Probability: Definition, Theorems, and Applications

Probability in Mathematics

Probability is a measure of how likely an event is to happen. With the help of probability, we can predict the chance of an event occurring. The value of probability ranges between $0$ and $1$, where $0$ represents the impossible event and $1$ represents the probability of events that are certain to happen. Now, let us look into the basic concepts of probability.

Types of Probability

• Classical Probability: Classical probability is based on the assumption that all outcomes of a random experiment are equally likely. It is calculated as the ratio of the number of favourable outcomes to the total number of possible outcomes. For example, the probability of rolling a $3$ on a fair six-sided die is $\frac{1}{6}$.
• Empirical Probability: Empirical probability is based on observed data rather than theoretical calculations. It is determined by conducting experiments or collecting data and then calculating the probability based on the relative frequency of an event occurring in the data. For example, if a coin is flipped 100 times and lands on heads $55$ times, the empirical probability of landing on heads is $\frac{55}{100}$ = $0.55$.
• Subjective Probability: Subjective probability is based on personal judgment, intuition, or experience rather than objective data or mathematical calculation. It reflects an individual's belief about the likelihood of an event occurring, which may vary from person to person. For example, someone might estimate a $70%$ chance of their favourite sports team winning a game based on their analysis and feelings about the team's performance.
• Axiomatic Probability: Axiomatic probability is a formal framework for probability theory based on a set of axioms or fundamental principles. It provides a rigorous foundation by defining probability functions and operations through axioms, such as the probability of the entire sample space being $1$ and the additivity of mutually exclusive events.

Probability Class 11 Terminologies

• Sample Space: The set of all possible outcomes in a random experiment is called a sample space. For example, the sample space for a coin toss is {Heads, Tails}.

• Outcome and Trial: A trial means doing or performing an experiment, and an outcome is the result of that experiment. For example, if rolling a dice is a trial, then getting a $3$ is an outcome.
• Event: An event is one of the specific outcomes of an experiment; in other words, an event is a subset of the sample space.
• Favourable Event: Favourable events are those events associated with the experiment which are of interest to us. When rolling a dice, if we are interested in even numbers, then getting a $2$ is called a favourable event.
• Non-favourable Event: Non-favourable events are those events associated with the experiment in which we are not interested. When rolling a die, if we are interested in even numbers, then getting a $3$ is called a non-favourable event.
• Exhaustive Events: Exhaustive events are all those events that take up all the possible outcomes of an experiment. When rolling a die, the exhaustive events are getting a $1, 2, 3, 4, 5,$ and $6$
• Mutually Exclusive Events: When some events can not happen at the same time or simultaneously, then they are called mutually exclusive events. When tossing a coin, getting a head or a tail is are mutually exclusive event.
• Independent Events: Independent events are those events where the occurrence of one event does not affect the other event's occurrence or vice versa. For example, flipping a coin and rolling a die can be considered independent events.
• Impossible Events: An impossible event is an event which cannot happen. For example, rolling a $7$ on a standard six-sided die is an impossible event.

Probability of an Impossible Event

The probability of an impossible event is always $0$, as it can not happen under any situation. If you roll a standard six-sided die, it is impossible to get a $7$, as there is no $7$ on the die. So in this case, the probability of rolling a $7$ is $0$.

Probability of a sure event

The probability of a sure event is always 1 because it is certain to occur. When we flip a coin, we are certain to get a head or a tail, so getting a head or a tail in this case is a sure event. Hence, the probability of getting either heads or tails is 1 when we toss a coin.

Probability Class 12 Terminologies

Random Variable

A random variable is a real-valued function whose domain is the sample space of a random experiment.

