantissa of log 325 = Value of 32 under 5 = 0.5119

(d) To find the mantissa of the logarithm of a four digit number

Let the number be 3257

Mantissa of log 3257 = Value of 32 under 5 plus the difference

log 3257 = 3.5128 [Note: The mantissa of the logarithms of all the numbers having the same significant digits is the same While tinding the mantissa, ignore the decimal point.]

For example:

[Note :

log 3257 = 3.5128 log 325.7 = 2.5128 log 3.257 = 0.5128 log 0.3257 = 1.5128 log 0.003257 = 3.5128 and so on]

(1) 3.4682 is equivalent to (-3 + 0.4682) and -3.4682

= -(3 + 0.4682) = -3 - 0.4682 i.e., in 3.4682, the mantissa is itive, while in -3.4682, the mantissa is negative

pos-(2) Remember, the mantissa should always be written positive

(3) To make the mantissa positive, subtract 1 from the integral part and add 1 to the decimal paJ;t, Thus,

[E] Antilogarithms

If log 5274 = 3.7221, then 5274 is called the antilogarithm of 3.722l

We write, antilog 3.7221 = 5274

To find an antilogarithm, the antilogarithm tables are used

The antilogarithm tables are used in the same way as the logarithm

tables The only difference between the two tables is that the column at the

extreme left of the log table contains all two digit numbers starting from 10

to 99; whereas an antilog table contains numbers from 00 to 99 (i.e., all fractional numbers with only two digits after decimal) in the extreme left column of it

is 368

From antilog table, the value of 36 under 8 is 2333

Since the characteristic of log of the number is 2

: The number has 2 + 1 = 3 digits in its integral part (i.e., 3 digits before the decimal point)

Antilog 2.368 = 233.3

Ex 2 : Find the antilog of 2.3536

Solution: From the antilog table, find the value of 35 under 3 and add

to it the mean difference under 6 The number thus obtained is 2257

Now place the decimal point so that the characteristic of its log is 2 Antilog 2.3536 = 0.02257

Problem 3: Explain differentiation with suitable examples

[I] Differentiation

respect to every change in x If the increment in x is denoted by cSx, increment

is called the difference quotient

But, if cSx -+ 0, then this ratio tends to a definite quantity This definite quantity is called the differential coefficient of f(x) with respect to x and it is denoted by

d1;) or ~ or f'(x) or y'

Therefore, differential coefficient of f(x) with respect to x,

!!l = !:lfJE = lim f(x + cSx) -fix)

The process of finding differential coefficient of f(x) with respect to x

is called 'differentiation' If Ox = It, then

!!l = lim f(x + It) - f(x)

[II] Some Standard Derivatives

(1) ! (xn) = nxn - 1, where n is a real number

For example, differential coefficient of ;?= 5x 5 - 1 = 5x 4

(6) dx (cosec x) = - cosec x cotx

(8) ~ (cos I1lx) = - m sin mx

[IV] Differentials of Inverse Functions

[VI] Differentiation of a Division

If u and v are any two functions of x, then

[VII] Partial Differentiation

If u == fix, y) be a continuous function of two independent variables

x and y, then the differential cofficient w.r.t x (taking y as constant) is called the partial derivative or partial differential coefficient of u w.r.t x and is represented by different symbols such as

Similarly, by keeping x constant and allowing y alone to vary, we can

define the partial derivative or partial differential coefficient of u w.r.t y It

is represented by anyone of the following symbols

au af oy' Oy,/y (x, y),/y

Symbolically, ou = lim f(x, y + By) - fix, y)

[I] Rules of Partial Differentiation

Rule (1) : (a) If u is a function of x, y and we are to differentiate partially w.r.t x, then y is treated as constant

(b) Similarly, if we are to differentiate u partially w.r.t y, then x is treated

as constant

(c) If u is a function of x, y, z, and we are to differentiate u w.r.t x, then

y and z are treated as constants

Problem 4 Determine thejirst order partial derivatives ofyx

Problem 6 Find the differential coefficients of the following:

Solution

(i) :.x (ax 4 ) = a :.x (x4) = 4ax 3

(ii) :.x (7 loglO x) = 7 :.x (loglO x) = 7 ~ loglo (e)

Problem 7 Find the differential coefficients oj the following:

(i) 4x 2 + ~ + 10 (ii) tan x 1-sin x

Problem 8 Find the differential coefficients of the following:

