Axioms of probability

AXIOMS OF PROBABILITY

The probability of an event has been defined, we can collect the assumptions that we have made concerning probabilities into a set of axioms that the probabilities in any random experiment must satisfy.

The axioms do not determine probabilities; the probabilities are assigned based on our knowledge of the system under study. However, the axioms enable us to easily calculate the probabilities of some events from knowledge of the probabilities of other events.

Axioms of probability

Probability is a number that is assigned to each number of a collection of events from a random experiment that satisfies the following properties : If S is the sample space and E is any event in a random experiment,

Axiom 1 : 0 ≤  P (E) ≤ 1

Axiom 2 : P (S) = 1

Axiom 3: For any sequence of mutually exclusive events E₁, E₂, …

(i.e., events for which E_i E_j = ϕ when i ≠ j),

We refer to P (E) as the probability of the event E nolaidmo er

Note 1: Axiom 1 states that the probability that the outcome of the experiment is an outcome in E is some number between 0 and 1.

Note 2: Axiom 2 states that, with probability 1, the outcome will be a point in the sample space S.

Note 3 : Axiom 3 states that for any sequence of mutually exclusive events the probability of atleast one of these events occurring is just the sum of their respective probabilities.

Theorem 1:

The probability of an impossible event is zero (or) The null event has probability 0 (i.e.,) p (ϕ) = 0

Proof :

If we consider a sequence of events E₁, E₂, .. where E₁ = S, E_i = ϕ

for i > 1, then the events are mutually exclusive and as

Theorem 2:

If A^c is the complementary event of A, P(A) = 1 - P(A) ≤ 1.

Proof : A and A^c are mutually exclusive events, such that

A ∪ A^c = S

P(A ∪ A^c) = P(S)

= 1 by Axiom (ii)

P (A) + P(A^c) = 1 by Axiom (iii)

P(A^c) = 1 - P(A) since P (A) ≥ 0, it followes that P(A^c) ≤ 1.

Theorem 3:

If B ⊂ A, P (B) ≤ P (A)

Proof :

B and A B^c are mutually exclusive events such that

Theorem 4 :Addition law of probability

If A and B are any two events, and are not disjoint, then

P (A ∪ B) = P (A) + P(B) - P (A ∩ B).

(or) P (A ∪ B) = P(A) + P(B)

Proof :

Theorem 5

If A, B and C are any three events then

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P (A ∩ B) - P (B ∩ C) - P (C ∩ A) + P(A ∩ B ∩ C)

 (OR)

P (A + B + C) = P (A) + P (B) + P(C) - P (AB) - P (BC) - P (CA) + P (ABC)

Proof: Using the above result we have

Theorem 6

If A₁, A₂, … A_n are n mutually exclusive events then the probability of the happening of one of them is

P (A₁ ∪ A₂ ∪... ∪ A_n) = P (A₁ +...+ A_n)

= P (A₁) + P(A₂) + ... + P (A_n)

Proof: Let N be the total number of mutually exclusive exhaustive and equally likely cases of which m1 are favourable to A₁, m₂ are favourable to A₂ and so on.

Probability of occurrence of A₁ = P (A₁) = m₁ / N

Probability of occurrence of A₂ = P (A₂) = m₂ / N

… … … … … … … … … … …

… … … … … … … … … … …

Probability of occurrence of A_n = P (A_n) = m_n / N

The events being mutually exclusive and equally likely, the number cases favourable to the event A₁ or A₂ or … or A_n is m_i + m₂ + … + m_n

The probability of occurrence of one of the events A₁, A₂, ... A_n is P

Note: Multiplication theorem :

If two events A and B are independent and can happen simultaneously, the probability of their joint occurrence

P (A ∩ B) = P(A) . P (B)

The theorem can be extended to three or more events.

Theorem 7:

If the events A and B are independent then

Sample spaces having equally likely outcomes.

If we assume that all outcomes of an experiment are equally likely to occur, then the probability of any event E equals the proportion of outcomes in the sample space that are contained in E.

P (E) = number of points in E / number of points in S

♦ TYPE 1(a) P(A) = n (A) / n (S)

 

Example 1.1.1

Find the probability that exactly one head appears in a single throw of a fair coin.

Solution :

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

Here A be the event of getting exactly one head in a single throw of a fair coin.

S - → Sample space = {H, T}

n (S) = 2

n (A) = 1

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

 

Example 1.1.2

If two dice are rolled, what is the probability that the sum of the upturned faces will be equal to 7 ?

Solution : The number of total outcomes is n (S)  = 36

Let A = {Sum of the upturned faces will equal 7}

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

n (A) = 6

It is assumed that all the 36 possible outcomes are equally likely.