Probability Distribution of a Random Variable

The probability distribution for a random variable describes how the probabilities are distributed over the values of the random variable.
The probability distribution of a random variable $X$ is the system of numbers

$
\begin{array}{rlllllll}
X & : & x_1 & x_2 & x_3 & \ldots & \ldots & x_n \\
P(X) & : & p_1 & p_2 & p_3 & \ldots & \ldots & p_n \\
& p_i \neq 0, & \sum_{i=1}^n p_i=1, & i=1,2,3, \ldots n
\end{array}
$

The real numbers $\underline{\underline{x_1},}, x_2, \ldots, x_n$ are the possible values of the random variable $X$ and $p_i(i=1,2, \ldots, n)$ is the probability of the random variable $X$ taking the value $x i$ i.e., $P\left(X=x_i\right)=p_i$

Types Of Probability Distribution:

1. Binomial distribution

2. Normal Distribution

3. Cumulative distribution frequency

Probability Concepts and Formulas

The probability of an event $A$ is written as $P(A)$, and it can easily calculated using the formula: $P(A) =\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

For example, while rolling a dice probability of getting a $6$ is $\frac{1}{6}$.

Steps to Calculate Probability

• Define the Experiment: Determine the process or action that generates outcomes.
• Identify the Sample Space: List all possible outcomes of the experiment.
• Determine the Favourable Outcomes: Identify which of these outcomes meet the criteria of the event of interest.
• Apply the Probability Formula: Use the formula to calculate the probability of the event.

Example: What is the probability of getting a number greater than $4$ when rolling a dice?

The experiment is rolling a single die.

The sample space $S$ consists of all possible outcomes: $S={1,2,3,4,5,6}$

The favourable outcomes are ${5,6}$

The required probability $= \frac{2}{6} = \frac{1}{3}$.

Probability of either of the two events occurring (Addition Rule)

The probability when either of two events occurs, or the addition rule, states that the probability of either of the two events occurring is the sum of their individual probabilities minus the probability of both events happening. The formula is:

$P(A \cup B) = P(A) + P(B) - P (A \cap B)$

Where $P(A \cup B)$ is the probability of either event $A$ or event $B$ occurring.

$P(A∩B)$ is the probability of both events $A$ and $B$ occurring.

$P(A)$ is the probability of event $A$ occurring.

$P(B)$ is the probability of event $B$ occurring.

Probability of a Complementary Event

Complementary events in probability are two such events which are mutually exclusive events which means they can not happen together. In other words, if one event occurs, then the other event can not occur. The sum of the probabilities of two complementary events is $1$. The formula:

$P(A′) = 1 - P(A)$

where $P(A)$ is the probability of the event $A$, and $P(A′)$ is the probability of the event $A′ $ (the complement of $A$).

Conditional Probability (Probability of $B$ if $A$ has already occurred)

Conditional probability is the probability of an event $B$ occurring given that another event $A$ has already happened. It is denoted by $P(B∣A)$. Conditional probability formula is:

$P(B∣A) = \frac{P(A \cap B)}{P(A)}$

Where $P(B∣A)$ is the probability of event $B$ given that $A$ has occurred.

$P(A∩B)$ is the probability of both events $A$ and $B$ occurring.

$P(A)$ is the probability of event $A$ occurring.

Probability when two events occur simultaneously

The probability of two events occurring simultaneously has two cases: one is when the events are independent, and the other is when the events are dependent.

• For independent events: If events $A$ and $B$ are independent, the probability of both events occurring simultaneously is the product of the probability of the individual events, it is given by: $P(A∩B) = P(A) × P(B)$
• For dependent events: When events $A$ and $B$ are dependent, the probability of both events occurring simultaneously is given by: $P(A∩B) = P(A) × P(B∣A)$, where $P(B∣A)$ is the conditional probability of event $B$ given that event $A$ has already occurred.

Theorems on Probability

Theorems on probability include Bayes' Theorem, Law of total probability, and some other theorems on probability.

Theorems on Probability Class 11

Theorem 1: The sum of the probability of happening of an event and not happening of an event is equal to $1$. $P(A) + P(A') = 1$.