Problem 10 Calculate the value of ~ from the following:

is called the integration of c1>(x) with respect to x

[II] Tables 01 Integral Formulae

(11) f tan x dx = loge sec x + C = - 10& cos u + C

(12) fcotx dx= 10& sin x + C = -loge cosec x + C

(13) fcosec x dx = loge (cosec x - cot u) + C = loge tan ~ + C

(14) f sin2 xdx= t x - t sin x cos x + C

(15) J cos2 x dx = t x + t sin x cos x + C

(16) Jsec2 x dx = tan x + C

(17) f cosec2 x dx = - cot x + C

(18) ftan2 x dx = tan x - x + C

(19) fcot 2 xdx=-cotx-x+C

(20) f sec x tan x dx = sec x

(21) Icosec xcotx dx= - cosec x

[III] Definite Integral

When any functionf(x) is integrated between the lower limit and upper limit of x, then it is called the definite integral

For f(x), if the lower and upper limits of x are a and b, respectively, then

Solution Multiplying numerator and denominator by (sec x + tan x)

f sec x + tan x sec x - tan x dx = f (sec x + tan (sec x - tan x) (sec x + tan x) x) (sec x + tan x) dx

-f (sec x + tanx)2 dx

- sec2 x - tan2 x

= f (sec x + tan X)2 dx (" sec2 x - tan2 x = 1)

= I see2 x + tan2 x + 2 sec x tan x dx

= I see2 x + (see2 x-I) + 2 sec x tan x dx

= I (2 see2 x-I + 2 sec x tan x) dx

= I 2 see2 x dx - I I dx + I 2 sec x tan x dx

= 2 f see2 x dx - f l dx + 2 f sec x tan x dx

=2tanx-x+2secx Problem 19 Evaluate the following integrals by substitution method:

or

Solution (i) In I (ax + bf dx, put ax + b = t, so that

adx=dt dx= dt

f cos (ax + b) dx = f cos t ~ = ~ f cos t dt

1 1 ( b)

= - SIn t = - SIn ax +

Problem 20 Evaluate the /oUowing integrals:

(i) f~dx (ii) f log x dx

Problem 21 Evaluate the following integrals:

Solution (i) Put x 2 + 3x + 2 = t

(ii) Put log x = t

Problem 22 Evaluate the following integrals:

(i) f log X dx (ii) f xex dx

(iii) f x log x dx (iv) f xsinxdx

Solution (i) f log x dx = f l log x dx

(iv) f x sin x dx = x f sin x dx - f { fx (x) f sin x dx } dx

= -x COSJ + sin x

Problem 23 Evaluate the integral eX sin x dx

Solution (i) Let 1= f e" sin x dx

= e' sm x - cos x e' x

or

or

til1le some or all elements of any set of things, thell such every arrangement

[II] Principle of Permutation

If one work can be done in m ways and second work in n ways, then

both the works can be done together in m X n ways

Important Formulae: The number of arrangements of taking r things out of II different things is

Factorial n :

II! = n (II - 1) (n - 2) 3.2.1 Here n ! is read as factorial Il and this is also represented by 1lL

When aU things are not different: If out of n things, p things are of

one type, q things are of second type and r things are of r type, then

Solution The number of permutations of n different objects taking all of

them at a time, is given by,

Solution First person can sit in five ways as all the five seats are lying vacant at the time of his entrance The second person can sit on anyone of the remaining four seats (one is already filled by first person) So the second person can sit in 4 different ways Similarly, thIrd person can SIt in 3 different ways on anyone of the three vacant seats (two already filled) Hence all the three can sit in 5 X 4 x 3 = 60 different ways

Problem 29 (i) How many words can be formed with the letters of the

word "DELHI"?

(ii) How many of these will begin with D?

(iii) How many of these will end at D?

(iv) How many of these will begin with D or L?

(v) How many of these will begin with D and end at L?

(vi) How many of these will begin with D or end at L?

(vii) How mallY of the vowels "E, I" occupy the even number of places?

(viii) How many of these will end at vowel only?

(ix) III how many of these vowels come together?

(x) How many of these will begin and end at vowel?

(xi) How many of these will begin and end with D or L?

