Example; If a coin is tossed, we can't say, whether head or tail will appear. So it is a random experiment.

Sample Space: The sent of all possible out-comes of an experiment is called the sample space. It is denoted by 'S' and its number of elements are n(s).

Example; In throwing a dice, the number that appears at top is any one of 1,2,3,4,5,6. So here:

S ={1,2,3,4,5,6} and n(s) = 6

Similarly in the case of a coin, S={Head,Tail} or {H,T} and n(s)=2.

The elements of the sample space are called sample point or event point.

Event: Every subset of a sample space is an event. It is denoted by 'E'.

Example: In throwing a dice S={1,2,3,4,5,6}, the appearance of an event number will be the event E={2,4,6}.

Clearly E is a sub set of S.

Simple event; An event, consisting of a single sample point is called a simple event.

Example: In throwing a dice, S={1,2,3,4,5,6}, so each of {1},{2},{3},{4},{5} and {6} are simple events.

Compound event: A subset of the sample space, which has more than on element is called a mixed event.

Example: In throwing a dice, the event of appearing of odd numbers is a compound event, because E={1,3,5} which has '3' elements.

Equally likely events: Events are said to be equally likely, if we have no reason to believe that one is more likely to occur than the other.

Example: When a dice is thrown, all the six faces {1,2,3,4,5,6} are equally likely to come up.

Exhaustive events: When every possible out come of an experiment is considered.

Example: A dice is thrown, cases 1,2,3,4,5,6 form an exhaustive set of events.

*Classical definition of
probability:*

If 'S' be the sample space, then the probability of occurrence of an event 'E' is defined as:

P(E) = n(E)/N(S) = __number of
elements in
'E'
__
number of elements in sample space 'S'

Example: Find the probability of getting a tail in tossing of a coin.

Solution:

Sample space S = {H,T} and n(s) = 2

Event 'E' = {T} and n(E) = 1

therefore P(E) = n(E)/n(S) = ½

Note: This definition is not true,
if

(a) The events are not equally likely.

(b) The possible outcomes are infinite.

Sure event: Let 'S' be a sample space. If E is a subset of or equal to S then E is called a sure event.

Example: In a throw of a dice, S={1,2,3,4,5,6}

Let E_{1}=Event of getting
a number less than '7'.

So 'E_{1}' is a sure
event.

So, we can say, in a sure event n(E) = n(S)

Mutually exclusive or disjoint event: If two or more events can't occur simultaneously, that is no two of them can occur together.

Example: When a coin is tossed, the event of occurrence of a head and the event of occurrence of a tail are mutually exclusive events.

Independent or mutually independent events: Two or more events are said to be independent if occurrence or non-occurrence of any of them does not affect the probability of occurrence or non-occurrence of the other event.

Example: When a coin is tossed twice, the event of occurrence of head in the first throw and the event of occurrence of head in the second throw are independent events.

Difference between mutually exclusive an mutually independent events: Mutually exclusiveness id used when the events are taken from the same experiment, where as independence is used when the events are taken from different experiments.

Complement of an event: Let 'S' be the sample for random experiment, and 'E' be an event, then complement of 'E' is denoted by 'E' is denoted by E'. Here E' occurs, if and only if E' doesn't occur.

__Part (1) : Foundation Level__

Random Experiment: An experiment is said to be a random experiment, if its out-come can’t be predicted with certainly.

Ex.: If a coin is tossed, we can’t say, whether head or rail will appear. So it is a random experiment.

Sample Space : The set of all possible out-comes of an experiment is called the sample – space.

I a dice is thrown, the number, that appears at top is any one of 1, 2, 3, 4, 5, 6,

So here :

S = { 1, 2, 3, 4, 5, 6, } and n(s) = 6

Similarly in the case of a coin, s = {H, T} and n (s) = 2.

The elements of the sample of the sample-space are called sample points or event points.

Ex.: if S = {H, T}, than ‘H’ and ‘T’ are sample points.

Event: Every subset of a sample space is an event. It is denoted by ‘E’.

Ex.: In throwing a dice S = {1, 2, 3, 4, 5, 6,}, the appearance of an even number will be the event E = {2, 4, 6}.

Clearly E S.

Important types of Events:

Simple or elementary event: An event, consisting of a single point is called a simple event.

Ex.: In throwing a dice s = {1, 2, 3, 4, 5, 6} so each of {1}, {2}, {3}, {4}, {5} and {6} is a simple event.