P (A) = n (A) / n (S) = 6 / 36 = 1 / 6

 

Example 1.1.3

From a pack of 52 cards two cards are drawn the first being replaced before the second is drawn. Find the probability that the first one is a diamond and second is a king.

Solution: Let A be an event of drawing a diamond.

P (A) = n (A) / n (S) = 13/ 52 = 1 / 4

Let B be an event of drawing a king.

P (B) = n (B) / n (S) = 4 / 52 = 1/13

A and B are independent events.

Hence P (A ∩ B) = P(A).P (B) = (1/4) (1/13) = 1/52

 

Example 1.1.4

A bag contains 5 white and 10 red balls. Three balls are taken out at random. Find the probability that all the three balls drawn red.

Solution:

Total number of balls = 15

S = {Three balls are taken out of 15}

n (S) = 15C₃ = 15.14.13 / 1.2.3 = 455

Number of red balls = 10

A = {Three balls which are red}

n (A) = 10 C₃  = 10.9.8 / 1.2.3 = 120

P(A) = n (A) / n (S) = 120 / 455 = 24 / 91

 

Example 1.1.5

If 3 balls are "randomly drawn" from a bowl containing 6 white and 5 black balls, what is the probability that one of the drawn balls is white and the other two black ?

Solution :

The total number of balls = 11

Among this 3 balls are randomly selected from the bowl.

Total number of possible outcomes = n(S) = 11C₃ = 11.10.9 / 1.2.3 = 165

The number of favourable outcomes

(i.e., 1 is white and the other two are black) = n(A) = 6C₁ × 5C₂

= 6 × 5.4/1.2 = 6 × 10 = 60

Hence the required probability P(A) = n(A) / n(S) = 60/165 = 4/11

 

Example 1.1.6

A lot of integrated circuit chips consists of 10 good, 4 with minor defects and 2 with major defects. Two chips are randomly chosen from the lot. What is the probability that atleast one chip is good? [A.U M/J 2017]

Solution :

P (atleast one is good) = n (A) / n (S) = (10C₁) (6C₁) + 10C₂ / 16C₂

= (10) (6) + 45 / 120 = 60 + 45 / 120 = 105 / 120 = 7/8

 

Example 1.1.7

A committee of 5 persons is to be selected randomly from a group of 5 men and 10 women. (a) Find the probability that the committee consists of 2 men and 3 women. (b) Find the probability that the committee consists of all women.

Solution:

(a) The number of total outcomes is given by

n(S) = 15C₅

It is assumed that "random selection" means that each of the outcomes is equally likely.

Let A = {The committee consists of 2 men and 3 women}

Then n(A) = (5C₂) (10C₃)

P(A) = n (A) / n (S) = (5C₂) (10C₃) / 15C₅ = 400 / 1001 = 0.4

(b) Let B = {the event that the committee consists of all women}

n(B) = (5C₀)  (10C₅)

P(B) = n (B) / n (S) = (5C₀) (10C₅) / 15C₅ = 36 / 429 = 0.084

 

Example 1.1.8

Four persons are chosen at random from a group containing 3 men, 2 women and 4 children. Show that the chance that exactly two of them will be children is 10 / 21 [A.U N/D 2006]

Solution :

Total number of persons = 9

4 persons can be chosen out of 9 persons = 9C₄ ways

= 9.8.7.6 / 1.2.3.4 = 126 ways.

The number of ways of choosing 2 children out of 4 children = 4C₂ ways

= 4.3/ 1.2 = 6 ways

The remaining two persons can be choosen from 5 persons (3 men + 2 women)

=  5C₂ ways

= 5.4 / 1.2 = 10 ways.

The number of favourable case = 4C₂ × 5C₂ ways

= 6 × 10 ways

= 60 ways.

Required probability = 60 / 126 = 10 / 21

 

Example 1.1.9

Four persons are chosen at random from a group containing 3 men, 2 women and 4 children. Show that the chance that exactly two of them will be children is 10 / 21'

Solution :

Total number of persons = 9

4 persons can be chosen out of 9 persons = 9C₄ ways

= 9.8.7.6 / 1.2.3.4 = 126 ways.

The number of ways of choosing 2 children out of 4 children = 4C₂ ways

= 4.3 / 1.2 = 6 ways

The remaining two persons can be choosen from 5 persons (3 men + 2 women) = 5C₂ ways

= 5.4 / 1.2 = 10 ways

The number of favourable case = 4C₂ ×  5C₂ ways

= 6 × 10 ways

= 60 ways.

