Probability: Formula, Types, Properties & Applications

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Probability, in simple terms, describes the likeness or occurrence of an event in numeric values. It is indicated by the number 0 or 1, where 0 represents the impossibility of the event and 1 indicates the likelihood of an event.

Probability is obtained by dividing the favourable outcomes by the total number of possible outcomes.
Accounting has become an effective tool in science, engineering, statistics, financing, and other mathematical branches.
Probabilities of events are less than equal to one or greater than equal to zero.
It is based on the concept of measure theory, which was invented by Andrey Kolmogorov.
The sum of all the events in the sample space is equal to one.
Tossing a coin is the most common example of the probability of an event.

Table of Content

What is Probability?
Probability Terms and Definitions
Formula for Probability
Types of Probability
Properties of Probability
Probability of an Event
Types of Event
Applications of Probability
Important Topics for JEE Main
Things to Remember
Sample Questions

Key Terms: Probability, Events, Die, Heads, Tail, Coin, Cards, Outcome, Sample Space, Experiment, Mutually Exclusive Events, Complimentary Events

What is Probability?

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Probability is defined as the extent to which an event is likely to occur. It is measured by the ratio of the favourable outcome to the whole number of possible outcomes.

Probability is used to define the regularities of complex systems.
It is derived from a Latin word that means probity.
To calculate the probability, first determine the total number of outcomes.
Chevalier de Mere introduced the concept of rolling a dice in the 16^th century.
The value of favourable events cannot be negative.

Solved Example of Probability

Example: There are 8 pillows in a bed, 4 are red, 2 are yellow and 2 is blue. What is the probability of picking a yellow pillow?

Ans: The probability is equal to the number of yellow pillows in the bed divided by the total number of pillows, i.e. 2/8 = 1/4.

Read More:

Chapter Related Concepts
Double Integral	Empirical Probability	Conditional probability
Geometric Mean	Integration	Rolle’s Theorem

Probability Terms and Definitions

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Here are the definitions of some important terms related to probability.

Sample space

The set of possible outcomes in a trial is referred to as the sample space. For example, when flipping a coin, the possible outcomes are heads or tails. Alternatively, when rolling a single die, the possible outcomes are 1, 2, 3, 4, 5, 6.

Sample Point

In a sample space, a sample point is one of the possible outcomes. When using a deck of cards, for example, a sample point would be the ace of spades or the queen of hearts.

Experiment

When the outcomes of a series of actions are always uncertain, this is referred to as an experiment or trial. The outcomes are uncertain when choosing a card from a deck, tossing a coin, or rolling a die.

Event

An event is a single outcome that occurs as a result of a trial or experiment. For example, getting a three on a die or an eight of clubs when selecting a card from a deck are occurrences of certain events. Read on Types of Events in Probability

Outcome

A possible outcome of a trial or experiment is referred to as an outcome. Tossing a coin, for example, could result in heads or tails. Here, the outcomes are heads or tails. Meanwhile, the outcomes of dice thrown are either 1, 2, 3, 4, 5, or 6.

Read More: Sure Event

Formula for Probability

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Mathematically, the probability of an event is given by -

P (E) = Number of favorable outcomes of E / total number of possible outcomes of E .

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

Where,
P(E) = Probability of an event E
n(A) = No. of favorable outcomes of E
n(S) = Total no. of events in Sample Space

Solved Example of Formula for Probability

Example: There are 11 pillows in a bed, 4 are red, 2 are yellow and 5 is blue. What is the probability of picking a blue pillow?

Ans: The probability is equal to the number of blue pillows in the bed divided by the total number of pillows, i.e. 5/11

Probability

Read More: Probability distribution

Types of Probability

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The different types of probability are as follows:

Classical or Theoretical Probability

According to the classical or theoretical probability, in an experiment with X equally likely outcomes and event Y having exactly Z of these outcomes, the probability of Y is Z/X or P (Y) = Z/X.

This is the first point of view that students encounter in formal education.
If you roll a fair die and there are six equally likely outcomes, each number has a 1/6 chance of being rolled.
The advantage is that it is conceptually simple for many situations.
It has limitations because many situations do not have a finite number of equally likely outcomes.