Compound or mixed event: A subset of the sample space which has more than one element is called a mixed event.

Ex.: In throwing a dice, the event of odd numbers appearing is a mixed event, because E = {1, 3, 5}, which has ‘3’ elements.

Equally likely events: Events are said to be equally likely, if we have no reason to believe that one is more likely to occur than the other.

Ex.: When a dice is thrown, all the six-faces {1, 2, 3, 4, 5, 6,} are equally likely to come-up.

Exhaustive events: When every possible outcome of an experiment is considered, the observation is called exhaustive events.

Ex.: When a dice is thrown, cases 1, 2, 3, 4, 5, 5 form an exhaustive set of events.

Mathematical or Classical definition of Probability:

It’ ‘S’ be the sample-space, then the probability of occurrence of an event ‘E’ is defined as:

P(E) = n (E) / n(S)

= number of elements in ‘E’ / number of elements in sample space

Ex: Find the probability of getting tails in tossing of a coin.

Sol.: Sample-space S = { H, T} n(S) = 2

Event ‘E’ = {T} n (E) = 1

P(E) = n (E) / n (S) = 1 / 2

Note: This definition is not true, if

(a) The events are not equally likely.

(b) The possible out comes are infinite.

Sure-event: Let “S” be a sample – space.

If S S, ‘S’ is an event, called sure event.

Ex.: In a throw of dice, S = {1, 2, 3, 4, 5, 6}

Let E_{1} = Event of getting a number less than
‘7’

Clearly each outcome is a number less than ‘7’.

So ‘E_{1}’ is a sure – event.

We can say, in a sure event n(E) = n(S).

Mutually Exclusive or Disjoint Events: If two or more events can’t occur simultaneously, i.e. no two of them can occur together.

So the event ‘A’ and ‘B’ are mutually exclusive if

A B = , so P (AB) = 0

Ex.: When a coin is tossed, the event of occurrence of a head and the event of occurrence of a tail are mutually exclusive events.

Pictorial Representation:

A B =

Independent or Mutually independent Events: Two or more events are said to be independent, if occurrence of non-occurrence of any of them does not affect the probability of occurrence or non-occurrence of other event.

Ex.: When a coin is tossed twice, the event of occurrence of head in the first throw and the event of occurrence of head in the second throw and independent events.

Difference between mutually exclusive and mutually Independent events: Mutually exclusiveness is used, when the events are taken from the same experiment, whereas the independence is used, when the events are taken from different experiments.

Complement of an event : let ‘S’ be the sample space for
a random experiment, and ‘E’ be an event, then
complement of ‘E’ is denoted by
‘E’’ or E or E^{c}.

Here E’ occurs, if and only if ‘E’ doesn’t occur.

Clearly n(E) + n(E’) = n(S)

__Theorems related t Probability__

Theorom1 : The probability of an event lies between ‘O’ and ‘1’.

i.e. O<= P(E) <= 1.

Proof: Let ‘S’ be the sample space and ‘E’ be the event.

Then

O <= n(E) <= n (S)

O / n(S) <= n(E)/ n(S) < = n(S) / n(S)

or O < =P(E) <= 1

The number of elements in ‘E’ can’t be less than ‘O’ i.e. negative and greater than the number of elements in S.

Theorem 2 : The probability of an impossible event is ‘O’ i.e. P () = O

Proof: Since has no element, n() = O

From definition of Probability:

P() = n () / n(S) = O / n(S)

P() = O

Theorem 3 : The probability of a sure event is 1. i.e. P(S) = 1. where ‘S’ is the sure event.

Proof : In sure event n(E) = n(S)

[ Since Number of elements in Event ‘E’ will be equal to the number of element in sample-space.]

By definition of Probability :

P(S) = n (S)/ n (S) = 1

P(S) = 1

Theorem 4: If two events ‘A’ and ‘B’ are such that A B, then P(A) < =P(B).

Proof: A B

n(A) < = n(B)

or n(A)/ N(S) < = n(B) / n(S)

P(A) < =P(B)

Since ‘A’ is the sub-set of ‘B”, so from set theory number of elements in ‘A’ can’t be more than number of element in ‘B’.

Theorem 5 :
If ‘E’ is any event and E^{1} be the
complement of event ‘E’, then P(E)

+ P(E^{1}) = 1.