Required probability = 60 / 126 = 10 / 21

 

Example 1.1.10

Two dice are thrown together. Find the probability that (a) the total of the numbers on the top face is 9 and (b) the top faces are same. [A.U. M/J 2006]

Solution :

 (a) Let A be the event which gives the sum of the top numbers as 9.

Favourable cases which gives the total as 9 are

(3, 6), (4, 5), (5, 4), (6, 3)

P(A) = 4 / 36 = 1/9

(b) Let B be the event which gives the top faces are same.

Favourable cases are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)

P(B) = 6 / 36 = 1 / 6

 

Example 1.1.11

Out of (2n + 1) tickets consecutively numbered three are drawn at random. Find the probability that the numbers on them are in arithmetic progression. [A.U. M/J 2006]

Solution: Out of (2n + 1) tickets, 3 tickets can be drawn in (2n+1) C₃ ways.

 total number of exhaustive cases

= (2n + 1) (2n) (2n - 1) / 3! = n (4n² - 1) / 3

To find the favourable number of cases, we give all possibilities in which the numbers on the drawn tickets are in A.P. with common difference d = 1, 2, 3, …, n - 1, n (say)

If d = 1, the possibilities are

If d = n, then there is only one case as (1, n + 1, 2n + 1)

the total number of favourable cases is + 5 + 3 + 1 which is an A.P. with common difference 2.

Hence favourable cases = n/2 [1 + (2n − 1)] = n²

= The no. of favourable of cases / The no. of exhaustive cases

Thus the required probability = n² /  [n (4n² - 1)/3] = 3n / 4n² - 1

 

Type 1. (b) Mutually Exclusive events (disjoint)

P(A ∪ B) = P(A) + P(B) (or) P(A + B) = P(A) + P(B)

Example 1.1.12

One card is drawn from a pack of 52 cards. What is the probability that it is either a king or a queen.

Solution:

A = {an event that the card drawn is king}

P(A) = n (A) / n (S) = 4/52 = 1/13

B = {an event that the card drawn is queen}

P(B) = n (B) / n (S) = 4/52 = 1/13

A ∪ B = {an event that the card to be either a king or a queen}

P ( A ∪ B ) = P(A) + P(B) [ A and B are mutually exclusive events]

= 1/13 + 1/13 =  2/13

 

Example 1.1.13

From a group of 5 first year, 4 second year and 4 third year students, 3 students are selected at random. Find the probability that they are first year or third year students.

Solution

Total number of students in the group = 13

Three students are selected at random n (S) = 13 C₃ = 13.12.11 / 1.2.3 = 286

A → The three students are from first year

i.e., n (A) = 5C₃ = 5.4.3 / 1.2.3 = 10

P(A) = n(A) / n(S) = 10 / 286

B → The three students are from third year

n (B) = 4 C₃ = 4

p (B) = n(B) / n(S) = 4/286

Here A and B are disjoint events since both cannot occur together.

P (A ∪ B) = P(A) + P(B)

= 10 / 286 + 4 / 286 = 14 / 286 = 7 / 143

 

Example 1.1.14

A bag contains 30 balls numbered from 1 to 30. One ball is drawn at random. Find the probability that the number of the ball drawn will be a multiple of (a) 5 or 7 and (b) 3 or 7.

Solution: Given: n (S) = 30

Let A = The probability of the number being multiple of 5.

P (A) = p(5, 10, 15, 20, 25, 30) = 6 / 30

Let B = The probability of the number being multiple of 7.

P (B) = p(7, 14, 21, 28) = 4/30

Let C = The probability of the number being multiple of 3.

P (C) = p(3, 6, 9, 12, 15, 18, 21, 24, 27, 30) = 10 / 30

(a) The events A and B are mutually exclusive the probability of the mix number being a multiple of 5 or 7 will be

= 6/30 + 4/30 = 10/30

(b) The events C and B are not mutually exclusive.

Here p (C  ∩ B) = p[21] = 1/30

p (C ∪ B) = = p (C) + P(B) -p (C ∩ B)

= 10 / 30 + 4 / 30 – 1 / 30 = 13 / 30

 

Example 1.1.15

What is the probability of picking a card that was red or black?

Solution:

Let A → The event of picking a red card.

B → The event of picking a black card.

P(A) = 26 / 52 ;

p (B) = 26 / 52

P(A ∪ B) = p (A) + p (B) = 26 / 52 +  26 / 52

Since A and B are mutually exclusive and exhaustive events.

