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NCERT Solutions for Class 10 Maths Chapter 15 Probability

NCERT Solutions for Class 10 Maths Chapter 15 Probability

Edited By Ramraj Saini | Updated on Mar 15, 2025 08:25 PM IST | #CBSE Class 10th
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Probability is a fundamental concept in Maths class 10 that helps in predicting the likelihood of an event occurring. It is a branch of mathematics that deals with studying random events and finding the chances of their occurrence. In this chapter, we explore different types of probability and how to calculate it using various methods. This article aims to simplify the chapter for students of class 10 by providing clear explanations and solutions. We will discuss the concepts and will resolve your doubts, such as 'What is the definition of probability?', 'how we calculate the probability of a single event or multiple events', 'What are the different types of probability?'.

This Story also Contains
  1. NCERT Solutions for Class 10 Maths Chapter 15 Probability PDF Free Download
  2. NCERT Solution for probability class 10 - Important Formuale
  3. NCERT Solutions for Class 10 Maths Chapter 14 Probability (Questions and Exercise)
  4. Probability Class 10 NCERT Maths - Topics
  5. Key Features of NCERT Solutions for Class 10 Maths Chapter 14
  6. NCERT Exemplar Solutions - Subject Wise
  7. NCERT Books and NCERT Syllabus
  8. Class 10 Solutions - Chapter Wise
NCERT Solutions for Class 10 Maths Chapter 15 Probability
NCERT Solutions for Class 10 Maths Chapter 15 Probability

NCERT Solutions for Class 10 Maths Chapter 14 Probability, created by experts at Careers360, offers detailed study material for students preparing for the CBSE Class 10 board exam. The solutions cover all NCERT Solutions for Class 10 and are readily available for download. With step-by-step explanations for various types of questions, these solutions will help students grasp the concepts of probability class 10 and gain confidence in solving related problems. The Class 10 Math Chapter 14 NCERT solutions are a valuable resource for students aiming to master this topic. Students can click on the above link for solutions to other classes and subjects. It is highly recommended for students to refer to the NCERT Class 10 Maths books and thoroughly study all the topics without missing any.

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NCERT Solutions for Class 10 Maths Chapter 15 Probability PDF Free Download

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NCERT Solution for probability class 10 - Important Formuale

Empirical Probability - Empirical probability gives us the likelihood of events based on real-world experiments.

Empirical Probability = (Number of Trials with Expected Outcome)/(Total Number of Trials)

Theoretical Probability -Theoretical probability deals with event likelihood we get from experiments. The formulation is as follows.

Theoretical Probability = (Number of Favourable Outcomes to Event E)/(Total Number of Possible Outcomes in the Experiment)

P(E)=n(E)n(S)

P(E) : The probability of an event
n(E) : The number of favorable outcomes
n(S) : The total number of outcomes

Free download NCERT Solutions for Class 10 Maths Chapter 14 Probability PDF for CBSE Exam.


NCERT Solutions for Class 10 Maths Chapter 14 Probability (Questions and Exercise)

NCERT Probability Class 10 Solutions:
Exercise: 14.1

Q1. Complete the following statements:
Q1 (i).Probability of an event E + Probability of the event ‘not E’ = ________

Answer:

Probability of an event E + Probability of the event ‘not E’ = 1

EE=S
So, P(EE)=P(S)=1

Q1 (ii). The probability of an event that cannot happen is ______ . Such an event is called ______.

Answer:

The probability of an event that cannot happen is 0. Such an event is called an impossible event.

When there is no favorable outcome, i.e., the number of outcomes is zero.

Q1 (iii). The probability of an event that is certain to happen is_____. Such an event is called _______.

Answer:

The probability of an event that is certain to happen is 1. Such an event is called a sure/certain event

When the number of favorable outcomes is the same as the number of all possible outcomes, it is a sure event.

Q1 (iv). The sum of the probabilities of all the elementary events of an experiment is _______.

Answer:

The sum of the probabilities of all the elementary events of an experiment is 1.

Q1 (v). The probability of an event is greater than or equal to and less than or equal to ________ .