Proof:

Let ‘S’ be the sample – space, then

n(E) + n(E^{1}) = n(S)

or
n (E) / n (S)
+ n
(E^{1}) / n (S) = 1

or
P(E) + P(E^{1})
= 1

Algebra of Events: In a random experiment, let ‘S’ be the sample – space.

Let A S and B S, where ‘A’ and ‘B’ are events.

Thus we say that :

(i) (A B), is an event occurs only when at least of ‘A’ and ‘B’ occurs. (A B) means (A or B).

Ex.: if A = { 2,4,6,} and B = {1, 6}, than the event ‘A’ or ‘B’ occurs, if ‘A’ or ‘B’ or both occur i.e. at least one of ‘A’ and ‘B’ occurs. Clearly ‘A’ or ‘B’ occur, if the out come is any one of the outcomes 1, 2, 4, 6. That is A B. (From set – theory).

(ii) (A B) is an event, that occurs only when each one of ‘A’ and ‘B’ occur (A B) means ( A and B).

Ex.: In the above example, if the out come of an experiment is ‘6’, then events ‘A’ and ‘B’ both occur, because ‘6’ is in both sets. That is A B.

(iii) A is an event, that occurs only when ‘A’ doesn’t occur – category of problems related to probability :

(1) Category A – When n(E) and n(S) are determined by writing down the elements of ‘E’ and ‘S’.

(2) Category B – When n(E) and n(S) are calculated by the use of concept of permutation and combination.

(3)
Category C – Problems based on P(E) + P(E^{1})
= 1

Q1: A coin is tossed successively three times. Find the probability of getting exactly one head or two heads.

Sol.: Let ‘S’ be the sample – space. Then,

S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

n (S) = 8

Let ‘E’ be the event of getting exactly one head or two heads.

Then:

E = { HHT, HTH, THH, TTH, THT, HTT }

n (E) = 6

Therefore:

P(E) = n (E)/ n (S) = 6 / 8 = 3 / 4

Q2: Three coins are tossed. What is the probability of getting (i) all heads, (ii) two heads, (iii) at least one head, (iv) at least two heads?

Sol.: Let ‘S’ be the sample – space. Then

S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }

(i)
Let ‘E_{1}’ = Event of getting all
heads.

Then E_{1} = { HHH }

n (E_{1}) = 1

P(E_{1})
= n (E_{1}) / n(S)
= 1 / 8

(ii)
Let E_{2} = Event of getting
‘2’ heads.

Then:

E_{2} = { HHT, HTH, THH }

n(E_{2}) = 3

P (E_{2})
= 3 / 8

(iii)
Let E_{3} = Event of getting at least one
head.

Then:

E_{3 } = { HHH, HHT, HTH, THH, HTT,
THT, TTH }

n(E_{3}) = 7

P (E_{3}) = 7
/ 8

(iv)
Let E_{4} = Event of getting at least one
head.

Then:

E_{4 } = { HHH, HHT, HTH, THH, }

n(E_{4}) = 4

P (E_{4}) =
4/8
= 1/2

Q3: What is the probability, that a number selected from 1, 2, 3, --- 2, 5, is a prime number, when each of the numbers is equally likely to be selected.

Sol.: S = { 1, 2, 3, ---- , 2, 5} n(S) = 25

And E = { 2, 3, 5, 7, 11, 13, 17, 19, 23 } n(E) = 9

Hence P(E) = n(E) / n(S) = 9 / 25

Q4: Two dice are thrown simultaneously. Find the probability of getting :

(i) The same number on both dice,

(ii) An even number as the sum,

(iii) A prime number as the sum,

(iv) A multiple of ‘3’ as the sum,

(v) A total of at least 0,

(vi) A doublet of even numbers,

(vii) A multiple of ‘2’ on one dice and a multiple of ‘3’ on the other dice.

Sol.: Here:

S = { (1,1), (1,2) ------, (1,6), (2,1), (2,2), ---- (2,6), (3,1), (3,2), -----, (3,6), (4,1), (4,2), -------- (4,6), (5,1), (5,2), ----- (5,6), (6,1,), (6,2), --------------- (6,6) }

n(S) = 6 x 6 = 36

(i)
Let E_{1} = Event of getting same number on both
side:

E_{1}
= { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }

n(E_{1}) = 6

P(E_{1}) = n(E_{1})/n(S)
= 6/36
= 1/6

(ii)
Let E_{2} = Event of getting an even number as
the sum.