Solution (i) There are five letters "D, E, L, H, I" in the word DELHI These five letters can be rcalTanged among themselves in 5 ! ::; 120 ways, (or say 5 P s• i.e., the number of permutations of these five letters taking all the five at a time) Hence, we can form 120 different words

(ii) All the arrangements in which D is 'in the beginning can be obtained

by fixing 'D' at the first place and then rearranging the remaining four letters Remaining four letters can be arranged in 4 ! ways Hence, in 4 ! = 24 ar-rangements the words will begin with D

Alternatively First place can be filled by Din 1 way only, the second place in 4 ways as any of the letters L H, E, I can be put there; 3rd place can

be filled in 3 ways, 4th place in 2 ways and 5th place in one way only Hence, all the five can be filled in 1 x 4 x 3 x 2 x 1 = 4 ! = 24 ways

(iii) The problem is similar to problem (ii) except that now we fix the letter D on the last place Hence, the remaining four letters can be arranged

in 4 ! ways Hence, the required numb~r of words, ending at D, is 4 ! = 24 (iv) As above there will be 4 ! arrangements starting with D, and also

4 ! an'angements will begin with L Hence, the total number of arrangements beginning with D or L are 4 ! + 4 ! = 48

AlternaHvely First place can be filled in two ways as any of the letters

D or L can be put there Then second place can be filled in 4 ways as anyone

of the remaining fOur letters can be put there The third place can be filled in three ways fourth place in two ways and fifth place can be filled in one way only Hence, all the five places can be filled in 2 x 4 x 3 x 2 x 1 = 48 ways

(v) In this we fix D at the beginning and L at the end The remaining three letters E, H, I can be rearranged in 3 ! ways Hence, 3 ! = 6 arrangements will begin with D and end at L

(vi) The arrangements which begin with Dare 4 ! and which end at L

are also 4 ! Hence total number of arrangements which begin with D or end

(viii) The arrangements ending at E or I give the arrangements ending

at vowels Such arrangements are 4 ! + 4 ! = 48 Proceed as in (iv)

(ix) We consider the two vowels forming as one letter say (E I) Thus there are four letters D, L, H, (E I) These can be rearranged in 4 ! ways Further, two vowels E, I can be rearranged among themselves in 2 ! ways Hence, total number of arrangements in which vowels come together is

2 ! x4! =48

(x) We want to put E or I in the beginning and at the end For the 3 !

arrangements will begin with E and end at I Again 3 ! arrangements will begin with I and end at E Hence, the number of required ways are

Solution: We consider three vowels forming as one letter say (AUE)

So, there are six letters DGHTR(AUE) These can be arranged in 6! ways Further three vowels, A, U E can be rearanged among thermselves in 3! ways So, total number of arrangements in which vowels come together is 6!

x 31 = 4320 ways

Problem 31 How mallY words call beformed with the letters of the word

"MEERUT"? In how many of these words vowels occupy only even places?

Solution: (i) There are 6 letters in the word "MEERUT", and the letter

E is repeated 2 times in this word Therefore, the total number of words that can be formed with the letters of the word "MEERUT"

_ 6! _ 6 x 5 x 4 x 3 x (2!)

= 6 x 5 x 4 x 3 = 360 (ii) There are three even places in the word "MEERUT", namely, second fourth and sixth places, respectively At these places the vowels

E, E and U are to be arranged So, the number of ways to arrange

E, E and U at these three places

= 3! =3 2!

Now at the odd places, the letters M, R, T are to be arranged; so the number of ways to arrange M, R, T at three odd places

=3! =6 Thus, the total number of words in which the vowels occupy even places

Now in this case, the total words formed by the letters of word

"BUSINESS" which begin with B and end with N are

6'

= ~ = 6 x 5 x 4 = 120 3!

Problem 33 Prove that nCr = nCn _ r

= 6 x 23 x 11 x 7

= 10626 Problem 35 Find the value ofr if20Cr _ 1 = 20 Cr + l'

Solution Since we know that if nCx = nCy then either x = y or x + y = n

Here 2OCr_ I = 20 Cr + I and r - 1 "* r + 1

Solution Number of ways of selecting cricket eleven

= Number of ways of selecting 11 players out of 15

(ii) Again since a player is never to be included i.e • always excluded

we are selecting 11 players out of 14 only This can be done in IOClI = 364 ways

Problem 37 How many triangles can be made by joining 12 points in a plane, given that 7 are in one line?

Solution The triangles can be formed by joining any three points But

7 points are in one line Hence with three points out of these 7 points in one line we cannot form a triangle Hence the required number of triangles is

Alternative Triangles can be formed in the following ways :

(i) Three points are taken from the five non-collinear points Number

Total number of lines which can be drawn by joining any two angular points = 16C2 = 120

But out of these lines, 16 will be sides Hence the required number of diagonals = 120 - 16 = 104

Problem 39 In how many ways can 12 things be divided equally among

Problem 40 There are six points on the circumference of a circle How many straight lines can be drawn through these points?

