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By Rohit RP
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Updated on 12 Aug 2026, 16:08 IST
The NCERT Solutions for Class 10 Maths Chapter 14 Probability provide clear and step-by-step solutions to the questions given in the current NCERT textbook. Chapter 14 introduces students to the theoretical or classical approach to probability and explains how to calculate the chance of an event when the possible outcomes are equally likely. The current CBSE Class 10 Maths 2026-27 syllabus focuses on the classical definition of probability and simple problems based on finding the probability of an event.
The current Class 10 Maths Chapter 14 Probability contains Exercise 14.1, which includes 25 main questions covering basic probability rules, equally likely outcomes, complementary events, coins, dice, playing cards, coloured objects, defective items, spinners, and other probability-based situations. Current-syllabus solution pages from major education publishers also organise the chapter around Exercise 14.1.
These Probability Class 10 NCERT Solutions are designed to help students understand not only the but also the method used to identify total outcomes, favourable outcomes, and the correct probability formula. Students can use the solutions for homework, revision, self-study, and exam preparation.
Students can use the NCERT Solutions Class 10 Maths Chapter 14 Probability PDF for offline study and quick revision. The PDF can include complete solutions to Class 10 Maths Chapter 14 Exercise 14.1, important probability formulas, coin and dice outcomes, playing-card facts, and common mistakes students should avoid.
Students can download the NCERT Solutions for Class 10 Maths Chapter 14 Probability PDF from the download option provided on this page for convenient offline practice.
Probability is a measure of how likely an event is to occur.
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For example, when a fair coin is tossed, there are two possible outcomes:
H = Head
T = Tail
Since both outcomes are equally likely:
P(Head) = 1/2
and:
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P(Tail) = 1/2
Similarly, when a fair die is thrown, the possible outcomes are:
{1, 2, 3, 4, 5, 6}
Each number has an equal chance of appearing.
The basic probability formula used throughout Class 10 Maths Chapter 14 Probability is:
P(E) = Number of favourable outcomes / Total number of equally likely outcomes
The value of probability always lies between 0 and 1:
0 ≤ P(E) ≤ 1
A probability of 0 represents an impossible event, while a probability of 1 represents a sure or certain event.
Most questions in Probability Class 10 Exercise 14.1 involve carefully identifying all equally likely possible outcomes and then counting the outcomes that satisfy the required event.
Probability tells us how likely an event is to happen.
If an experiment has equally likely outcomes, then the probability of an event E is:
P(E) = Number of favourable outcomes / Total number of equally likely outcomes
For example, suppose a bag contains:
3 red balls
and:
5 black balls
Total balls:
3 + 5 = 8
If one ball is chosen at random, then:
P(Red) = 3/8
because 3 of the 8 equally likely balls are red.
An outcome is one possible result of an experiment.
For example, when a coin is tossed, the possible outcomes are:
H and T
When a die is thrown, the possible outcomes are:
1, 2, 3, 4, 5, 6
An event is a condition involving one or more outcomes.
For example, when a die is thrown, the event of getting an odd number contains the outcomes:
{1, 3, 5}
Outcomes are called equally likely when each outcome has the same chance of occurring.
For a fair die:
{1, 2, 3, 4, 5, 6}
are equally likely outcomes.
For a fair coin:
{H, T}
are equally likely outcomes.
The standard Class 10 probability formula based on counting outcomes should be used when the outcomes under consideration are equally likely.
An elementary event contains only one basic outcome.
For example, when a die is thrown:
The probabilities of all elementary events of an experiment add up to 1.
Therefore:
Sum of probabilities of all elementary events = 1
P(E) = Number of favourable outcomes / Total number of equally likely outcomes
0 ≤ P(E) ≤ 1
A probability can never be negative or greater than 1.
A sure event is certain to happen.
P(S) = 1
An impossible event cannot happen.
P(I) = 0
If E is an event and "not E" represents the event that E does not happen:
P(E) + P(not E) = 1
Therefore:
P(not E) = 1 - P(E)
This formula is especially useful in questions asking for:
Possible outcomes:
{H, T}
Total outcomes:
2
Therefore:
P(H) = 1/2
P(T) = 1/2
Possible outcomes:
{HH, HT, TH, TT}
Total outcomes:
4
Notice that:
HT and TH are different elementary outcomes.