E_{2} = { (1,1), (1,3), (1,5), (2,2), (2,4),
(2,6), (3,1), (3,3), (3,5), (4,2), (4,4),

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

n(E_{2}) = 18 hence P(E_{2})
= n(E_{2})/n(S)
=
18/36
= 1/2

(iii)
Let E_{3} = Event of getting a prime number as
the sum..

E_{3} = { (1,1), (1,2), (1,4), (1,6), (2,1),
(2,3), (2,5), (3,2), (3,4), (4,1), (4,3),

(5,2), (5,6), (6,1), (6,5),}

n(E_{3}) = 15

P(E_{2})
= n(E_{3}) /
n(S) =
15/36
= 5/12

(iv)
Let E_{4} = Event of getting a multiple of
‘3’ as the sum.

E_{4} = { (1,2), (1,5), (2,1), (2,4), (3,3),
(3,6), (4,2), (4,5), (5,1), (5,4),

(6,3), (6,6),}

n(E_{4}) = 12

P(E_{4})
= n(E_{4})/n(S)
=
12/36
= 1/3

(v)
Let E_{5} = Event of getting a total of at least
10.

E_{5} = { (4,6), (5,5), (5,6), (6,4), (6,5),
(6,6), }

n(E_{5}) = 6

P(E_{5})
= n(E_{5})/n(S)
= 6/36
=
1/6

(vi)
Let E_{6} = Event of getting a doublet of even
numbers.

E_{6} = { (2,2), (4,4), (6,6), }

n(E_{6}) = 3

P(E_{6})
= n(E_{6})/
n(S) =
3/36
=
1/12

(vii) Let E_{7}
= Even of getting a multiple of ‘2” on one dice
and a multiple of ‘3’ on the other dice.

E_{7} = { (2,3), (2,6), (4,3), (4,6), (6,3),
(3,2), (3,4), (3,6), (6,2), (6,4) }

n(E_{7}) = 11

P(E_{7})
= n(E_{7}) /
n(S) = 11/36

Q5.: What is the probability, that a leap year selected at random will contain 53 Sundays?

Sol.: A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days.

The remaining 2 days may be any of the following :

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

For having 53 Sundays in a year, one of the remaining 2 days must be a Sunday.

n(S) = 7

n(E) = 2

P(E) = n(E) / n(S) = 2 / 7

__ __

__CATEGORY
– B__

__ __

Problems based on fundamental principal of counting and permutations and combinations :

Q1. A bag contains ‘6’ red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability, that

(i) ‘1’ is red and ‘2’ are white, (ii) ‘2’ are blue and 1 is red, (iii) none is red.

Sol.: We have to select ‘3’ balls, from 18 balls (6+4+8)

n(S) = ^{18}C_{3} = 18! / (3!
x 15!) = (18x17x16) / (3x2x1) = 816

(i)
Let E_{1} = Event of getting ‘1’ ball
is red and ‘2’ are white

Total number of ways = n(E_{1})
= ^{6}C_{1} x ^{4}C_{2}

= 6! / (1! x 5!) x 4! / (2! x 2!)

= 6 x 4 / 2

= 36

P (E_{1}) = n (E_{1}) /
n(S) = 36/816 = 3/68

(ii)
Let E_{2} = Event of getting ‘2’ balls
are blue and ‘1’ is red.

= Total no. of ways n(E_{2}) = ^{8}C_{2}
x ^{6}C_{1}

= 8! / (2! x 6!) x 6! / (1! x 5!)

= (8 x 7) / 2 x 6 / 1 = 168

P(E_{2}) = 168 /
816 = 7/34

(iii)
Let E_{3 }= Event of getting ‘3’
non – red balls. So now we have to choose all the
three balls from 4 white and 8 blue balls.

Total number of ways :

n(E_{3}) = ^{12}C_{3}
= 12! / (3! x 9!) = (12x11x10) /
(3x2x1) = 220

P(E_{3}) = n(E_{3}) /
n(S) = 220 / 816 = 55/204

Q: A box contains 12 bulbs of which ‘4’ are defective. All bulbs took alike. Three bulbs are drawn randomly.

What is the probability that :

(i) all the ‘3’ bulbs are defective?

(ii) At least ‘2’ of the bulbs chosen are defective?

(iii) At most ‘2’ of the bulbs chosen are defective?