Theorem 2: The probability of an impossible event or the probability of an event not happening is always equal to $0$. $P(ϕ) = 0$.

Theorem 3: The probability of a sure event is always equal to $1$. $P(A) = 1$

Theorem 4: The probability of happening of any event always lies between $0$ and $1$, i.e. $0 \leq P(A) \leq 1$.

Theorems on Probability Class 12

Theorem 1: If there are two events $A$ and $B$, we can apply the formula of the union of two sets, and we can derive the formula for the probability of happening of event $A$ or event $B$ as follows.

$P(A∪B) = P(A) + P(B) - P(A∩B)$

Also for two mutually exclusive events $A$ and $B$, we have $P( A U B) = P(A) + P(B)$

Bayes’ Theorem for Conditional Probability

Bayes’ Theorem provides a way to update the probability of an event based on new evidence. It is expressed as $P(A∣B) = \frac{P(B∣A). P(A)}{P(B)}$
where $P(A∣B)$ is the probability of event $A$ given that event $B$ has occurred, $P(B∣A)$ is the probability of observing $B$ given $A$, $P(A)$ is the prior probability of $A$, and $P(B)$ is the total probability of $B$

Law of Total Probability

The Law of Total Probability states that the probability of an event can be found by considering all possible ways that the event can occur through a partition of the sample space. It is expressed as $P(B) = \sum_{i} P(B|A_i). P(A_i)$, where $A_i$ are mutually exclusive and exhaustive events that partition the sample space. This law allows us to compute $P(B)$ by summing the probabilities of B occurring with each $A_i$ weighted by the probability of $A_i$.

Important Points

• The probability ranges from $0$ to $1$.
• The sum of probabilities of all possible outcomes in a sample space is $1$.
• For mutually exclusive events, the probability of either event occurring is the sum of their individual probabilities.

Probability tree

A probability tree is a diagram that represents all possible outcomes of an event and their probabilities. It is useful for calculating the probabilities of combined events.

Application of probability

Probability has a wide variety of applications in real life. Some of the common applications that we see in our everyday lives while checking the results of the following events:

• Choosing a card from the deck of cards
• Flipping a coin
• Throwing a dice in the air
• Pulling a red ball out of a bucket of red and white balls
• Winning a lucky draw

Understanding the deck of cards

A standard deck has $52$ cards with $4$ suits in two colours [hearts, diamonds, clubs, spades] each containing $13$ ranks $(2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace)$.

Tossing of a coin, Tossing of two or more coins

Each coin toss has $2$ possible outcomes. Tossing two coins can result in $HH, HT, TH,$ or $TT$.

Rolling a dice

A standard die has $6$ faces numbered $1$ to $6$. Each face has an equal probability of $\frac{1}{6}$.

Important Formulae for Probability

We have provided below the important formulae, along with the key note that specifies why it is used. With these important formulae, you will get to know how to use them while solving different problems.

Probability in Mathematics: Solved Previous Year Questions

Question 1:

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is $\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{n}^2-\mathrm{m}^2$ is equal to :

Answer:

Let $U$ : let unbiased coin is drawn, $H$ : head turns up

$\begin{aligned}
& \Rightarrow P(U / H)=\frac{P(U) \cdot P(H / U)}{P(U) P(H / U)+P(\bar{U}) \cdot P(H / U)} \\
& =\frac{\frac{19}{20} \cdot \frac{1}{2}}{\frac{19}{20} \cdot \frac{1}{2}+\frac{1}{20} \cdot\left(\frac{2}{2}\right)}=\left(\frac{19}{21}\right)=\frac{m}{n} \\
& n^2-m^2=21^2-19^2=(21+19)(21-19)=80
\end{aligned}$

Hence, the correct answer is 80.