Therefore:
P(Two heads) = 1/4
P(Two tails) = 1/4
P(One head and one tail) = 2/4 = 1/2
This distinction is important in Exercise 14.1 Question 25.
Possible outcomes:
{1, 2, 3, 4, 5, 6}
Total outcomes:
6
Prime numbers on a die:
{2, 3, 5}
Odd numbers:
{1, 3, 5}
Even numbers:
{2, 4, 6}
When two dice are thrown, each outcome can be represented as an ordered pair:
(First die, Second die)
Total outcomes:
6 × 6 = 36
For example:
(1, 2)
and:
(2, 1)
are different outcomes.
The sums 2, 3, 4, ... , 12 are not equally likely, because different sums can be obtained in different numbers of ways. Exercise 14.1 Question 22 directly tests this idea.
A standard deck contains 52 cards.
Students should remember the following facts:
| Card Information | Number |
| Total cards | 52 |
| Suits | 4 |
| Cards in each suit | 13 |
| Red cards | 26 |
| Black cards | 26 |
| Hearts | 13 |
| Diamonds | 13 |
| Clubs | 13 |
| Spades | 13 |
| Kings | 4 |
| Queens | 4 |
| Jacks | 4 |
| Face cards | 12 |
| Red face cards | 6 |
| Red kings | 2 |
The face cards are:
Jack, Queen and King
There are 3 face cards in each suit.
Therefore:
Total face cards = 3 × 4 = 12
The following section covers the current Exercise 14.1 question set through Question 25. The questions are paraphrased for clarity, while the mathematical methods and s follow the exercise structure.
We know:
P(E) + P(not E) = 1
1
An event that cannot happen is called an impossible event.
Its probability is:
P(Impossible event) = 0
0; impossible event
An event that is certain to happen is called a sure event.
Its probability is:
P(Sure event) = 1
1; sure event
The probabilities of all elementary events together cover all possible outcomes.
Therefore:
Sum = 1
1
Probability cannot be less than 0 or greater than 1.
Therefore:
0 ≤ P(E) ≤ 1
0 and 1
The chance of a car starting depends on factors such as its condition, fuel, battery, and mechanical state.
Therefore, the two outcomes cannot automatically be assumed to have equal chances.
Not equally likely
The chance depends on the player's ability, position, conditions, and other factors.
Therefore:
Not equally likely
For the theoretical situation in this exercise, the two outcomes are treated as equally likely.
Equally likely
For the theoretical probability model used in this question, the two possibilities are treated as equally likely.
Equally likely
A fair coin has two possible outcomes:
H and T
Both outcomes have the same probability:
P(H) = 1/2
P(T) = 1/2
Therefore, both teams receive an equal chance.
A fair coin toss is considered fair because heads and tails are equally likely outcomes.
Which of the following cannot represent a probability?
A probability must satisfy:
0 ≤ P(E) ≤ 1
Check each value:
2/3 ≈ 0.667
Valid.
-1.5
Negative, so invalid.
15% = 15/100 = 0.15
Valid.
0.7
Valid.
-1.5 cannot be the probability of an event.
If:
P(E) = 0.05
find:
P(not E)
Using:
P(not E) = 1 - P(E)
we get:
P(not E) = 1 - 0.05
P(not E) = 0.95
0.95
A bag contains only lemon-flavoured candies.
There are no orange candies.
Therefore:
P(Orange) = 0
0
Every candy is lemon flavoured.
Therefore:
P(Lemon) = 1
1
0.992
Find the probability that they have the same birthday.
The two events are complementary.
Therefore:
P(Same birthday) + P(Not same birthday) = 1
So:
P(Same birthday) = 1 - 0.992
P(Same birthday) = 0.008
0.008
A bag contains:
3 red balls
5 black balls
Total balls:
3 + 5 = 8
Favourable outcomes:
3
Total outcomes:
8
Therefore:
P(Red) = 3/8
3/8
Balls that are not red are black.