Sol.: We have to select ‘3’ bulbs from 12 bulbs.

n(S) = ^{12}C_{3}
= 12! / (3! x 9!) = (12x11x10) /
(3x2x1) = 220

(i)
Let E_{1} = All the ‘3’
bulbs are defective.

All bulbs have been chosen, from ‘4’ defective bulbs.

n(E_{1}) = ^{4}C_{3}
= 4! / (3! x 1!) = 4

P(E_{1}) = n(E_{1}) /
n(S) = 4 /220 = 1/55

(ii)
Let E_{2} = Event drawing at least 2
defective bulbs. So here, we can get ‘2’
defective and 1 non-defective bulbs or 3 defective bulbs.

n(E_{2}) = ^{4}C_{2}
x ^{8}C_{1} + ^{4}C_{3}
[Non-defective bulbs = 8]

= 4! / (2! x 2!) x 8! / (1! x 7!) + 4! / (3! x 1!)

= 4x3 / 2 x 8/1 + 4/1 = 48+4

n(E_{2}) = 52

P(E_{2}) = n(E_{2}) / n(S) =
52/220 = 13/55

(iii)
Let E_{3} = Event of drawing at most
‘2’ defective bulbs. So here, we can get no
defective bulbs or 1 is defective and ‘2’ is
non-defective or ‘2’ defective bulbs.

n(E_{3}) = ^{8}C_{3}
+ ^{4}C_{1} x ^{8}C_{2} +
^{4}C_{2} x ^{8}C_{1}

= 8? / (3? x 5?) + 4? / (1? x 3?) x 8? / (2? x 6?) + 4? / (2? x 2?) x 8? / (1? x 7?)

= (8x7x6) / (3x2x1) + 4 x (8x7)/2 + (4 x 3) / 2 + 8/1

= 216

P(E_{3}) = n(E_{3}) /
n(S) = 216 / 220 = 54 / 55

Q: In a lottery of 50 tickets numbered from ‘1’ to ‘50’ two tickets are drawn simultaneously. Find the probability that:

(i) Both the tickets drawn have prime number on them,

(ii) None of the tickets drawn have a prime number on it.

Sol.: We want to select ‘2’ tickets from 50 tickets.

n(S) = ^{50}C_{2} = 50? / (2?
x 48?) = (50x49) / 2 = 1225

(i)
Let E_{1} = Event that both the tickets have
prime numbers Prime numbers between ‘1’ to
‘50’ are :

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47.

Total Numbers = 15.

We have to select ‘2’ numbers from these 15 numbers.

n(E_{1}) = ^{15}C_{2}
= 15? / (2? x 13?) = (15x47) / 2 = 105

P(E_{1}) = n(E_{1}) / n(S) =
105/1225 = 21/245

(ii) Non prime numbers between ‘1’ to ‘50’ = 50-15 = 35

Let E_{2} = Event that both the
tickets have non-prime numbers.

Now we have to select ‘2’ numbers, from ‘35’ numbers.

n(E_{2}) = ^{35}C_{2}
= 35? / (2? x 33?) = (35x34) / 2 = 595

P(E_{2}) = n(E_{2}) / n(S) =
595 / 1225 = 17/35

Q.: A bag contains 30 tickets, numbered from ‘1’ to ‘30’. Five tickets are drawn at random and arranged in ascending order. Find the probability that the third number is 20.

Sol.: Total number of
ways of selecting ‘5’ tickets from 30 tickets =
^{30}C_{5}

n(S) = ^{30}C_{5} = 30? /
(5? x 25?) = (30 x 29 x 28 x 27 x 26) / (5 x 4 x 3 x 2 x
1)

n(S) = 29 x 27 x 26 x 7

Suppose the ‘5’ tickets are a1, a2,20, a4, a5

They are arranged in ascending order.

a1, a2 {1, 2, 3, ------- , 19} and a4, a5 { 21, 22, 23, -----, 30}

We have to select ‘2’ tickets from first ‘19’ tickets and ‘2’ tickets from last 10 tickets.

n(E) = ^{19}C_{2}
x ^{10}C_{2}

= 19? / (2? x 17?) = 10? / (2? x 8?) = (19 x 18) / 2 = (10 x 9) / 2

= 19 x 9 x 5 x 9

P(E) = n(E) / n(S) = (19x9x5x9) / (29x27x26x7) = 285 / 5278

Odds is Favour and Odds against an Event:

Let ‘S” be the sample space and ‘E’ be an event. Let ‘E’ devotes the complement of event ‘E’, then.

