# Concept of Probability

#### Random Experiment:

• If an experiment has more than one possible results which are known in advance and it is not possible to predict which one is going to occur, then the experiment is called a random experiment.
• Examples: Tossing a fair coin, drawing a card from a well-shuffled pack of cards.

#### Outcome:

• The result of a random experiment is called an outcome.

#### Sample Space:

• A sample space of an experiment is the set of all possible distinct outcomes of the experiment and it is denoted by ‘S’.
• Example: A sample space for different events are as follows

A fair coin is tossed:

S  = { H, T}

∴ Total number of outcomes = 2   ∴ n (S) =  2

Two fair coins are tossed or a Fair coin is tossed twice:

S =  { H, T} × { H, T}  =  {HH, HT, TH, TT}

∴ Total number of outcomes =  4  = 2²   ∴n (S) =  4

Three fair coins are tossed or a Fair coin is tossed thrice:

S =  { H, T} × { H, T} × { H, T}  =  {HH, HT, TH, TT} × { H, T} =  {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

∴ Total number of outcomes =  8  = 2³  ∴ n (S) =  8

Notes:

• An element of a sample space is called a sample point.
• If the number of elements in a sample space is finite then the sample space is called a finite sample space.
• If n coins are tossed then the number of outcomes is 2n.
• A fair coin is tossed twice is equivalent to two fair coins are tossed.
• A fair coin is tossed three times is equivalent to three fair coins are tossed.
• A fair coin is tossed ‘n’ times is equivalent to ‘n’ fair coins are
• A word fair is equivalent to unbiased.

A fair dice is tossed:

• S  = {1,2,3,4,5,6}    ∴ n (S) =  6.

Two fair dice are tossed

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

∴ n (S) =  36

Notes:

• The sum of the two numbers on two dice is called the score on two dice.
• The minimum score on two dice is 2 and the maximum score on two dice is 12.
• The cases favourable to a particular score can be read along the diagonal of that score.

#### Event:

• Any subset of a sample space is called an event. An event is denoted by a capital letter.
• Example – 1: A = an event, that a card selected from a pack of cards is a Diamond.

A Diamond card can be obtained in   13C= 3 ways

Number of cases favourable to the event A = 13 ∴ n(A) = 13

• Example – 2: B = an event, that 2 cards selected are consisting of one King and other a Queen.

A King can be obtained in 4C1 = 4 ways.

After the selection of a King in any ‘one of four ways,

the selection of a queen can be made in 4C1= 4 ways.

∴ n(B) = 4 x 4   =  16

• Example – 3: C = an event, that a die shows a number greater than 3

∴ C = { 4, 5, 6}    ∴n(C) = 3

• Example – 4: D = an event that the score on two dice is 4

∴ D = {(1, 3), (2, 2), (3, 1)}    ∴ n(D) = 3

• Example – 5: E = an event that the score on two dice is a prime number  i.e. 2,3,5,7,11

∴ E = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2). (4, 1), (1, 6), (2, 5), (3, 4),

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

∴ n(E) = 15

#### Impossible Event:

• If an event is an empty set, then the event is called an impossible event and is denoted by Φ.
• Example – 1:  A = an event that the score on two dice is 15. ∴ A = Impossible event.
• Example – 2: B = an event of having a quadratic equation of three roots. ∴ B = Impossible event.

#### Certain Event:

• If an event is the same as the sample space of the experiment then the event is called a certain event or sure event.
• Example – 1: A = an event that a card selected from a pack of 52 cards is either a red card or a black card.

∴ n(A) = 52 = n(S). and A = S

• Example – 2: B = an event that a die shows a number which is odd or even.

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

#### Elementary event:

• If an event contains only one sample point, then the event is called an elementary event or a simple event.
• Example – 1:  A = an event, card selected is a queen of hearts.

A queen of hearts can be obtained in one way.

∴ A is an elementary event. ∴ n(A) = 1.

• Example – 2: B = an event that the score on two dice is 12.

B  =  {(6,6)} \ n(B) = 1

∴ B is an elementary event.

#### The complement of an event A:

• Let A be an event of a sample space S then the event consisting of all the cases of the samples space which are not favourable to the event is called the complement of the event A and is denoted by A’ or ∴ A’  = {x | x  ∈ S, but x  ∉ A}
• Examples:

A = an event that card selected is a spade.

∴  A’ = an event that card selected is not spade

B = an event that the score on the two dice is greater than 4.

∴  B’ = an event that the score on the two dice is less than or equal to 4.

C = an event that, the room is lit.

∴  C’ = an event that the room is not lit.

D = an event that, India wins at least one game.

∴  D’  = an event that, India does not win any game.

E = an event, that the room has at least one fan.

∴  E’ = an event, that the room has no fan.

Notes :

1. If A = an event consisting of at least one then A’ = an event consisting of none.
2. If S contains n sample points and A contains m sample points. Then A’ will contain (n – m) sample points.

Important Results:

 (a)  Φ’ = S (b)   S’ = Φ (c)   A U A’ = S (d)   A  ∩ A’ = Φ (e)   n(A) + n(A’) = n(S)

#### Probability:

• If A is an event of sample space S’ then the probability of event A denoted by P(A) is defined as

P(A) = n(A)/n(S)

#### Important Relations and Their Proofs:

• Prove that P(Φ) = 0
• Proof :

Since Φ is an impossible event. Hence Φ is an empty set.

∴  n(Φ) = 0

∴ P(Φ) = n(Φ)/n(S) = 0/n(S) = 0

∴ P(Φ) = 0  is proved

• Prove that P(S) = 1
• Proof :

Since S is a certain event.

∴  n(Φ) = n(S)

∴ P(S) = n(S)/n(S) = 1

∴ P(S) = 1  is proved

• Prove that 0 < P(A)  < 1
• Proof :

If A is an event of sample space S. Then we have

0 <  n(A) < n (S)

Dividing by n(S)

0/n(S) <  n(A)/n(S) < n (S)/n(S)

∴ 0   <   P(A)  < 1       (proved)

• Prove that P(A) = 1 – P(A’)
• Proof :

If S contains n sample points and A contains m sample points. Then A’ will contain (n – m) sample points.

Where A’ is a complement of the set A.

∴ A U A’ = S  and A ∩ A’ = Φ

∴  n(A) + n(A’) = n(S)

Dividing by n(S)

∴  n(A)/n(S) + n(A’)/n(S) = n(S)/n(S)

∴  P(A)  +   P(A’)  =  1

∴  P(A) = 1 – P(A’)  is proved

#### Explanation of  the Phrases :

i) odds in favour of an event A  and (ii) odds against an event A

• If x cases are favourable to an event A and y cases are not favourable to the event A then we say odds in favour of A are x : y  OR odds against A are y : x,