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# NCERT Solutions for Class 10 Maths Chapter 14 Probability

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Maths Chapter 14 Probability Class 10 - Free PDF Download

Class 10 Maths NCERT Solutions for Chapter 14 helps students understand the concept of set theory, like unions and intersections, which can be applied to represent events and analyse their probabilities. Probability provides a way of quantifying uncertainty and helps in making informed decisions based on the likelihood of various events. Understanding and applying probability concepts is essential for solving real-world problems effectively.

Table of Content
1. NCERT Solutions for Maths Chapter 14 Probability Class 10 - Free PDF Download
2. Glance on Maths Chapter 14 Class 10 - Probability
3. Access Exercise wise NCERT Solutions for Chapter 14 Maths Class 10
4. Exercises under NCERT Solutions for Class 10 Maths Chapter 14 Probability
5. Access NCERT Solutions for Class 10 Maths Chapter 14 – Probability
6. Overview of Deleted Syllabus for CBSE Class 10 Maths Probability
7. Other Study Material for CBSE Class 10 Maths Chapter 14
8. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

The free PDF of Class 10 Maths Chapter 14 Solutions PDF download is available on Vedantu, giving students a better understanding of the problems. It covers solutions to every exercise in this chapter and is updated according to the latest CBSE syllabus

## Glance on Maths Chapter 14 Class 10 - Probability

• Chapter 14 of Class 10 Maths deals with the foundation by defining probability as the measure of how likely an event is to occur. It introduces the concept of outcomes, events, and sample spaces.

• You'll learn how to identify favourable outcomes (those that satisfy a specific condition) and calculate the probability of an event by dividing the number of favourable outcomes by the total number of possible outcomes.

• Probability in class 10 explains Fundamental theorems like the addition and multiplication rules of probability are introduced. These rules help you calculate the probability of combined events.

• There is one exercise (25 fully solved questions) in class 10th maths chapter 14 Probability.

## Access Exercise wise NCERT Solutions for Chapter 14 Maths Class 10

 Current Syllabus Exercises of Class 10 Maths Chapter 14 NCERT Solutions of Class 10 Maths Probability Exercise 14.1
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## Exercises under NCERT Solutions for Class 10 Maths Chapter 14 Probability

• Exercise 14.1: This exercise involves solving word problems by paying close attention to key terms like event, sample space, and favourable outcome definitions. Break down complex problems into simpler events and apply the relevant rules.

## Access NCERT Solutions for Class 10 Maths Chapter 14 – Probability

### Exercise 14.1

1. Complete the following statements:

i. Probability of an event E + Probability of the event ‘not E’ = _____.

Ans:

If the probability of an event be $p$, then the probability of the ‘not event’ will be, $1-p$ . Thus, the sum will be, $p+1-p=1$.

ii. The probability of an event that cannot happen is _____. Such an event is called _____.

Ans:

The probability of an event that cannot happen is always $0$.

iii. The probability of an event that is certain to happen is _____. Such an event is called _____.

Ans:

The probability of an event that is certain to happen is $1$ . Such an event is called, sure event.

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

Ans:

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

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

equal to _____.

Ans:

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

2. Which of the following experiments have equally likely outcomes? Explain.

i. A driver attempts to start a car. The car starts or does not start.

Ans:

Equally likely outcomes defined as the outcome when each outcome is likely to occur as the others. So, the outcomes are not equally likely outcome.

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

Ans:

The outcomes are not equally likely outcome.

Ans:

The outcomes are equally likely outcome.

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

Ans:

The outcomes are not equally likely outcome.

3. 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?

Ans:

We already know the fact that a coin has only two sides, head and tail. So, when we toss a coin, it will either give us the result head or tail. There is no chance of the coin landing on his edge. And on the other hand, the chances of getting head and tail are also just the same. So, it can be concluded that the tossing of a coin is a fair way to decide the utcome, as it can not be biased and both teams will have the same chance of winning.