(i)
Odds in favour of event ‘E’ = n(E) / n(E^{1})

(ii)
Odds in against of an event ‘E’ = n(E^{1})
/ n(E)

Note : Odds in favour of ‘E’ =
n(E) / n(E^{1})

= [n(E) / n(S)] / [n(E^{1}) /
n(S)] = P(E) / P(E^{1})

Similarly odds in against of
‘E’ = P(E^{1}) / P(E)

Ex.: The odds in favours of an event are 3:5 find the probability of the occurrence of this event.

Sol.: Let ‘E’ be an event.

Then odds in favour of E = n(E) / n(E^{1})
= 3 / 5

n(E) = 3 and n(E^{1}) = 5

Total number of out-comes n(S) = n(E) +
n(E^{1}) = 3+5 = 8

P(E) = n(E) / n(S) = 3 / 8

Q.: If ‘12’ persons are seated at a round table, what is the probability that two particulars persons sit together?

Sol.: We have to arrange 12 persons along a round table.

So if ‘S” be the sample – space, then n(S) = (12-1)? = 11?

n(S) = 11?

Now we have to arrange the persons in away, such that ‘2’ particulars person sit together.

Regarding that 2 persons as one person, we have to arrange 11 persons.

Total no. of ways = (11-1)? = 10? ways.

That ‘2’ persons can be arranged among themselves in 2? ways.

So, total no. of ways, of arranging 12 persons, along a round table, so that two particular person sit together : = 10? x 2?

n(E) = 10? x 2?

P(E) = n(E) / n(S) = (10? x 2?) / 11? = 2 / 11

Q.: 6 boys and 6 girls sit in a row randomly, find the probability that all the ‘6’ girls sit together.

Sol.: We have to arrange ‘6’ boys and ‘6’ girls in a row.

n(S) = 12?

Now, we have to arrange ‘6’ girls in a way, such that all of them should sit together.

Regarding all the 6 girls as one person, we have to arrange 7 person in a row.

Total no. of ways = 7?

But 6 girls can be arranged among themselves in 6? ways.

n(E) = 7? x 6?

P(E) = n(E) / n(S) = (7? x 6?) / 12? = (6x5x4x3x2x1) / (12x11x10x9x8)

P(E) = 1 / 132

Q: If from a pack of ‘52’ playing cards one card is drawn at random, what is the probability that it is either a kind or a queen?

Sol.: n(S) = Total number of ways of selecting 1 card out of 52 cards.

= ^{52}C_{1} = 52

n(E) = Total number of selections of a card, which is either a kind or a queen.

= ^{4}C_{1} + ^{4}C_{1}
= 4 + 4 = 8

P(E) = n(E) / n(S) = 8 / 52 = 2 / 3

Q.: From a pack of 52 playing cards, three cards are drawn at random. Find the probability of drawing a king, a queen and a jack.

Sol.: Here n(S) = ^{52}C_{3}
= 52? / (3? x 49?) = (52x51x50) / (3x2x1)

= 52x17x25

n(E) = ^{4}C_{1}. ^{4}C_{1}.
^{4}C_{1}

= 4? / (1? x 3?) x 4? / (1? x 3?) = 4? / (1? x 3?)

n(E) = 4 x 4 x 4

P(E) = n(E) / n(S) = (4x4x4) / (52x17x25) = 16 / 5525

__CATEGORY
– C__

Problems based on finding P(E^{1}),
by the use of P(E^{1}) – 1 – P(E) :

Note : When an event has a lot of out comes, then we use this concept.

Ex.: What is the probability of getting a total of less than ‘12’ in the throw of two dice?

Sol.: Here n(S) = 6x6x = 36

It is very difficult to find out all the cares, in which we can find the total less then ‘12’.

So let E = The event, that the sum of numbers is ‘12’.

Then E = { 6, 6}

n(E) = 1

P(E) = n(E) / n(S) = 1/36

Required probability, P(E^{1}) =
1-P(E)

= 1 – 1/36

P(E1) = 35 /36

Ex.: There are ‘4’ envelopes corresponding to ‘4’ letters. If the letters are placed in the envelopes at random, what is the probability that all the letters are not placed in the right envelopes?

Sol.: We have to place ‘4’ letters in 4 envelopes.

n(S) = 4!