Number of black balls:
5
Therefore:
P(Not red) = 5/8
Alternatively:
P(Not red) = 1 - 3/8 = 5/8
5/8
A box contains:
5 red marbles
8 white marbles
4 green marbles
Total marbles:
5 + 8 + 4 = 17
P(Red) = 5/17
5/17
P(White) = 8/17
8/17
Green marbles:
4
Therefore:
P(Green) = 4/17
Using the complement:
P(Not green) = 1 - 4/17
= 13/17
13/17
A piggy bank contains:
100 coins of 50 paise
50 coins of ₹1
20 coins of ₹2
10 coins of ₹5
Total coins:
100 + 50 + 20 + 10 = 180
Favourable coins:
100
Therefore:
P(50 paise) = 100/180
= 5/9
5/9
Number of ₹5 coins:
10
Therefore:
P(₹5) = 10/180
= 1/18
Using complement:
P(Not ₹5) = 1 - 1/18
= 17/18
17/18
A tank contains:
5 male fish
8 female fish
Total fish:
5 + 8 = 13
Favourable outcomes:
5
Total outcomes:
13
Therefore:
P(Male fish) = 5/13
5/13
1, 2, 3, 4, 5, 6, 7, 8
All outcomes are equally likely.
Total outcomes:
8
Only one outcome is 8.
P(8) = 1/8
1/8
Odd numbers are:
1, 3, 5, 7
Favourable outcomes:
4
Therefore:
P(Odd) = 4/8
= 1/2
1/2
Numbers greater than 2:
3, 4, 5, 6, 7, 8
Favourable outcomes:
6
Therefore:
P(Number > 2) = 6/8
= 3/4
3/4
All eight numbers are less than 9.
Therefore:
P(Number < 9) = 8/8
= 1
1
Possible outcomes:
{1, 2, 3, 4, 5, 6}
Total outcomes:
6
Prime numbers on the die are:
2, 3, 5
Favourable outcomes:
3
Therefore:
P(Prime) = 3/6
= 1/2
1/2
Numbers strictly between 2 and 6 are:
3, 4, 5
Therefore:
P(Between 2 and 6) = 3/6
= 1/2
1/2
Odd outcomes are:
1, 3, 5
Therefore:
P(Odd) = 3/6
= 1/2
1/2
There are two red kings:
Therefore:
P(Red king) = 2/52
= 1/26
1/26
There are 12 face cards.
Therefore:
P(Face card) = 12/52
= 3/13
3/13
There are 6 red face cards:
3 in Hearts and 3 in Diamonds.
Therefore:
P(Red face card) = 6/52
= 3/26
3/26
There is only one Jack of Hearts.
Therefore:
P(Jack of Hearts) = 1/52
1/52
There are 13 spades.
Therefore:
P(Spade) = 13/52
= 1/4
1/4
There is only one Queen of Diamonds.
Therefore:
P(Queen of Diamonds) = 1/52
1/52
Five cards are used:
Total cards:
5
Number of queens:
1
Therefore:
P(Queen) = 1/5
1/5
Remaining cards:
4
There is one ace.
Therefore:
P(Ace) = 1/4
1/4
The only queen has already been removed.
Therefore:
P(Queen) = 0/4
= 0
0
132 good pens
12 defective pens
Total pens:
132 + 12 = 144
P(Good pen) = 132/144
Divide numerator and denominator by 12:
P(Good pen) = 11/12
11/12
A lot contains:
20 bulbs
of which:
4 are defective
Therefore:
16 are good
P(Defective) = 4/20
= 1/5
1/5
Originally:
16 good bulbs
One good bulb has been removed.
Good bulbs left:
15
Total bulbs left:
19
Therefore:
P(Second bulb is good) = 15/19
15/19
1 to 90
Total discs:
90
Two-digit numbers from 1 to 90 are:
10 to 90
Count:
90 - 10 + 1 = 81
Therefore:
P(Two-digit number) = 81/90
= 9/10
9/10
Perfect squares up to 90 are:
1, 4, 9, 16, 25, 36, 49, 64, 81
Number of favourable outcomes:
9
Therefore:
P(Perfect square) = 9/90
= 1/10
1/10
Numbers divisible by 5 from 1 to 90 are:
5, 10, 15, ... , 90
Number of such values:
90/5 = 18
Therefore:
P(Divisible by 5) = 18/90
= 1/5
1/5
Total faces:
6
A appears on 2 faces.
Therefore:
P(A) = 2/6
= 1/3
1/3
D appears on 1 face.
Therefore:
P(D) = 1/6
1/6
3 m × 2 m
Inside the rectangle is a circle of diameter:
1 m
Area of rectangle:
3 × 2 = 6 m²
Diameter of circle:
1 m
Radius:
1/2 m
Area of circle:
π × (1/2)²
= π/4 m²
Assuming the die is equally likely to land anywhere within the rectangular region:
Probability = Area of circle / Area of rectangle
Therefore:
P(Inside circle) = (π/4) / 6
= π/24
π/24
A lot contains:
144 ball pens
Defective pens:
20
Good pens:
144 - 20 = 124
Nuri buys the pen if it is good.