Experimental Probability

Empirical or experimental probability defines probability through experiments or statistics. The formal definition of this viewpoint is P(A) = the limit B / C, as C approaches infinity. Where A is the probability of the event, B is the number of times the event A occurs, and C is the number of times the process, such as rolling a die or tossing a coin, is repeated.

Solved Example of Experimental Probability

Example: When you throw a die, you can estimate the probability of each outcome by rolling the die an enormous number of times and calculating the proportion of times the die gives the desired outcome - statistical method.

Read More: Types of Events

Axiomatic Probability

The axiomatic probability perspective is a unifying perspective in which the coherent conditions used in theoretical and experimental probability demonstrate subjective probability.

Kolmogorov's set of rules or axioms is applied to all types of probability.
They are known as Kolmogorov's three axioms by mathematicians.
You can use axiomatic probability to calculate the likelihood of an event occurring or not occurring.
The three axioms are applicable to all other probability perspectives.

The viewpoint is defined as the probability of any function from numbers to events that are satisfied by the three axioms listed below:

The least possible probability is zero, and the greatest possible probability is one.
A certain event has a probability of one.
Two mutually exclusive events cannot occur at the same time, but the union of events states that only one of them can.

Probability

Read More: Probability density function

Properties of Probability

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The various properties of probabilitiy are as follows:

Probability of a sure event or certain event is 1.
It is a number P(E) such that 0 ≤ P (E) ≤ 1.
The probability of an impossible event is 0.
It is always a positive number.
If A and B are two events that are mutually exclusive, then P (A â< ƒ B) = P (A) + P (B).
An elementary event is an event having only one outcome.
The sum of probabilities of an event and its complementary event is 1: P(A) + P(A’) = 1.
P(A â< ƒ B) = P(A) + P(B) – P(A â< ƒ B).
If A₁, A₂, A₃ ,………, A_n are mutually exclusive events, then P (A1 â< ƒ A2 â< ƒ A3… â< ƒ A_n) = P(A₁) + P(A₂ ) + ……… + P(A_n).

Read More:

Class 12 Maths Concepts
Mean of Grouped Data	Types of Sets	Geometric distribution
Uniform distribution	T distribution formula	Probability and statistics symbols

Probability of an Event

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Consider an event E, which occurs in r number of ways. The total number of events is equal to m. Then the probability of occurring an event is given by:

P(E) = r / m

The probability that an event will take place is given by
P(E’) = (m-r) / m
P(E’) = 1 – r/m
P(E’) = 1 – P(E)
P(E’) + P(E) = 1

Probability

Read More: Continuity and Differentiability

Types of Event

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The various types of events are as follows:

Complementary Events

The two events are said to be complimentary if one of the two events in the total set of outcomes takes place.

Solved Example of Complementary Events

Example: The student will pass in an examination or not.

Example: A person will win the lottery or not.

Independent Events

Two events, A and B, are said to be independent when the probability of occurrence of an event A is not dependent on the probability of occurrence of an event B.

Mutually Exclusive Events

Two events are said to be mutually exclusive if they do not occur at the same time. There is no common point between mutually exclusive events.

Solved Example of Mutually Exclusive Events

Example: Weather cannot be hot and chilly at the same time.

Read More: Sum of Probabilities

Applications of Probability

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The various applications of probability of an event are as follows:

Probability is used to forecast the weather conditions of a particular area.
Politicians can use it to predict the outcome of an election.
The insurance company uses probability to determine a person’s chance of living or death.
Share companies use the concept to determine the share prices.

Read More: Calculus Formula

Important Topics for JEE Main

As per JEE Main 2024 Session 1, important topics included in the chapter probability are as follows:

Formula for Probability
Types of Probability
Properties of Probability
Probability of an Event

Some memory based important questions asked in JEE Main 2024 Session 1 include:

Bag A contains 7 white balls & 3 red balls. Bag B contains 3 white balls & 2 red balis. A ball is chosen randomly & found to be red then find the probability that it is taken from bag A.
In a paper there are 3 sections A, B and C which has 8, 6 and 6 questions each. A student have to attempt 15 questions such that they have to attempt atleast 4 questions out of each sections, then number of ways of attempting these questions are
An urn contains 15 red. 10 white, 60 orange, and 15 green balls. If 2 balls are taken with replacement, find the probability 1 ball is red and the other ball is white.
If the mean of 15 observations is 12 and the standard deviation is 3. If 12 is replaced by 10 in data, then the new mean is µ and variance is o² then what is the value of 15(μ + μ² + σ²) = ?