Now:

Let E = The event, that all the 4 letters are placed in the corresponding envelopes.

So E^{1} = The event that all the
‘4’ letters are not placed in the right
envelope.

Here n(E) = 1

P(E) = n(E) / n(S) = 1 / 4! = 1 / 24

Required probability, P(E^{1})
= 1- P(E)

= 1 – (1/24)

P(E^{1}) = 23 / 24

Some information’s about playing cards:

(1) A pack of 52 playing cards has 4 suits :

(a) Spades, (b) Hearts, (c) Diamonds, (d) Clubs.

(2) Spades and clubs are black and Hearts and Diamonds are red faced cards.

(3) The aces, kings, queens, and jacks are called face cards or honours – cards.

__Part
– 2 : (Total Probability)__

Theorem – 1 : If ‘A’ and ‘B’ are mutually exclusive events then P(AB) = 0 or P ( A and B) = 0

Proof : If ‘A’ and ‘B’ are mutually exclusive events then A B =

P(AB) = P()

= n() / n(S) [ By definition of probability]

= o / n(S) [Since the number of elements in a null – set is ‘0’]

P(AB) = 0

(2) Addition Theorem of Probability : If ‘A’ and ‘B’ by any two events, then the probability of occurrence of at least one of the events ‘A’ and ‘B’ is given by:

P(A or B) = P(A) + P(B) – P (A and B)

P(AB) = P(A) + P(B) – P (AB)

From set theory, we have :

n(AB) = n(A) + n(B) – n(AB)

Dividing both sides by n(S) :

n(AB) / n(S) = n(A) / n(S) + n(B) / n(S) - n(AB) / n(S)

or P(AB) = p(A) + P(B) – P(AB)

Corollary : If ‘A’ and ‘B’ are mutually exclusive events,

Then P(AB) = 0. [ As we have proved]

In this case :

P(AB) = p(A) + P(B)

Addition theorem for ‘3’ events ‘A’, ‘B’ and ‘C’ :

P(ABC) = P(A) + P(B) - P(AB) – P(BC) – P(AC) + P(ABC)

Proof : P(ABC) = P[(AB) C]

= P(Ab) + P(C) – P[(AB) C] [ By addition theorem for two events]

= P(AB) - P(C) – [P(AC) + P(BC) – P(ACBC)]

= P(A) + P(B) – P(AB) + P(C) – P(AC) – P(BC) + P(ABC)

P(ABC) = P(A) + P(B) + P(C) – P(AB) – P(BC) – P(AC) + P(ABC)

Corollary : If ‘A’, ‘B’ and ‘C’ are mutually exclusive events, then P(AB) = 0, P(BC) = 0, P(AC) = 0 and P(ABC) = 0.

In this case :

P(ABC) = P(A) + P(B) + P(C)

General Form of Addition Theorem of Probability:

n

P(A_{1} A_{2} ----- A_{n})
= P(A_{i}) - P(A_{i}
A_{j})

i=1 i<j

+ P(A_{i} A_{j}
A_{k})
----- (-1)^{n-1} P(A_{1} A_{2}
---- A_{n})

i<j<k

Corollary : For any number of mutually
exclusive events, A_{1}, A_{2}, ------ ,
A_{n} :

P(A_{1} A_{2} ---- A_{n})
= P(A_{1}) + P(A_{2}) + ------ +
P(A_{n})

Theorem – 3 : For any two events ‘A’ and ‘B’

P(A-B) = P(A) – P(AB) =
P(AB^{1})

From the figure:

(A-B) (AB) = ---------------> (i)

and

(A-B) (AB) = A

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

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

[From (i) (A-B) (AB) = i.e. These events are mutually exclusive]

P(A-B) = P(A) – P(AB)

or P(AB) = P(A) – P(AB)

Similarly P(AB) = P(B) – P(AB)

Proof of P(E) + P(E^{1}) = 1, by
the addition theorem of probability:

We know that :

P(AB) = P(A) + P(B) – P(AB)

Putting A = E and B = E^{1}

P (EE^{1}) = P(E) + P(E^{1})
– P (EE^{1}) ----------------> (1)

From set theory : E E^{1}
= S

And E E^{1} =

From:

P(S) = P(E) + P(E^{1}) – P()

1 = P(E) + P(E^{1})
– 0

or P(E) + P(E^{1}) = 1

[P(S) = 1, P() = 0]

__EXAMPLES__

Problems based on addition theorem of probability:

Working rule :

(i) A B denotes the event of occurrence of at least one of the event ‘A’ or ‘B’

(ii) A B denotes the event of occurrence of both the events ‘A’ and ‘B’.

