Equally Likely Events: Formulas & Examples

Content Writer

Equally Likely Events is the most important concept included in NCERT Class 10 Mathematics Probability. It is defined as two or more events that have an equal chance of occurrence.

Equally Likely Events use the concept of probability to solve these types of problems.
Event refers to a possible set of outcomes of an experiment.
These types of events are random.
An event consisting of only a single experiment in the sample space is called a Simple event.
Both events have the same set of probabilities.
Most common example of equally likely events is cricket matches during coin toss in which you are equally likely to get heads or tails.
It can mathematically be represented as:

Probability of equally likely event: 1 / size of sample space

Table of Content

Equally Likely Events
What is Probability?
Solved Examples of Equally Likely Events
Things to Remember
Sample Questions

Key Terms: Equally Likely Event, Events, Probability, Outcomes, Sample Space, Mutually Exclusive Events, Exhasutive Events, Experiments, Compound Events

Equally Likely Events

[Click Here for Sample Questions]

Equally Likely Events are type of events that are equally likely to happen. These type of event randomly in set of outcomes. It will determine how likely an event will take place.

The set of outcome in a random experiments are called sample space.
The probability of occurence of any type of event is in between 0 and 1.
Event is said to be the subset of sample space to which a probability value is assigned.
There are chances that an event will be equally like as well as mutually exclusive.
The probability of an event can be mathematically represented as follows:

P(E) = Number of Favourable Outcomes/ Total Number of Outcomes

Example of Equally Likely Events

Example 1: In our everyday life we see many examples like-

There are chances to get selected in exams.
Most probably, it will rain today.
There are 50% chances to get 1st chance to play and so on.

Example 2: Concept of tossing two coins: Two coins tossed 500 times simultaneously and will get-

Two heads: 105 times
One head: 275 times
No head: 120 times

Calculate the probability of all these events?

Solution: We will find out the probability of all these events one by one by putting the values in the formulae-

Number of times getting two heads/one heads or No heads/Total number of times the coins are tossed

Probability of getting two heads: 105/500 = 0.21

Probability of getting One head: 275/500 = 0.55

Probability of getting No head: 120/500 = 0.24

Conclusion:

The probability values of each event lie between 0 and 1.
The sum of all probability values is 1.
These probability results cover all the possible outcomes of tossing two coins.

Equally Likely Events

Read More:

Chapter Related Concepts
Conditional Probability	Geometric Probability	Conditional probability
Multiplication Theorem on Probability	Sum of Probabilities	Rolle’s Theorem

What is Probability?

[Click Here for Sample Questions]

Probability is the branch of mathematics that determine chance of occurrence of an event in terms of mathematical expression. If the value of probability is high then the value of occurence of an event is also high.

The events can be explained with the use of probability tree diagram.
The concept of probability was first introduced by Andrey Kolmogorov.
It is used in the field of science, engineering, and statistics.
There are various types of events used in probability.
Mutually exclusive and exhaustive events are two most important types of events.
It determines the occurence or likeness of an event.

Example of What is Probability?

Example: Let's perform an activity and find out the equally likely events with respect to that event: Take a coin toss it 10 times, 20 times, 30 times and try to find out how many times head and tail comes, record the observations for calculating probability and you can write down the values as-

Number of times a head we get/Total number of times the coin is tossed

Number of times a tail we get/Total number of times the coin is tossed

Conclusion:

You will find that as you increase the number of tosses your value comes close to 0.5 which indicates 50% chances of getting head or tail at the time of performing coin toss.

Solved Examples of Equally Likely Events

[Click Here for Sample Questions]

Various solved examples of equally likely event are as follows:

Example 1: Concept of throwing a dice:

The possible outcomes 1,2,3,4,5,6 respectively of throwing a dice are given in table below-

Outcome	1	2	3	4	5	6
Frequency	179	150	157	149	175	190

Find out the probability of different possible outcomes-

1- Getting 1 as possible outcome

2- Getting 2 as possible outcome

3- Getting 3 as possible outcome

4- Getting 4 as possible outcome

5- Getting 5 as possible outcome

6- Getting 6 as possible outcome

Solution:

Formulae:

Probability of getting the frequency 1or2or3or4or5or6/Total number of times the dice is thrown

1- As per the data given in the table

The frequency of getting outcome 1 is 179-

179/1000 = 0.179

2- The frequency of getting 2 as outcome is 150

150/1000 = 0.150

3-The frequency of getting 3 as outcome is- 157

157/1000 = 0.157

4-The frequency of getting 4 as outcome- 149

149/1000 = 0.149

5-The frequency of getting 5 as outcome- 175

175/1000 = 0.175

6-The frequency of getting 5 as outcome- 190

190/1000 = 0.190

Example 2: Concept of calculation of probability of percentage of marks

Ques: Here the marks obtained by the students of a class in their unit tests are given below-

Unit tests	I	II	III	IV	V
Percentage of marks obtained	42%	76%	65%	94%	84%

Calculate the probability of-

1- Students get more than 70% in a unit test

2- Students get more than 80% in a unit test

3- Students get more than 90% in a unit test

Solution:

The total number of unit tests held is 5.