Things To Remember

Probability is defined as the extent to which an event is likely to occur.
It is measured by the ratio of the favourable outcome to the whole number of possible outcomes.
The probability of an event E is a number P(E) such that 0 ≤ P (E) ≤ 1.
It is always a positive number.
There are three major types of probability- experimental probability, theoretical probability, and axiomatic probability.

Read More: Coin toss probability formula

Sample Questions

Ques: A box consists of 200 watches of which 190 are good, 7 have minor defects and 3 have major defects. One watch is drawn at random from the box. What is the probability that the drawn watch is good? Also, find the probability of the drawn watch which doesn't have major defects? (2 marks)

Ans: A watch is drawn at random from 200 watches. So, there are 200 equally likely outcomes.

The probability that the drawn watch is good = 190 / 200 or 95%

The probability of the drawn watch which doesn't have major defects is

= ( 190+7 ) / 200 = 197 / 200 = 98.5%

Ques: Two dice are thrown simultaneously. What is the probability of obtaining a total of 10? (2 marks)

Ans: The possible outcomes are for a sum 10 are -

{ 4, 6 }, { 5, 5 } and { 6, 4 }

The total no. of outcomes are 36

So, the probability of obtaining a total of 10 is 3 / 36 = 1 / 12 or 8.33%

Ques: Two coins are flipped 20 times simultaneously. What is the probability of both coins landing on heads? (2 marks)

Ans: The possible outcomes - (H, H), (H, T), (T, H), and (T, T).

The no. of possible outcomes of both coins landing on heads is 1

So, the probability will be 1/4 or 25%.

Ques: Find the probability of getting an even number when a die is tossed? (2 marks)

Ans: When a die is tossed there are 6 possible outcomes, S = { 1, 2, 3, 4, 5, 6 }

According to the question, favorable events of getting an even number is { 2,4,6 }

Therefore, no. of favorable event = 3

And the total no. of outcomes = 6

Therefore, the probability of getting an even number when a die is tossed is 3 / 6 = 1 / 2

Ques: The blood groups of 30 class 8 students are recorded as follows - (2 marks)

A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, AB, A, O, O, AB, B, A, B, O

You were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random,
(A) has blood group AB
(B) has blood group AB

Ans: The total number of students in the class is = 30

From the given data the number of students having blood group AB = 4

So, the probability of a student whose blood group is AB = 4 / 30 = 2 / 15

Ques: What is probability? Mention 2 of its important property? (3 marks)

Ans: Probability is defined as the extent to which an event is likely to occur. It is measured by the ratio of the favorable outcome to the whole number of possible outcomes.

P (E) = Number of favorable outcomes of E / total number of possible outcomes of E.

Or, P(E) = n(A) / n(S)

Where, P (E) = Probability of an event E; n(A) = No. of favorable outcomes of E; n(S) = Total no. of events in Sample Space

Two properties of probability are -

The probability of an event E is a number P(E) such that 0 ≤ P (E) ≤ 1. Probability is always a positive number.
The probability of a sure event or certain event is 1. Whereas, the probability of an impossible event is 0.

Ques: In a Survey 1500 random families were selected in a certain demographic, and a number of female or girl children were recorded. Find the probability of - (3 marks)
(A) No girl child
(B) At least 1 girl child
(C) At most 2 girls

No. of girl child in the family	0	1	2	More than 2
No. of families	211	814	394	81

Ans: Consider E i denote the event of getting the outcome of no. of girl child, where i = 0, 1, 2, 3 ( 3 represents more than 2 girls child in the family

(A) The probability of the outcome of no girl child is given by,

P ( no girl child) = No. of no girl child in the family/Total no. of families in the survey

Or, P ( no girl child) = 211 / 1500= 0.1406

(B) The probability of the outcome of at least 1 girl child in the family is given by,