Question 2:

A card from a pack of 52 cards is lost. From the remaining 51 cards, $n$ cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$, the n is equal to:

Answer:

Let the event $L$ be: "Lost card is a spade"
Let the event $D$ be: "All $n$ drawn cards are spades"

We are given:

$
P(L \mid D) = \frac{11}{50}
$

We will use Bayes' Theorem:

$
P(L \mid D) = \frac{P(D \mid L) \cdot P(L)}{P(D)}
$

First, compute the prior probabilities:
Number of spades in a full deck = 13
Probability that the lost card is a spade:

$
P(L) = \frac{13}{52} = \frac{1}{4}
$

Probability that the lost card is not a spade:

$
P(L') = \frac{39}{52} = \frac{3}{4}
$

Now calculate: If the lost card is a spade, then the number of spades left = $13 - 1 = 12$, and the total cards left = 51.

So, $
P(D \mid L) = \frac{{^{12}C_n}}{{^{51}C_n}}
$

If the lost card is not a spade, then all 13 spades are still there.

So, $
P(D \mid L') = \frac{{^{13}C_n}}{{^{51}C_n}}
$

Now apply Bayes' Theorem:

$
\frac{11}{50} = \frac{\frac{{^{12}C_n}}{{^{51}C_n}} \cdot \frac{1}{4}}{\frac{{^{12}C_n}}{{^{51}C_n}} \cdot \frac{1}{4} + \frac{{^{13}C_n}}{{^{51}C_n}} \cdot \frac{3}{4}}
$

Cancel common factor ${^{51}C_n}$ in numerator and denominator:

$
\frac{11}{50} = \frac{{^{12}C_n} \cdot \frac{1}{4}}{{^{12}C_n} \cdot \frac{1}{4} + {^{13}C_n} \cdot \frac{3}{4}}
$

Multiply numerator and denominator by 4:

$
\frac{11}{50} = \frac{{^{12}C_n}}{{^{12}C_n} + 3 \cdot {^{13}C_n}}
$

Now, cross-multiply:

$
11 \left( {^{12}C_n} + 3 \cdot {^{13}C_n} \right) = 50 \cdot {^{12}C_n}
$

$
11 \cdot {^{12}C_n} + 33 \cdot {^{13}C_n} = 50 \cdot {^{12}C_n}
$

$
33 \cdot {^{13}C_n} = 39 \cdot {^{12}C_n}
$

Divide both sides by 3: $
11 \cdot {^{13}C_n} = 13 \cdot {^{12}C_n}
$

Use identity: $
{^{13}C_n} = \frac{13}{13 - n} \cdot {^{12}C_n}
$

So: $
11 \cdot \frac{13}{13 - n} \cdot {^{12}C_n} = 13 \cdot {^{12}C_n}
$

Cancel ${^{12}C_n}$: $
\frac{143}{13 - n} = 13
$

Now solve: $
143 = 13 \cdot (13 - n)
$

$
143 = 169 - 13n
$

$
13n = 169 - 143 = 26
$

$n = 2$

Hence, the correct answer is 2.

Question 3:

The probability of forming a 12-person committee from 4 engineers, 2 doctors, and 10 professors containing at least 3 engineers and at least 1 doctor is:

Answer:

3 engineering +1 doctor +8 Professors $\rightarrow{ }^4 \mathrm{C}_3 \cdot{ }^2 \mathrm{C}_1 \cdot{ }^{10} \mathrm{C}_8$ $=360$
3 engineering +2 doctors +7 Professors $\rightarrow{ }^4 \mathrm{C}_3$. ${ }^2 \mathrm{C}_2 \cdot{ }^{10} \mathrm{C}_7$ $=480$
4 engineering +1 doctor +7 Professors $\rightarrow{ }^4 \mathrm{C}_4 \cdot{ }^2 \mathrm{C}_1 \cdot{ }^{10} \mathrm{C}_7$ $=240$
4 engineering +2 doctors +6 Professors $\rightarrow{ }^4 \mathrm{C}_4$. ${ }^2 \mathrm{C}_2 \cdot{ }^{10} \mathrm{C}_6$ $=210$

Total $=1290$
Required probability $=\frac{1290}{{ }^{16} \mathrm{C}_{12}}$

$=\frac{1290}{1820}$

$=\frac{129}{182}$

Hence, the correct answer is $\frac{129}{182}$.