Answer:

The probability of an event is greater than or equal to 0 and less than or equal to 1.

Q2. Which of the following experiments have equally likely outcomes? Explain.
Q2 (i). A driver attempts to start a car. The car starts or does not start.

Answer:

It is not an equally likely event since it depends on various factors like there is no fuel, engine malfunctioning etc, that are not alike for both outcomes.

Q2 (ii). A player attempts to shoot a basketball. She/he shoots or misses the shot.

Answer:

It is not an equally likely event because it depends on the ability and amount of practice of the player. If he is a professional player, he will more likely have a successful shot. Whereas an amateur player will more likely miss the shot.

Q2 (iii). A trial is made to answer a true-false question. The answer is right or wrong.

Answer:

It is an equally likely event. The only options are true or false, and only one of them is correct.

Q2 (iv). A baby is born. It is a boy or a girl.

Answer:

It is an equally likely event. The only possibilities of gender are boy and girl. Hence, if not boy, then girl and vice versa.

Q3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Answer:

The tossing of the coin is considered to be a fair way of deciding because the only possible outcomes are heads and tails. Hence, they are equally likely events.

Q4. Which of the following cannot be the probability of an event?

(A) 23

(B) –1.5

(C) 15%

(D) 0.7

Answer:

We know that the probability of an event is either greater than or equal to 0 and always less than or equal to 1. Hence, the probability of an event can never be negative.

Therefore, (B) 1.5 cannot be the probability of an event.

Also, (A) : 23=0.67

(C): 15%=15100=0.15

(D): 0.7

Hence, (A), (C), and (D) all lie between 0 and 1.

Q5. If P(E) = 0.05, what is the probability of ‘not E’?

Answer:

Given, P(E)=0.05

We know,

P(not E)=1P(E)

P(not E)=10.05=0.95

Hence, the probability of 'not E' is 0.95

Q6 . A bag contains lemon-flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange-flavored candy?

(ii) a lemon-flavored candy?

Answer:

(i) According to the question, the bag contains only lemon-flavored candies. It does not contain any orange-flavored candy. Hence, every time, only lemon-flavored candy will come out. Therefore, P(an orange-flavored candy) = 0. i.e., the event of taking out an orange-flavored candy is impossible.
(ii) According to the question, the bag contains only lemon-flavored candies. So the event that Malini takes out a lemon-flavored candy is sure. Therefore, P(a lemon-flavored candy) = 1.

Q7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Answer:

Given,

The probability of two students not having the same birthday P(E)=0.992

Probability of two students having the same birthday = P(E)=1P(E)

=10.992=0.008

Hence, the probability that the 2 students have the same birthday is 0.008.

Q8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red?
(ii) not red?

Answer:

Total number of balls in the bag = 8
No. of red balls = 3
No. of black balls = 5

(i) Let E be the event of getting a red ball

n(E) = No. of red balls = 3
n(S) = No. of total balls = 8

Probability of the ball drawn to be red =
P=n(E)n(S)
=38

(ii) We know,
P(not E)=P(E)=1P(E)
where E and E are complementary events.

Probability of not getting the red ball
=1 - Probability of getting a red ball
=138=58

Q9. A box contains 5 red marbles, 8 white marbles, and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red?
(ii) white?
(iii) not green?

Answer:

Given,

Total number of balls in the bag = 5 + 8 + 4 = 17

(i) Let R be the event that the ball taken out is red

The number of possible outcomes = 17

The number of outcomes favorable to the event R = 5

P(R)=favourable outcomestotal outcomes=517

(ii) Let W be the event that the ball taken out is white

The number of possible outcomes = 17

The number of outcomes favorable to the event W = 8

P(W)=favourable outcomestotal outcomes=817

(iii) Let G be the event that the ball taken out is green

The number of possible outcomes = 17

The number of outcomes favorable to the event G = 4

P(G)=favourable outcomestotal outcomes=417

P(not G)=P(G)=1P(G)=1417=1317

The required probability of not getting a green ball is 1317.