For at least a girl child we sum E 1 , E 2, and E 3

Or, no. of at least a girl child = 814 + 394 + 81 = 1289

P ( At Least 1 girl child ) = No. of at least a girl child in the family/Total no. of families in the survey

Or, P ( At Least 1 girl child ) = 1289 /1500 = 0.859

For at least a girl child we sum E 0 , E 1, and E 2

Or, no. of at most two girl child = 211 + 814 + 394 = 1419

P ( at most two girl child ) = No. of at most two girl child in the family / Total no. of families in the survey

Or, P ( at most two girl child ) =1419 / 1500 = 0.946

Ques: An organization selected 2400 families at random and surveyed them to determine a relation between income level and the number of vehicles in a family. The information gathered is listed in a tabular manner? (3 marks)

Monthly Income ( in Rs)	Vehicle per Family
0	1	2	Above 2
Less than 7000	10	105	25	0
7000 - 10000	0	305	27	2
10000 - 13000	1	535	29	1
13000 - 16000	2	465	59	25
More than 16000	1	579	82	88

Suppose a family is chosen. Find the probability that the family chosen is -
(A) earning Rs 10000-13000 per month and owning exactly 2 vehicles.
(B) earning Rs 16000 or more per month and owning exactly 1 vehicle.
(C) earning less than Rs 7000 per month and does not own any vehicle.

Ans: Here, the total number of families surveyed is given to be 2400

(A) From the table the no. of families earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles is 29

Therefore, the probability of a family earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles is = 29 / 2400

(B) From the table the no. of families earning Rs. 16000 or more per month and owning exactly 1 vehicle is 579

Therefore, the probability of a family earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579 / 2400

Therefore, the probability of a family earning less than Rs. 7000 per month and does not own any vehicle = 10 / 2400

Ques: A die is thrown 1000 times and the frequency of the outcomes of 1, 2, 3, 4, 5, and 6 are noted in the given table. Find the probability of each outcome? (4 marks)

Outcome	1	2	3	4	5	6
Frequency	178	151	157	180	149	159

Ans: Consider E i denote the event of getting the outcome i , where i = 1, 2, 3, 4, 5, 6

The probability of the outcome of 1 is given by,

P (E 1) = [ Frequency of desired outcome ( 1 ) ] / [ Total no. of trials or times the die is thrown ]

Or, P (E 1) = 178 / 1000 = 0.178

Therefore, the probability of outcome 1 is 0.178

Similarly for other outcomes, we get,

Or, P (E 2 ) = 151 / 1000= 0.151

Or, P (E 3 ) = 157 / 1000 = 0.157

Or, P (E 4 ) = 180 / 1000 = 0.180

Or, P (E 5 ) = 149 / 1000 = 0.149

Or, P (E 6 ) = 159 / 1000 = 0.159

Therefore, the probability of outcomes 1, 2, 3, 4, 5, and 6 is 0.178, 0.171, 0.157, 0.180, 0.149 , and 0.159 respectively.

Ques: In a Survey 1800 random families were selected in a certain demographic, and a number of male or boy children were recorded. Find the probability of - (3 marks)
(A) No boy child
(B) At least 1 boy child
(C) At most 2 boys

No. of boy child in the family	0	1	2	More than 2
No. of families	211	814	394	81

Ans: Consider E i denote the event of getting the outcome of no. of boy child, where i = 0, 1, 2, 3 ( 3 represents more than 2 girls child in the family

(A) The probability of the outcome of no boy child is given by,

P ( no boy child) = No. of no girl child in the family/Total no. of families in the survey

Or, P ( no boy child) = 211 / 1800= 0.117

(B) The probability of the outcome of at least 1 boy child in the family is given by,

For at least a boy child we sum E 1 , E 2, and E 3

Or, no. of at least a boy child = 814 + 394 + 81 = 1289

P ( At least 1 boy child ) = No. of at least a boy child in the family/Total no. of families in the survey

Or, P ( At least 1 boy child ) = 1289 /1800 = 0.716

For at least a boy child we sum E 0 , E 1, and E 2

Or, no. of at most two boy child = 211 + 814 + 394 = 1419

P ( at most two boy child ) = No. of at most two boy child in the family / Total no. of families in the survey

Or, P ( at most two boy child ) =1419 / 1800 = 0.7883

Ques: What is the probability of getting a sum of 9 when two dice are thrown? (3 marks)

Ans: There are 36 possibilities when we throw two dice.