The probability of getting different percentages are-

1- 3 students get more than 70% marks in unit tests

3/5 =0.6

2- 2 students get more than 80% marks

2/5=0.4

3- 1 student gets more than 90% marks

1/5=0.2

Read More:

Class 10 Maths Concepts
Mean of Grouped Data	Types of Sets	Empirical Probability
Types of events	Probability — A Theoretical Approach	Probability and statistics symbols

Things to remember:

Equally likely events is a type of elementary or compound events.
If any event consists of more than one possible outcome of a sample space is called compound event.
Any event which consists of only one possible outcome is called elementary events.
Elemementary events are also known as simple events.
The probability of a single outcome of an event lies between 0 and 1.
The sum of all the outcomes of a single event comes 1.

Sample Questions:

Ques. Here 11 bags of wheat flour are given which have marked 5 kg, but the actual weight of flour (in kg) are-
5.05, 5.08, 5.00, 4.96, 5.03, 5.06, 5.08, 5.07, 5.04, 5.00, 4.98
Calculate the probability of any of the bags randomly having weight more than 5 kg of wheat flour? (1 mark)

Ans: The probability of any of the bags randomly selected and having weight more than 5 kg is given as: 7/11

Ques: The tyre manufacturing company performs 1000 cases to record the distance covered before a tyre is needed to be replaced shows in table given below-

Distance (in km)	Less than 4000	4000 to 9000	9001-14000	More than 14000
Frequency	20	210	325	445

You have to buy a tyre and find out the probability of-
(A) The tyre needs to be replaced before it has covered less than 4000km.
(B) Tyres will last more than 9000 km.
(C) Tyre will be replaced when it has covered the distance between 4000 km - 14000 km? (3 marks)

Ans: (A) The tyre needs to be replaced before it has covered less than 4000km: 0.02

(B) Tyres will last more than 9000 km: 0.77

Ques: A weather station records the weather forecasts in 250 consecutive days, and they were correct 175 times. Find out the probability of-
(A) A given day when the weather forecast was correct?
(B) A given day when it was not correct? (2 marks)

Ans: (A) A given day when the weather forecast was correct: 0.7

(B) A given day when it was not correct: 0.3

Ques. In a basketball competition a player goal 6 balls out of 10 times he wishes to do so. Calculate the probability of the time he failed to score the basketball in the goal? (1 mark)

Ans: The probability of the time he failed to score the basketball in the goal: 6/10=0.6

Ques: Here the marks obtained by the students of a class in their unit tests are given below:

Unit tests	I	II	III	IV	V
Percentage of marks obtained	41%	75%	64%	93%	83%

Calculate the probability of-
(A) Students get more than 71% in a unit test
(B) Students get more than 81% in a unit test
(C) Students get more than 91% in a unit test ? (3 marks)

Ans: The total number of unit tests held is 5.

The probability of getting different percentages are-

(A) 3 students get more than 71% marks in unit tests: 3/5 =0.6

(B) 2 students get more than 81% marks: 2/5=0.4

(C) 1 student gets more than 91% marks: 1/5=0.2

Ques: Find the probability of getting a number less than 5 in a single throw of a die? (2 marks)

Ans: Possible outcome = {1, 2, 3, 4}

∴ P (Getting a number < 5) = 4/6 = 2/3

Ques. If P(E) = 0.05, what is the probability of 'not E'? (2 marks)

Ans: It is given that P(E) = 0.05

P(E) + P (not E) = 1
0.05 + P (not E) = 1 ⇒ P (not E) = 1 – 0.05
0.95

Thus, probability of 'not E' = 0.95.

