An experiment is said to be a random experiment, if it's out-come can't be predicted with certainty.

The set 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).

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.

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

Clearly E is a sub set of S.

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

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

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

When every possible out come of an experiment is considered.

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

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

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

therefore

Note: This definition is not true, if

(a) The events are not equally likely.

(b) The possible outcomes are infinite.

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

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)

If two or more events can't occur simultaneously, that is no two of them can occur together.

Pictorial Representation:

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.

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.

Let 'S' be the sample for random experiment, and 'E' be an event, then
complement of 'E' is is denoted by E^{1}.
Here E occurs, if and only if E^{1} doesn't occur.

Clearly

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

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

If 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.

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

Clearly

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

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

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

When every possible outcome of an experiment is considered, the observation is called exhaustive 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).

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

(iii)
A is an event, that occurs only when 'A'
doesn't occur.

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

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

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

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

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

(ii) Let E

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

n(E(iii) Let E

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

n(E(iv) Let E

(v) Let E

(vi) Let E

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

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

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

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

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?

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

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

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

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

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

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)

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

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

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

= ^{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

= 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

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

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

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

P(AB) = P()

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

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

P(AB) = 0

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)

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

= 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)

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})

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(A B) = P(A) + P(B) – P(AB)

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

P (E
E^{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__

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'.

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

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

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

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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