P(Good) = 124/144
Divide by 4:
= 31/36
31/36
She does not buy a defective pen.
Therefore:
P(Not buy) = 20/144
= 5/36
Alternatively:
P(Not buy) = 1 - 31/36
= 5/36
5/36
Total equally likely ordered outcomes:
6 × 6 = 36
| Sum | Number of Outcomes | Probability |
| 2 | 1 | 1/36 |
| 3 | 2 | 2/36 = 1/18 |
| 4 | 3 | 3/36 = 1/12 |
| 5 | 4 | 4/36 = 1/9 |
| 6 | 5 | 5/36 |
| 7 | 6 | 6/36 = 1/6 |
| 8 | 5 | 5/36 |
| 9 | 4 | 4/36 = 1/9 |
| 10 | 3 | 3/36 = 1/12 |
| 11 | 2 | 2/36 = 1/18 |
| 12 | 1 | 1/36 |
For example, a sum of 7 can be obtained in six ways:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Therefore:
P(Sum = 7) = 6/36 = 1/6
No.
Although there are 11 possible values for the sum, these values are not elementary outcomes and they do not occur equally often.
For example:
A sum of 2 occurs only as:
(1,1)
Therefore:
P(Sum = 2) = 1/36
A sum of 7 occurs in six ways.
Therefore:
P(Sum = 7) = 6/36 = 1/6
Since these probabilities are different, each sum cannot have probability 1/11.
No, the argument is incorrect because the sums are not equally likely.
A fair coin is tossed three times.
Possible outcomes:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
Total outcomes:
8
Hanif wins if all three results are the same.
Winning outcomes:
HHH, TTT
Number of winning outcomes:
2
Losing outcomes:
8 - 2 = 6
Therefore:
P(Lose) = 6/8
= 3/4
Alternatively:
P(Win) = 2/8 = 1/4
Therefore:
P(Lose) = 1 - 1/4
= 3/4
3/4
A fair die is thrown twice.
Total ordered outcomes:
6 × 6 = 36
For each throw, any value except 5 is allowed:
1, 2, 3, 4, 6
There are 5 possibilities for the first throw and 5 for the second.
Favourable outcomes:
5 × 5 = 25
Therefore:
P(No 5 either time) = 25/36
25/36
Use the complementary event:
P(At least one 5) = 1 - P(No 5)
Therefore:
P(At least one 5) = 1 - 25/36
= 11/36
11/36
No.
The elementary outcomes are:
HH, HT, TH, TT
Each elementary outcome has probability:
1/4
Now:
P(Two heads) = 1/4
P(Two tails) = 1/4
"One of each" includes:
HT and TH
Therefore:
P(One of each) = 2/4 = 1/2
The three events do not have equal probability.
The argument is incorrect.
The elementary outcomes are:
1, 2, 3, 4, 5, 6
Odd outcomes:
1, 3, 5
Number of odd outcomes:
3
Therefore:
P(Odd) = 3/6
= 1/2
The even outcomes also number 3.
So in this case, odd and even are equally likely.
The argument is correct, and P(Odd) = 1/2.
Students can use the following method for most Probability Class 10 Exercise 14.1 questions.
Ask what is happening.
Examples:
Count the complete set of possible outcomes.
For example:
One die:
6 outcomes
Two coins:
4 outcomes
Two dice:
36 outcomes
Determine exactly what the question asks.
Examples:
Count the outcomes that satisfy the event.
P(E) = Favourable outcomes / Total equally likely outcomes
For example:
12/52 = 3/13
The must satisfy:
0 ≤ P(E) ≤ 1
If the answer is negative or greater than 1, something has gone wrong.
For questions involving:
it may be quicker to use:
P(not E) = 1 - P(E)
When two coins are tossed, students may write:
and assume each has probability 1/3.
This is incorrect.
The elementary outcomes are:
HH, HT, TH, TT
"One of each" contains two outcomes.
Although two dice can produce sums from 2 to 12, these sums do not occur equally often.
For example:
Sum 2: 1 outcome
Sum 7: 6 outcomes
Therefore:
P(Sum 2) ≠ P(Sum 7)
Always remember:
0 ≤ P(E) ≤ 1
Values such as:
-0.3
or:
1.4
cannot be probabilities.