Q10. A piggy bank contains a hundred 50p coins, fifty Rs 1 coins, twenty Rs 2 coins, and ten Rs 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down,
(i) What is the probability that the coin will be a 50 p coin?

(ii) Will not be a Rs 5 coin?

Answer:

Total number of coins in the piggy bank =100+50+20+10=180

(i) Let E be the event of getting a 50p coin.

Number of possible outcomes = 180

Number of outcomes favorable to event E = 100

P(E)=favourable outcomestotal outcomes=100180

=59

Therefore, the probability of getting a 50p coin is 59

(ii) Let F be the event of getting a Rs. 5 coin.

Number of possible outcomes = 180

Number of outcomes favorable to event F = 10

P(F)=favourable outcomestotal outcomes=10180

=118

P(not getting a Rs. 5 coin) =1P(F)=1118=1718

Therefore, the probability of not getting a Rs. 5 coin is 1718.

Q11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 14.4). What is the probability that the fish taken out is a male fish?

16360856852881742045265039

Answer:

Total number of fishes in the tank = 5 (male) + 8 (female) = 13

Let E be the event that the fish taken out is a male fish.

Number of possible outcomes = 13

Number of outcomes favorable to E = 5

P(E)=favourable outcomestotal outcomes=513

Therefore, the probability that the fish taken out is a male fish is 513.

Q12. A game of chance consists of spinning an arrow that comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig.), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?

(ii) an odd number?
(i) a number greater than 2?
(i) a number less than 9?

16360861630731742045265086

Answer:

Total possible outcomes = {1,2,3,4,5,6,7,8}

Number of possible outcomes = 8

(i) Let E be the event of getting 8.

P(E)=favourable outcomestotal outcomes=18

Therefore, the probability that it will point at 8 is 18.

(ii) Let E be the event of pointing at an odd number.

Total number of odd numbers = n({1,3,5,7}) = 4

P(E)=favourable outcomestotal outcomes=48

=12
Therefore, the probability of getting an odd number is 12.

(iii) Let E be the event of pointing at a number greater than 2

Number of favorable outcomes= n({3,4,5,6,7,8}) = 6

P(E)=favourable outcomestotal outcomes=68=34
Therefore, the probability of pointing at a number greater than 2 is 34.

(iv) Let E be the event of pointing at a number less than 9

Since all the numbers on the wheel are less than 9, this is a sure event.

Number of favorable outcomes = 8

P(E)=favourable outcomestotal outcomes=88=1
Therefore, the probability of pointing at a number less than 9 is $1.

Q13. A die is thrown once. Find the probability of

(I) getting a prime number.

(ii) a number lying between 2 and 6.

(iii) getting an odd number.

Answer:

Possible outcomes when a die is thrown = {1,2,3,4,5,6}

Number of possible outcomes once = 6

(i) Let E be the event of getting a prime number.

Prime numbers on the die are = {2,3,5}

Number of favorable outcomes = 3

P(E)=favourable outcomestotal outcomes=36

=12

Therefore, the probability of getting a prime number is 10012

(ii) Let F be the event of getting a number lying between 2 and 6

Numbers lying between 2 and 6 on the die are = {3,4,5}

Number of favorable outcomes = 3

P(F)=favourable outcomestotal outcomes=36

=12

Therefore, the probability of getting a number lying between 2 and 6 is 12

(iii) Let O be the event of getting an odd number.

Odd numbers on the die are = {1,3,5}

Number of favorable outcomes = 3

P(O)=favourable outcomestotal outcomes=36

=12

Therefore, the probability of getting an odd number is 12.

Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(I) a king of red color.

(ii) a face card.

(iii) a red-face card

(iv) the Jack of hearts

(v) a spade

(vi) the queen of diamonds

Answer:

The total number of cards in a well-shuffled deck = 52

Hence, the total possible outcomes = 52

(i) Let E be the event of getting a king of red color.

There are only red color kings: Hearts and diamonds

Hence, the number of favorable outcomes = 2

P(E)=favourable outcomestotal outcomes=252

=126

Therefore, the probability of getting a king of red color is 126.

(ii) Let E be the event of getting a face card.