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

(A) $\frac{2}{3}$

Ans:

The probability of an event have to always be in the range of $[0,1]$ .

Now, let us the check the given values.

We can see, $\frac{2}{3}=0.67$ . This is in the given range. It can be a probability of an event.

(B) $-1.5$

Ans:

We can see, $-1.5$ , which is a negative number and not inside the given range. It can not be a probability of an event.

(C) $15%$

Ans:

We can see, $15%=\frac{15}{100}=0.15$ . This is in the given range of $[0,1]$ . It can be a probability of an event.

(D) $0.7$

Ans:

We can see, $0.7$ , which is in the given range. It can be a probability of an event.

5. If $P(E)=0.05$ , what is the probability of an event ‘not $E$’?

Ans:

The sum of the probabilities of all events in always $1$ .

Thus, if $P(E)=0.05$ , the probability of the event ‘not E’ is, $1-0.05=0.95$.

6. 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?

Ans:

There is no orange candy available in the bag, so, the probability of taking out an orange flavored candy is $0$.

(ii) a lemon flavored candy?

Ans:

All the candies in the bag are lemon flavored candies only. Thus, any candy Malini takes out will be a lemon flavored candy.

So, the probability of taking out a lemon flavored candy is $1$.

7. 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?

Ans:

It is provided to us that, probability of 2 students not having the same birthday is, $0.992$ .

So,$P(2\text{ students }\text{having }\text{the}\text{ same}\text{ birthday})+P(\text{2}\text{ students }\text{not }\text{having }\text{the}\text{ same}\text{ birthday})=1$

$\Rightarrow P(\text{2}\text{ students}\text{ having}\text{ the }\text{same}\text{ birthday})+0.992=1$

Simplifying further,

$P(2 \text{students having the same birthday})=0.008$.

8. 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 ?

Ans:

The bag is having $3$ red balls and $5$ black balls.

Now, the probability of getting a red ball will be, $\frac{number\,of\,red\,balls}{total\,number\,of\,balls}$ .

Putting the values, we get, $\frac{3}{3+5}=\frac{3}{8}$ .

(ii) not red?

Ans:

And, the probability of getting a red ball will be, $1-\frac{number\,of\,red\,balls}{total\,number\,of\,balls}$.

Again, putting the values, $1-\frac{3}{8}=\frac{5}{8}$.

9. A box contains $5$ red marbles, $8$ white marbles and 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 ?

Ans:

The box is containing, $5$ red marbles, $8$ white marbles and green marbles.

The probability of getting a red marble will be, $\frac{number\,of\,red\,marbles}{total\,number\,of\,marbles}$

Putting the values, $\frac{5}{5+8+4}=\frac{5}{17}$.

(ii) white ?

Ans:

Again, the probability of getting a white marble will be, $\frac{number\,of\,white\,marbles}{total\,number\,of\,marbles}$

Putting the values, $\frac{8}{5+8+4}=\frac{8}{17}$.

(iii) not green?

Ans:

And, the probability of getting a green marble will be, $\frac{number\,of\,green\,marbles}{total\,number\,of\,marbles}$.

Putting the values, $\frac{4}{5+8+4}=\frac{4}{17}$ .

So, the probability of the marble not being green will be, $1-\frac{4}{17}=\frac{13}{17}$.

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

Ans:

We are provided with the fact that, the piggy bank contains, hundred $50$ p coins, fifty $1$ rs coins, twenty $2$ rs coins and ten $5$ rs coins.

So, the total number of coins, $100+50+20+10=180$ .

Thus, the probability of drawing a $50$ p coin, $\frac{100}{180}=\frac{5}{9}$.

ii. will not be a  $5$  coin?

Ans:

Similarly, the probability of drawing a $5$ rs coin, $\frac{10}{180}=\frac{1}{18}$ .

Thus, the probability of not getting a $5$ rs coin, $1-\frac{1}{18}=\frac{17}{18}$ .

11.

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. 15.4). What is the probability that the fish taken out is a male fish?