Solution: Here no three tJoints lie on a line and through any two points, a straight line can be drawn Therefore the total number of lines that can be drawn from 6 points

=6C2

=~= 6x5 = 15 2! 4! 2 xl

Problem 41 Out of 6 teachers and 4 students, a committee of 5 is to be formed How many such committees can be formed including (i) at least one student (ii) 3 teachers and 2 students?

Solution: (i) In this case we have to select at least one student; this means that from one to all students are to be selected So there can be the following formations, of a committee of 5 persons :

(a) 1 student and 4 teachers (b) 2 students and 3 teachers

(c) 3 students and 2 teachers (d) 4 students and 1 teacher

Tota ways III case a) = I X 4= I! 3! x 4! 2!

= 4 x 15 = 60 Total ways in case (b) = 4C2 x 6C3

4! 6!

::: 2! 2! x 3! 3! = 6 x 20 = 120 Total ways in case (c) = 4C3 X 6C2

4! 6!

= 3!1! x 2! 4! = 60

Total ways in case (d) = 4C 4 x 6C1

=lx6=6

Hence, the total ways = 60 + 120 + 60 + 6 = 246

(ii) In this case the committee has 3 teachers and 2 students So the total ways to form this committee

Problem 42 : What is probability? Give its definition also

In our daily life, we generally come across with the following ments:

state-(i) Most probably Amit will stand first in his class

(ii) It is quite probable that Amit may stand first in his class (iii) It is least expected that Amit may stand first in his class

(iv) It is impossible that Amit will stand first in his class

In all these above statements we have tried to express the chances of Amit for standing first in his class qualitatively This is an event which may and may not happen But we are predicting the result of the event with some

uncertainty This uncertainty associated with the event may be lesser or

greater, i.e., it may vary In mathematics we measure this uncertainty in terms

of number quantitatively which we call probability or chance With the help

of probability we can predict the outcome of any random experiment by associating some probability to that outcome

Probability: If an event E can happen in m ways and fails (cannot happen) in n ways, all the ways are equally likely to occur, then the probability

of happening of the event E, denoted by pee), is given by

P(E)=~

Note that 0 :5 peE) :5 1

If we denote the event of "not happening of event E" by symbol E or by

E', then according to the above definition

or peE) + peE) = 1 - peE)

In equation (1) note that (m + II) are the total number of ways (outcomes)

in which a trial or an experiment may end Out of these (m + n) ways in m

ways, event E happens or say m ways are favourable to the event E Therefore, the probability of event E is also given by

P(E) = Favourable number of ways to event E

Total number of ways of the experiment Problem 43 : Find the chance of throwing more than 4 in one throw of cubic dice marked 1 to 6 its six faces,

Solution: Here are 6 equalJy likely cases of which only 2 are favourable because we want 5 or 6 on the upper face of the cubical dice Hence, the required probability of throwing more than 4 in one throw with one dice

2 1

=6=3"

Problem 44 In a single throw with two dices, what is the probability of throwing 9?

Solution The number of the first dice may appear in 6 ways Similarly,

on the second dice also the number may appear in 6 ways Hence, the two dices may appear in 6 x 6 ways namely,

(5,5), (6,5),

(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Out of these 36 ways, those which give desired sum of 9 are (3, 6), (4, 5), (5, 4) and (6, 3), i.e., only 4 favourable ways

: The probability of throwing 9 = 4/36 = 119

Problem 45 A bag contains 5 white, 8 black and 3 red balls If three balls are drawn at random from the bag, then find the probability of the event, (i) that all the balls may be white,

bag

(ii) that one ball may be black and the other two white

Solution Total number of balls = 5 + 8 + 3 = 16

Total number of ways of drawing three balls from the 16 balls in the

16

C3 = ~ = 16 15 14 = 560 3!13! 3.2.1 (i) Total number of ways of drawing three white balls out of 5

(ij) Total number of ways of drawing two white balls out of 5 and one black ball out of 8

(i) aU the three will be kings,

(ii) the cards are a king, a queen and ajack

Solution (i) Total number of ways of drawing 3 cards from a pack of

52 cards

3 3! 49! 3 x 2 x 1 Number of kings in the packet is 4, so the favourable number of ways

of drawing three kings

t e reqUIre pro a I Ity = 1326 = 663·

Problem 48 Two cards are drawn from a full pack of 52 cards What is

the chance that (i) both are aces of different colours (li) one is red and other

is black?