(iii) P(AB) or P(A+B) denotes the probability of occurrence of at least one of the event ‘A’ or ‘B’.

(iv) P(B) or P(AB) denotes the probability of occurrence of both the event ‘A’ and ‘B’.

--------------------x-------------------x---------------------x--------------------x---------------x-----

Ex.: The probability that a contractor will get a contract is ‘2/3’ and the probability that he will get on other contract is 5/9 . If the probability of getting at least one contract is 4/5, what is the probability that he will get both the contracts ?

Sol.: Here P(A) = 2/3, P(B) = 5/9

P(Ab) = 4/5, (P(AB) = ?

By addition theorem of Probability:

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

= 4/5 = 2/3 + 5/9 - P(AB)

or 4/5 = 11/9 – P(AB)

or P(AB) = 11/9 – 4/5 = (55-36) / 45

P(AB) = 19/45

Ex2.: Two cards are drawn at random. Find the probability that both the cards are of red colour or they are queen.

Sol.: Let S = Sample – space.

A = The event that the two cards drawn are red.

B = The event that the two cards drawn are queen.

AB = The event that the two cards drawn are queen of red colour.

n(S) = ^{52}C_{2},
n(A) = ^{26}C_{2}, n(B) = ^{4}C_{2}

n(AB) = ^{2}C_{2}

P(A) = n(A) / n(S) = ^{26}C_{2}
/ ^{52}C_{2} , P(B) = n(B) /
n(S) = ^{4}C_{2} / ^{52}C_{2}

P(AB) = n(AB) /
n(S) = ^{2}C_{2} / ^{52}C_{2}

P(AB) = ?

We have P(AB) = P(A) + P(B) – P(AB)

= ^{26}C_{2} / ^{52}C_{2}
+ ^{4}C_{2} / ^{52}C_{2}
– ^{2}C_{2} / ^{52}C_{2}

= (^{26}C_{2} + ^{4}C_{2}
– ^{2}C_{2}) / ^{52}C_{2}

= (13X25+2X3-1) / (26X51)

P(AB) = 55/221

Ex.3: A bag contains ‘6’ white and ‘4’ red balls. Two balls are drawn at random. What is the chance, they will be of the same colour?

So.: Let S = Sample space

A = the event of drawing ‘2’ white balls.

B = the event of drawing ‘2’ red balls.

AB = The event of drawing 2 white balls or 2 red balls.

i.e. the event of drawing ‘2’ balls of same colour.

n(S) = ^{10}C_{2} = 10!
/ (2! x 8!) = 45

n(A) = ^{6}C_{2} =
6! / ((2! x 4!) = (6 x 5 ) / 2 = 15

n(B) = ^{4}C_{2} =
4! / (2! x 2!) = (4x3) / 2 = 6

P(A) = n(A) / n(S) = 15/45 = 1/3

P(B) = n(B) / n(S) = 6/45 = 2/15

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

= 1/3 + 2/15 = (5+2) / 15

P(AB) = 7/15

Ex.: For a post three persons ‘A’, ‘B’ and ‘C’ appear in the interview. The probability of ‘A’ being selected is twice that of ‘B’ and the probability of ‘B’ being selected is thrice that of ‘C’, what are the individual probability of A, B, C being selected?

Sol.: Let ‘E_{1}’,
‘E_{2}’, ‘E_{3}’ be
the events of selections of A, B, and C respectively.

Let P(E_{3}) = x

P(E_{2}) = 3. P(E_{3})
= 3x

and P(E_{1}) = 2P(E_{2})
= 2 x 3x = 6x

As there are only ‘3’ candidates ‘A’, ‘B’ and ‘C’ we have to select at least one of the candidates A or B or C, surely.

P( E_{1} E_{2} E_{3})
= 1

and E_{1}, E_{2}, E_{3}
are mutually exclusive.

P(E_{1} E_{2} E_{3})
= P(E_{1}) + P(E_{2}) + P(E_{3})

1 = 6x + 3x + x

10x – 1 or x = 1/10

P(E_{3}) = 1/10, P(E_{2})
= 3/10 and P(E_{1}) = 6/10
= 3/5

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