Face cards: (J, Q, K) of each four suits

Hence, the number of favorable outcomes = 12

P(E)=favourable outcomestotal outcomes=1252

=313

Therefore, the probability of getting a face card is 313.

(iii) Let E be the event of getting a red face card.

Face cards: (J, Q, K) of hearts and diamonds

Hence, the number of favorable outcomes = 3×2=6

P(E)=favourable outcomestotal outcomes=652

=326

Therefore, the probability of getting a red face card is 326.

(iv) Let E be the event of getting the jack of hearts

Hence, the number of favorable outcomes = 1

P(E)=favourable outcomestotal outcomes=152

Therefore, the probability of getting the jack of hearts is 152.

(v) Let E be the event of getting a spade.

There are 13 cards in each suit.

Hence, the number of favourable outcomes = 13

P(E)=favourable outcomestotal outcomes=1352

=14

Therefore, the probability of getting a spade is 14.

(vi) Let E be the event of getting the queen of diamonds

Hence, the number of favorable outcomes = 1

P(E)=favourable outcomestotal outcomes=152

Therefore, the probability of getting the queen of diamonds is 152.

15. Five cards—the ten, jack, queen, king, and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random. What is the probability that

(i) the card is the queen?

(ii) the second card picked up is (a) an ace? (b) a queen?

Answer:

Total number of cards = 5

Hence, the total possible outcomes = 5

(i) There is only one queen.

Hence, favorable outcome = 1

P(getting a queen)=15

(ii-a) There is only one ace.

Hence, favorable outcome = 1

P(getting an ace)=14

Therefore, the probability of getting an ace is 0.25.

(ii-b) Since there is no queen left.

Hence, favorable outcome = 0

P(getting a queen)=04=0

Therefore, the probability of getting a queen is 0. Thus, it is an impossible event.

Q16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen is taken out is a good one.

Answer:

Total number of pens = 132 (good) + 12 (defective)

Hence, the total possible outcomes = 144

Number of good pens = number of favorable outcomes = 132

P(getting a good pen) =favourable outcometotal outcome=132144=1112

17. (i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now, one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer:

(i) Total number of bulbs = 20

Hence, the total possible outcomes = 20

Number of defective bulbs = 4

Hence, the number of favorable outcomes = 4

P(getting a defective bulb)=favourable outcomestotal outcomes=420

=15

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced.

Number of remaining bulbs =201=19
Number of remaining non-defective bulbs =161=15
The probability that this bulb is not defective =1519.

Q18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box. find the probability that it bears

(i) a two-digit number.

(ii) a perfect square number.

(iii) a number divisible by 5.

Answer:

(i) Total number of discs = 90

Number of discs having a two-digit number between 1 and 90 = 81

P(getting a two-digit number) =favourable outcomestotal outcomes=8190=910

(ii) Total number of discs = 90

Perfect square numbers from 1 to 90 are {1, 4, 9, 16, 25, 36, 49, 64, 81}

Therefore, the total number of discs having perfect squares = 9.
P(getting a perfect square number) =favourable outcomestotal outcomes=990=110

(iii) Total number of discs = 90

Numbers between 1 and 90 that are divisible by 5 are (5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90)

Therefore, the total number of discs having numbers that are divisible by 5 = 18.

P(getting a number divisible by 5) =favourable outcomestotal outcomes=1890=15

Q19. (i) A child has a die whose six faces show the letters as given below:

16360883256521742045265117

The die is thrown once. What is the probability of getting

(i) A?

(ii) D?

Answer:

The six faces of the die contain: {A, B, C, D, E, A}

Total number of letters = 6

(i) Since there are two A's,

Number of favorable outcomes = 2
P(getting A) =favourable outcomestotal outcomes=26

=13

Therefore, the probability of getting A is 13.

(ii) Since there is only one D,

Number of favorable outcomes = 1

P(getting D) =favourable outcomestotal outcomes=16

Therefore, the probability of getting D is 16.

Q20. Suppose you drop a die at random on the rectangular region shown in Fig. (15.6). What is the probability that it will land inside the circle with a diameter of 1m?