Question 4:

A board has 16 squares as shown in the figure :

Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is:

Answer:

Total ways for selecting any two squares $={ }^{16} \mathrm{C}_2$

$
=120
$

Total ways for selecting common side squares

$=\underbrace{3 \times 4}_{\text {Horizontal side }}+\underbrace{3 \times 4}_{\text {vertical side }}$

$
=24
$

So required probability

$
\begin{aligned}
& =1-\frac{24}{120} \\
& =\frac{4}{5}
\end{aligned}
$

Hence, the correct answer is $\frac{4}{5}$.

Question 5:

One die has two faces marked 1, two faces marked 2, one face marked 3, and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3, and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both dice are thrown together, is:

Answer:

Dice 1 : 1, 1, 2, 2, 3, 4
Dice 2 : 1, 2, 2, 3, 3, 4

For Sum $=4$

$
\begin{aligned}
& (1,3) \Rightarrow 2 \times 2=4 \\
& (3,1) \Rightarrow 1 \times 1=1 \\
& (2,2) \Rightarrow 2 \times 2=4
\end{aligned}
$

Total ways $=9$

$
\text { Sum = } 5
$
$
\begin{aligned}
& (1,4) \Rightarrow 2 \times 1=2 \\
& (4,1) \Rightarrow 1 \times 1=1 \\
& (2,3) \Rightarrow 2 \times 2=4 \\
& (3,2) \Rightarrow 1 \times 2=2
\end{aligned}
$
Total ways $=9$

$
\text { Probability }=\frac{18}{36}=\frac{1}{2}
$

Hence, the correct answer is $\frac{1}{2}$.

List of Topics related to probability according to NCERT/JEE Mains

This section covers all the important topics in probability that are part of the Class 11 and Class 12 NCERT syllabus and are frequently asked in JEE Main exams.

Probability in Different Exams

The chapter Probability deals with measuring the likelihood of the occurrence of events and helps students understand uncertainty in a mathematical way. It is an important part of Class 11 and Class 12 Mathematics and is frequently tested in board as well as competitive examinations. Questions from this chapter assess a student’s ability to apply probability rules, analyse events logically, and solve real-life situation–based problems. Regular practice of NCERT questions and a clear understanding of concepts help in scoring well in this chapter.

Important Books and Resources for Probability

Here you will find the best reference books and study materials that explain probability formulas, concepts, and problem-solving methods in detail.

NCERT Resources for Probability

This section highlights key NCERT textbooks and chapters that build a strong foundation in probability for board exams and competitive exams.

• NCERT Maths Notes for Class 12th Chapter 13 Probability

• NCERT Maths Solutions for Class 12th Chapter 13 Probability

• NCERT Maths Exemplar Solutions for Class 12th Chapter 13 Probability

NCERT Subjectwise Resources

Here, we explore subject-wise NCERT resources such as notes, solutions, and exemplar solutions to help you prepare effectively.

Practice Questions based on Probability

This part includes a set of practice questions designed to strengthen your understanding of probability concepts and improve problem-solving speed.

Conclusion

The chapter Probability helps students understand and quantify uncertainty in a logical and systematic way. By learning the rules, formulas, and theorems of probability, students develop strong analytical and reasoning skills that are essential for solving real-life and exam-based problems. This chapter is an important scoring area in Class 12 Mathematics and forms a foundation for advanced studies in statistics, data science, economics, and research. With regular practice of NCERT problems and a clear understanding of concepts like conditional probability and Bayes’ theorem, students can confidently master Probability for board and competitive examinations.