The desired outcome is 9. To get 9, we can have three favorable outcomes.

{(4,5),(6,3),(5,4),(3,6)}

Probability of an event = number of favorable outcomes/ sample space

Probability of getting number 9 = 4/36 =1/9

Ques: Find the probability of ‘getting 5 on rolling a die’? (2 marks)

Ans: Sample Space = S = {1, 2, 3, 4, 5, 6}

Total number of outcomes = n(S) = 6
Let A be the event of getting 5.
Number of favourable outcomes = n(A) = 1
i.e. A = {5}
Probability, P(A) = n(A)/n(S) = 1/6
Hence, P(getting 5 on rolling a die) = 1/6

Ques: A vessel contains 4 blue balls, 4 red balls and 10 white balls. If three balls are drawn from the vessel at random, what is the probability that the first ball is red, the second ball is blue, and the third ball is white? (4 marks)

Ans: Given, the probability to get the first ball is red or the first event is 4/18.

Since we have drawn a ball for the first event to occur, then the number of possibilities left for the second event to occur is 18 – 1 = 17.

Hence, the probability of getting the second ball as blue or the second event is 4/17.

Again with the first and second event occurring, the number of possibilities left for the third event to occur is 17 – 1 = 16.

And the probability of the third ball is white or the third event is 10/16.

Therefore, the probability is 4/18 x 4/17 x 10/16 = 160/4896 = 0.032.

Or we can express it as: P = 3.2%.

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Probability: Formula, Types, Properties & Applications

What is Probability?

Solved Example of Probability

Probability Terms and Definitions

Sample space

Sample Point

Experiment

Event

Outcome

Formula for Probability

Solved Example of Formula for Probability

Types of Probability

Classical or Theoretical Probability

Experimental Probability

Solved Example of Experimental Probability

Axiomatic Probability

Properties of Probability

Probability of an Event

Types of Event

Complementary Events

Solved Example of Complementary Events

Independent Events

Mutually Exclusive Events

Solved Example of Mutually Exclusive Events

Applications of Probability

Important Topics for JEE Main

Things To Remember

Sample Questions

CBSE X Related Questions

1.
Assertion (A) : The system of linear equations $3x - 5y + 7 = 0$ and $-6x + 10y + 14 = 0$ is inconsistent.
Reason (R) : When two linear equations don't have unique solution, they always represent parallel lines.

2.
The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

4.
A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

5.
If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

6.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is $3/5$, then find the number of yellow balls.

Similar Mathematics Concepts

Comments

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Probability: Formula, Types, Properties & Applications

What is Probability?

Solved Example of Probability

Probability Terms and Definitions

Sample space

Sample Point

Experiment

Event

Outcome

Formula for Probability

Solved Example of Formula for Probability

Types of Probability

Classical or Theoretical Probability

Experimental Probability

Solved Example of Experimental Probability

Axiomatic Probability

Properties of Probability

Probability of an Event

Types of Event

Complementary Events

Solved Example of Complementary Events

Independent Events

Mutually Exclusive Events

Solved Example of Mutually Exclusive Events

Applications of Probability

Important Topics for JEE Main

Things To Remember

Sample Questions

CBSE X Related Questions

1.Assertion (A) : The system of linear equations $3x - 5y + 7 = 0$ and $-6x + 10y + 14 = 0$ is inconsistent. Reason (R) : When two linear equations don't have unique solution, they always represent parallel lines.

2.The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

3.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

4.A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

5.If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

6.A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is $3/5$, then find the number of yellow balls.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Assertion (A) : The system of linear equations $3x - 5y + 7 = 0$ and $-6x + 10y + 14 = 0$ is inconsistent.
Reason (R) : When two linear equations don't have unique solution, they always represent parallel lines.

2.
The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

4.
A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

5.
If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

6.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is $3/5$, then find the number of yellow balls.