16360885692831742045265140

Answer:

Here, the Total outcome is the area of the rectangle, and the favorable outcome is the area of the circle.

Area of the rectangle = l×b=3×2=6 m2

Area of the circle = πr2=π(12)2=π4 m2

P(die will land inside the hole)=Area of circleTotal area

=π46=π24

Q21. A lot consists of 144 ball pens, of which 20 are defective and the others are good. Nuri will buy a pen if it is good but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?

(ii) She will not buy it?

Answer:

Total number of pens = 144

Total number of defective pens = 20

Number of good pens = 144 - 20 = 124

She will buy it if the pen is good.

(i) Therefore, the probability that she buys = probability that the pen is good =

P(getting a good pen)=number of good penstotal pens=124144

=3136

(ii) The probability that she will not buy = probability that the pen is defective

P(getting a defective pen)=no. of defective penstotal pens=20144

=536

Q22. Refer to Example 13. (i) Complete the following table:


Event: 'Sum on 2 dice'

23456789101112

1/36




5/36


1/36
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Answer:

Number of possible outcomes to get the sum as 2 = (1,1) = 2

Number of possible outcomes to get the sum as 3 = (2,1), (1, 2) = 2

Number of possible outcomes to get the sum as 4 = (2, 2), (1, 3), (3,1) = 3

Number of possible outcomes to get the sum as 5 = (3, 2), (2, 3), (4,1), (1, 4) = 4

Number of possible outcomes to get the sum as 6 = (5,1), (1, 5), (3, 3), (4, 2), (2, 4) = 5

Number of possible outcomes to get the sum as 7 = (4, 3), (3, 4), (6,1), (1, 6), (5, 2), (2, 5) = 6

Number of possible outcomes to get the sum as 8 = (4, 4), (6, 2), (2, 6), (5, 3), (3, 5) = 5

Number of possible outcomes to get the sum as 9 = (5, 4), (4, 5), (6, 3), (3, 6) = 4

Number of possible outcomes to get the sum as 10 = (5, 5), (6, 4), (4, 6) = 3

Number of possible outcomes to get the sum as 11 = (6, 5), (5, 6) = 11

Number of possible outcomes to get the sum as 12 = (6, 6) = 1

The table becomes:


The sum of two dice23456789101112
Probability1/361/181/121/95/361/65/361/91/121/181/36
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22. (ii) A student argues that ‘there are 11 possible outcomes: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. Therefore, each of them has a probability of 1/11. Do you agree with this argument? Justify your answer.

Answer:

A student argues that "there are 11 possible outcomes: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. Therefore, each of them has a probability of 1/11. We do not agree with this argument because there are a different number of possible outcomes for each sum. We can see that each sum has a different probability.

Q23. A game consists of tossing a one-rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer:

The possible outcomes when a coin is tossed 3 times (Same as 3 coins tossed at once)

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

Number of total possible outcomes = 8

For Hanif to win, there are only two favorable outcomes: {HHH, TTT}

Number of favorable outcomes = 2

P(Hanif will win) =favourable outcomestotal outcomes=28

=14

P(Hanif will lose) =114

=34

Therefore, the probability that Hanif will lose is 34.

Q24 (i) A die is thrown twice. What is the probability that 5 will not come up either time?

Answer:

When a die is thrown twice, the possible outcomes =

{(x,y):x,y{1,2,3,4,5,6}}

Total number of possible outcomes = 6×6=36

The outcomes when 5 comes up at least once =

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

Number of such favorable outcomes = 11

P(5 will come up atleast once)=1136

P(5 will not come up either time)=11136=2536

Therefore, the probability that 5 will not come either time is 2536.

24 (ii) A die is thrown twice. What is the probability that 5 will come up at least once?

Answer:

When a die is thrown twice, the possible outcomes =

{(x,y):x,y{1,2,3,4,5,6}}

Total number of possible outcomes = 6×6=36

The outcomes when 5 comes up at least once =

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

Number of such favorable outcomes = 11

P(5 comes up at least once)=1136

Therefore, the probability that 5 comes at least once is 1136.