Ques: Consider an event which can take place in 999 ways. If a trial is carried out in this event then it will have a probability of 99%, determine the number of favourable events? (2 marks)

Ans: As we known the formula of the probability of an equally likely event. If any event E is carried out and P(E) represents the probability of this event then it can be written as:

P (E) = (number of events in favour of E)/(Total number of possible events)
Total number of required possible events = 999.
The probability determined to be 99% or .99.
Therefore we can write it as:
P (E) = (number of events in favour of E) / 999 or .99 = (number of events in favour of E) / 999
Therefore, we can determine the number of events in favour of E = 999×.99 = 989.01 ≈ 989

Ques: Consider when an individual is tossing two coins. The two coins are tossed 1000 times simultaneously and will us possible sample spaces:

Two heads: 105 times
One head: 275 times
No head: 120 times

Calculate the probability of all these events? (3 marks)

Ans:We will find out the probability of all these events one by one by putting the values in the formulae-

Number of times getting two heads/one heads or No heads/Total number of times the coins are tossed

Probability of getting two heads: 105/1000 = 0.105
Probability of getting One head: 275/1000 = 0.275
Probability of getting No head: 120/1000 = 0.120

Ques: Consider an individual is throwing a dice whose outcomes are tabulated below:

Outcome	1	2	3	4	5	6
Frequency	170	100	150	140	170	200

Find out the probability of different possible outcomes-
(A) Getting 1 as possible outcome
(B) Getting 2 as possible outcome
(C) Getting 3 as possible outcome
(D) Getting 4 as possible outcome
(E) Getting 5 as possible outcome (F) Getting 6 as possible outcome? (3 marks)

Ans: The formula to calculate the equally likely of an event is given as: Probability of getting the frequency 1or2or3or4or5or6/Total number of times the dice is thrown

(A) As per the data given in the table

The frequency of getting outcome 1 is 179-

170/1000 = 0.17

(B) The frequency of getting 2 as outcome is 150

100/1000 = 0.1

150/1000 = 0.15

(D) The frequency of getting 4 as outcome- 149

140/1000 = 0.14

(E) The frequency of getting 5 as outcome- 175

170/1000 = 0.17

(F) The frequency of getting 5 as outcome- 190

200/1000 = 0.2

Ques: Here the marks obtained by the students of a class in their unit tests are given below: (3 marks)

Unit tests	I	II	III	IV	V
Percentage of marks obtained	45%	78%	69%	98%	79%

Calculate the probability of-
(A) Students get more than 72% in a unit test
(B) Students get more than 82% in a unit test
(C) Students get more than 92% in a unit test? (3 marks)

Ans: The total number of unit tests held is 5.

The probability of getting different percentages are-

(A) 3 students get more than 72% marks in unit tests

3/5 =0.6

(B) 1 students get more than 82% marks

1/5=0.2

(C) 1 student gets more than 92% marks

1/5=0.2

Check-Out:

Study Guides
NCERT Solutions Class 6 to 12	NCERT Solutions for Class 7 Science	NCERT Solutions for Class 7 Maths
Biology Study Notes	Biology MCQs	Topics for Comparison in Physics
Physics Study Notes	Choice-based questions in physics	Topics with relation in physics
Comparison Topics in Biology	SI units in Physics	Important Derivations in Physics
Physics MCQs	Formulas in Physics	Important Physics Constants and Units
Chemistry Study Notes	Comparison topics in Chemistry	Chemistry MCQs
Maths MCQs	Important Chemistry Formulas	Important Maths Formulas

Equally Likely Events: Formulas & Examples

Equally Likely Events

Example of Equally Likely Events

What is Probability?

Example of What is Probability?

Solved Examples of Equally Likely Events

Things to remember:

Sample Questions:

CBSE X Related Questions

1.
Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.

2.
Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

3.
In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

4.
The value of p for which roots of the quadratic equation $x^2 - px + 6 = 0$ are rational, is

5.
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.

6.
In the given figure, point D divides the side BC of $\Delta ABC$ in the ratio $1 : 2$. Find length AD.

Comments

SUBSCRIBE TO OUR NEWS LETTER

Equally Likely Events: Formulas & Examples

Equally Likely Events

Example of Equally Likely Events

What is Probability?

Example of What is Probability?

Solved Examples of Equally Likely Events

Things to remember:

Sample Questions:

CBSE X Related Questions

1.Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.

2.Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

3.In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

4.The value of p for which roots of the quadratic equation $x^2 - px + 6 = 0$ are rational, is

5.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.

6.In the given figure, point D divides the side BC of $\Delta ABC$ in the ratio $1 : 2$. Find length AD.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Use graphical method to solve the system of linear equations : $x = -3$ and $5x - 2y = -5$.

2.
Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

3.
In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

4.
The value of p for which roots of the quadratic equation $x^2 - px + 6 = 0$ are rational, is

5.
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.

6.
In the given figure, point D divides the side BC of $\Delta ABC$ in the ratio $1 : 2$. Find length AD.