### Random Experiment

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

#### 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 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).

#### 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.

### Elements

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:

#### 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

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.Pictorial Representation:

### 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 is denoted by E^{1}.
Here E occurs, if and only if E^{1} doesn’t occur.

Clearly

### Random Experiment

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

#### Example

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.

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.

#### Example

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’.

#### Example

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

## Important types of Events

### Simple or elementary event

An event, consisting of a single 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} 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.

#### Example

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.

#### 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 outcome of an experiment is considered, the observation is called exhaustive events.

#### Example

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

#### Example

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).

#### Example

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.

## Categories of Problems Related to Probability

#### Category A

When n(E) and n(S) are determined by writing down the elements of ‘E’ and ‘S’.#### Category B

When n(E) and n(S) are calculated by the use of concept of permutation and combination.#### Category C

Problems based on P(E) + P(E^{1}) = 1

### Category A: When n(E) and n(S) are determined by writing down the elements of ‘E’ and ‘S’

#### Question

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

Let ‘S’ be the sample – space. Then, S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} n (S) = 8Let ‘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

#### Question

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?

#### Solution

Let ‘S’ be the sample – space. ThenS = { 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

#### Question

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.

#### Solution

S = { 1, 2, 3, —- , 2, 5} > n(S) = 25And E = { 2, 3, 5, 7, 11, 13, 17, 19, 23 } n(E) = 9

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

#### Question

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.#### Solution

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.

_{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

#### Question

What is the probability, that a leap year selected at random will contain 53 Sundays?#### Solution

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

#### Question

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.

#### Solution

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

#### Question

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?

#### Solution

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!)

= 4×3 / 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 (8×7)/2 + (4 x 3) / 2 + 8/1

= 216

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

#### Question

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.

#### Solution

We want to select ‘2’ tickets from 50 tickets.
n(S) = ^{50}C_{2} = 50? / (2?
x 48?) = (50×49) / 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?) = (15×47) / 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?) = (35×34) / 2 = 595

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

#### Question

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.

#### Solution

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 in 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)

#### Example

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

#### Solution

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

#### Question

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

#### Solution

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

#### Question

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

#### Solution

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

#### Question

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?#### Solution

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

#### Question

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.

#### Solution

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

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

#### Question

What is the probability of getting a total of less than ’12’ in the throw of two dice?#### Solution

Here n(S) = 6x6x = 36It 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

#### Question

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?

#### Solution

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

## 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]

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

P(AB) = 0

### 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(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__

### 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’.

#### Question

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 ?

#### Solution

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

#### Question

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

#### Solution

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

#### Question

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?

#### Solution

Let S = Sample spaceA = 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!) = (4×3) / 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

#### Question

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?

#### Solution

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