Q25. (i) Which of the following arguments are correct, and which are not correct? Give reasons for your answer. (i) If two coins are tossed simultaneously, there are three possible outcomes—two heads, two tails, or one of each. Therefore, for each of these outcomes, the probability is 1/3

Answer:

The possible outcomes when two coins are tossed = {HH, HT, TH, TT}

Total number of possible outcomes = 4

P(getting two heads)=favourable outcomestotal outcomes=14

Hence, the given statement is not correct. This is because one of each can occur in two different ways. Hence the mentioned events are not equally likely.

Q25 (ii). Which of the following arguments are correct and which are not correct? Give reasons for your answer. (ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 12.

Answer:

The possible outcomes when a die is thrown= {1,2,3,4,5,6}

Total number of possible outcomes = 6

Number of odd numbers, {1,3,5} = 3

And the number of even numbers {2,4,6} = 3

Hence, both these events are equally likely

P(getting an odd)=favourable outcomestotal outcomes=36=12

Similarly, P(getting an odd)=favourable outcomestotal outcomes=36=12.

So, this argument is correct.


Probability Class 10 NCERT Maths - Topics

We have provided the solutions for the type of questions given below:

  • The basic formula for finding out the probability

  • Questions based on tossing a coin multiple times

  • Cards based questions

  • Question related to probability when a dice is rolled

  • Word problems related to real-life situations

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

NCERT solutions are also available for classes from 6 to 12.

Also Read,

Key Features of NCERT Solutions for Class 10 Maths Chapter 14

Comprehensive Coverage: These class 10 math chapter 14 solutions offer comprehensive coverage of Chapter 14, ensuring that all important topics and concepts are thoroughly covered.

Clarity and Accuracy: The probability solution class 10 is presented with clarity and accuracy, making it easier for students to understand and apply mathematical principles.

Step-by-Step Approach: Each solution for class 10th chapter probability follows a step-by-step approach, guiding students through the problem-solving process.

In-Depth Explanation: Complex concepts are explained in-depth, helping students grasp the underlying principles and logic in class 10th chapter probability solutions.

CBSE Guidelines: The solutions adhere to the CBSE curriculum and guidelines, making them a reliable resource for exam preparation.

NCERT solutions of class 10 - Subject Wise

How to Use NCERT Solutions for Class 10 Maths Chapter 14 Probability?

  • First of all, go through the NCERT Class 10 Maths syllabus of chapter 14 Probability.

  • Go through the theory given in the chapter from the NCERT Class 10 textbook.

  • Then, come to the practice exercises and list down those questions with which you are having trouble in class 10th probability.

  • Now, go through Chapter 14 Probability Class 10 ncert Solutions to know the solution to those questions.


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Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Solutions for Class 10 Maths Chapter 14?

Chapter 14 of Class 10 Maths is about Probability, and the topics we covered were the introduction to probability, theoretical probability, experiments and outcomes, complementary events, and some basic theorems of probability. Also, we have covered questions based on:
1. Cards
2. Dice
3. Coins
4. Numbers

2. How to solve probability questions in Class 10 Maths Chapter 14?

When solving probability questions, we have to remember some steps, such as identifying the experiment, defining the total outcomes, and using the probability formula.
Use this formula:
P(E)=n(E)n(S).
Where n(E) = number of favourable events.
and, n(S) = number of total outcomes.

3. Where can I find NCERT Solutions for Class 10 Maths Chapter 14 PDF?

You can find the NCERT Solutions for Class 10 Maths Chapter 14 PDF by clicking on the given download link where Career 360 provides all solutions to the given question of ex 14.1 Download.

4. What are the basic formulas used in Class 10 Probability?

In class 10, we only use the given basic formulae for given problems, which are:
1. P(A)= Number of favorable outcomes for event A  Total number of possible outcomes 
2. P( not A) = 1- P(A)

5. How to find the probability of an event in Class 10 Maths?

To find the probability of an event, first, we determine the total number of outcomes known as sample space [n(S)]. We calculate the number of outcomes of the event [n(E)]. At last, by applying the given formula, we get our answer.
P(E)=n(E)n(S)

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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