The denominator of:
P(E) = Favourable outcomes / Total outcomes
should contain all equally likely possible outcomes, not only the outcomes related to the event.
When the question asks for "not E", it may be easier to use:
P(not E) = 1 - P(E)
For example:
If:
P(Defective) = 1/5
then:
P(Not defective) = 1 - 1/5 = 4/5
Remember:
Total cards = 52
Cards in each suit = 13
Face cards = 12
Red cards = 26
Black cards = 26
If an item is removed and not returned, the total number of objects changes.
For example, in Question 17:
Original bulbs:
20
After one bulb is removed:
19
The second probability must therefore use 19 as the new denominator.
For two coin tosses:
HT ≠ TH
They are different ordered outcomes.
Write the probability in its simplest form whenever possible.
For example:
12/52 = 3/13
The NCERT Solutions for Class 10 Maths Chapter 14 Probability can help students understand the chapter systematically and improve accuracy while solving probability problems.
Each solution shows how to identify total outcomes and favourable outcomes before applying the formula.
Students can understand concepts such as equally likely outcomes, complementary events, sure events, and impossible events.
The solutions provide practice with common probability models involving coins, dice and playing cards.
Students can compare their own calculations with the detailed solutions to find counting or simplification mistakes.
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Probability measures how likely an event is to occur.
For equally likely outcomes:
P(E) = Number of favourable outcomes / Total number of equally likely outcomes
The main formula is:
P(E) = Number of favourable outcomes / Total number of equally likely outcomes
The current chapter is organised around Exercise 14.1. Current-syllabus NCERT solution resources list Exercise 14.1 as the chapter exercise.
Equally likely outcomes are outcomes that have the same chance of occurring.
For example, when a fair die is rolled:
1, 2, 3, 4, 5 and 6
are equally likely outcomes.
An elementary event contains one basic outcome.
For example, when a fair die is thrown, getting 4 is an elementary event.
The probability of an event always satisfies:
0 ≤ P(E) ≤ 1
An impossible event has probability:
0
Therefore:
P(Impossible event) = 0
A sure event has probability:
1
Therefore:
P(Sure event) = 1
An event E and the event "not E" are complementary.
Their probabilities satisfy:
P(E) + P(not E) = 1
The formula is:
P(not E) = 1 - P(E)
The outcomes are:
HH, HT, TH, TT
Therefore:
Total outcomes = 4
Favourable outcomes are:
HT and TH
Therefore:
P(One head and one tail) = 2/4
= 1/2
Each die has 6 outcomes.
Therefore:
Total outcomes = 6 × 6
= 36
No.
For example:
A sum of 2 occurs in only one way:
(1,1)
A sum of 7 occurs in six ways.
Therefore, different sums have different probabilities.
A standard deck contains:
52 cards
There are:
4 suits
with:
13 cards in each suit
Each suit contains:
Jack
Queen
King
Therefore:
3 face cards per suit
With 4 suits:
3 × 4 = 12
Therefore:
Total face cards = 12
Face cards:
12
Total cards:
52
Therefore:
P(Face card) = 12/52
= 3/13
There are two red kings in a standard deck.
Therefore:
P(Red king) = 2/52
= 1/26
Use these steps:
Identify all equally likely outcomes.
Count total outcomes.
Identify the required event.
Count favourable outcomes.
Apply the probability formula.
Simplify the answer.
Check that the final value lies between 0 and 1.
Complementary probability is useful when it is easier to calculate the opposite event.
Use:
P(not E) = 1 - P(E)
It is particularly useful in questions involving phrases such as:
not
none
at least once
A probability of 0 means an event cannot happen, while a probability of 1 means it is certain to happen.
All other events have probabilities somewhere between these two values.
Therefore:
0 ≤ P(E) ≤ 1
HT means:
Head on the first toss and Tail on the second toss
TH means:
Tail on the first toss and Head on the second toss
Because the order is different, they are separate elementary outcomes.
Probability becomes easier when students focus on correctly counting total and favourable outcomes. Most questions in Chapter 14 use the same central formula, but careful reading is important because the difficulty often comes from identifying outcomes correctly rather than from complicated calculations.
Students can use the NCERT Solutions Class 10 Maths Chapter 14 Probability PDF provided by Infinity Learn website for offline practice, Exercise 14.1 solutions, formulas, and quick revision.