Type (1) (c) Not mutually exclusive, independent events

(i) P (A ∪ B) = P(A) + P(B) – P (A ∩ B )

(ii) P(A ∩ B) = P(A). P (B)

 

Example 1.1.16

A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, what is the probability of getting 2 tails and 1 head. Solution: The sample space

Solution :

The sample space

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

Since a coin is biased and a head is twice as likely to occur as a tail,

P(H) = 2/3 , and P(T) = 1 / 3

Let A be the event of getting 2 tails and 1 head in the 3 tosses of the coin.

Then A = {TTH, THT, HTT}

The outcomes of the 3 tosses are independent,

 

Example 1.1.17

A can hit a target in 4 out of 5 shots and B can hit the target in 3 out of 4 shots. Find the probability that (i) the target being hit when both try. (ii) the target being hit by exactly one person.

Solution:

Let A, B the events

A hit the target P(A) = 4/5

B hit the target P(B) = 3/4

(i) The events A and B are not mutually exclusive because both of them hit the target.

 

Example 1.1.18

A is known to hit the target in 2 out of 5 shots whereas B is known to hit the target in 3 out of 4 shots. Find the probability of the target being hit when they both try?

Sol. : A → be the event that 'A' hits the target

B → be the event that 'B' hits the target

Given: P(A) = 2/5 ; P(B) = ¾

To find P (A ∪ B)

 

Example 1.1.19

One card is drawn from a deck of 52 cards. What is the probability of the card being either red or a king.

Solution:

Let A = {an event that the card drawn is red}

B = {an event that the card drawn is king}

A ∪ B = {an event that a card to be either red or a king}

P (A) = n (A) / n (S) = 26 / 52 = 1/2

P (B) = n (B) / n (S) = 4 / 52 = 1/13

There are two red coloured king cards.

So P (A ∩ B) = 2/52 = 1/26

P (A ∪ B) = P(A) + P(B) - P(A ∩ B)

 [ A and B are not mutually exclusive]

= 1/2 + 1/13 – 1/26= 7/13

 

Example 1.1.20

A total of 36 members of a club play tennis, 28 play squash, and 18 play badminton. Furthermore, 22 of the members play both tennis and squash 12 play both tennis and badminton, 9 play both squash and badminton, and 4 play all the three sports. How many members of this club play atleast one of these sports ?

Solution :

Let N → the number of members of the club

C → any subset

P(C) = number of members in C / N

Now

T → set of members that plays tennis

S → set of members that plays squash

B → set of members that plays badminton

P (T ∪ S ∪ B) = P(T) + P(S) + P(B) - P(TS) - P(TB) - P (SB) + P (TSB)

= 36 + 28 + 18 - 22 – 12 – 9 + 4 / N = 43 / N

Hence we can conclude that 43 members play atleast one of the sports.

 

Example 1.1.21

If A and B are events with P(A) = 3/8, P(B) = 1/2 and P(A ∩ B ) = 1/4, find P(A^c  ∩ B^c) [A.U N/D 2006]

Solution: We know that

P(A ∪ B) = P(A) + P(B) – P (A ∩ B)

= 3/8 + 1/2 – 1/4 = 3 + 4 – 2 / 8 = 5/8

P(A^c  ∩ B^c) = 1 – P (A ∪ B)

= 1 – 5 / 8 = 3 / 8

 

Example 1.1.22

If A and B are independent events with P(A) = 0.4 and P(B) = 0.5 find P(A ∪ B). [A.U. Dec, 96]

Solution:

Since A and B are independent

P(A ∩ B) = P(A)  P (B)

P(A ∪ B) = P(A)  + P (B) – P(A ∩ B)

= P(A) + P(B) - P(A)  P(B)

= 0.4  + 0.5 – (0.4) (0.5)

= 0.9 - 0.2 = 0.7

 

Example 1.1.23

Let events A and B be independent with P(A) = 0.5 and P(B) = 0.8. Find the probability that neither of the events A nor B occurs. [A.U. May 2000]

Solution:

Since A and B are independent

P(A ∩ B) = P(A) P(B)

 

Example 1.1.24

Event A and B are such that P (A+B) = P(AB) = 1/4 AND  find P (B).

Solution :

 

Example 1.1.25

If P(A) = 0.4, P(B) = 0.7 and P(A ∩ B) = 0.3 find the probability that neither A nor B occurs.

Solution :

 

Example 1.1.26

If you twice a flip a balanced coin, what is the probability of getting atleast one head ?

Solution:

When we flip a balanced coin, the sample space will be,

S = {HH, HT, TH, TT}

P (atleast one head) = 1-P (no head)

= 1 – 1/4 = 3/4

 

Example 1.1.27

Prove that for any event A in 

Solution :