Ans:

The total number of fishes in the tank, $5+8=13$ .

Thus, the probability of getting a male fish, $\frac{no\,of\,male\,fishes}{total\,no\,of\,fishes}=\frac{5}{13}$.

12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers $1,2,3,4,5,6,7,8$  (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at

(i)$8?$

Ans:

We can see, there are $8$  numbers on spinner then the total number of favorable outcome is $8$ .

Thus, the probability of getting the number $8$ is, $\frac{no\,of\,digit\,8\,on\,the\,spinner}{total\,number\,of\,digits}=\frac{1}{8}$.

(ii) $\text{an odd number?}$

Ans:

There are 4 odd digits, $1,3,5,7$ .

Thus, the probability of getting an odd number is,

$\frac{no\,of\,odd\,digits\,}{total\,number\,of\,digits}=\frac{4}{8}=\frac{1}{2}$ .

(iii) $\text{a number greater than 2?}$

Ans:

There are 6 numbers greater than 2, say $3,4,5,6,7,8$ .

Thus, the probability of getting a number greater than 2, $\frac{no\,of\,digits\,greater\,than\,2\,on\,the\,spinner}{total\,number\,of\,digits}=\frac{6}{8}=\frac{3}{4}$.

(iv) $\text{a number less than 9}$$?$

Ans:

As we can see, every number in the spinner is less than $9$ , thus, we get,

The probability of getting a number less than $9$ , will be, $1$.

13. A die is thrown once. Find the probability of getting (i) a prime number;

Ans:

There is $6$ results can be obtained from a dice.

There are 3 prime numbers, $2,3,5$ among those results.

Thus, the probability of getting a prime number,$=\frac{\text{no }\text{of }\text{prime }\text{numbers }\text{in }\text{a}\text{ dice}}{\text{total}\text{ numbers}\text{ on }\text{dice}}=\frac{3}{6}=\frac{1}{2}$

(ii) a number lying between $2$ and $6$

Ans:

There are 3 numbers between 2 and 6, 3,4,5.

Thus, the probability of getting a number between 2 and 6,

$=\frac{\text{no }\text{of }\text{numbers }\text{between }\text{2}\text{ and}\text{ 6 }\text{in}\text{ a}\text{ dice}}{\text{total}\text{ numbers }\text{on}\text{ dice}}=\frac{3}{6}=\frac{1}{2}$

(iii) an odd number.

Ans:

There are 3 odd numbers among the results, 1,3,5.

Thus, the probability of getting a odd number,

$=\frac{\text{no }\text{of}\text{ odd }\text{numbers}\text{ between}\text{ in }\text{a }\text{dice}}{\text{total}\text{ numbers }\text{on}\text{ dice}}=\frac{3}{6}=\frac{1}{2}$

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

(i) a king of red color

Ans:

We know there are 52 numbers in the deck.

There are 2 kings of red color in the deck.

Thus, the probability,

$=\frac{total\,number\,of\,red\,kings}{total\,number\,of\,cards}=\frac{2}{52}=\frac{1}{26}$

(ii) a face cards

Ans:

There are 12 face cards in the deck.

Thus, the probability,

$=\frac{total\,number\,of\,face\,cards}{total\,number\,of\,cards}=\frac{12}{52}=\frac{3}{13}$

(iii) a red face cards

Ans:

There are 6 red face cards in the deck.

Thus, the probability,

$=\frac{total\,number\,of\,red\,face\,cards}{total\,number\,of\,cards}=\frac{6}{52}=\frac{3}{26}$

(iv) the jack of hearts

Ans:

There are 1 jack of hearts card in the deck.

Thus, the probability,

$=\frac{total\,number\,of\,red\,face\,cards}{total\,number\,of\,cards}=\frac{1}{52}$

Ans:

There are 13 spade cards in the deck.

Thus, the probability,

$=\frac{total\,number\,of\,spade\,cards}{total\,number\,of\,cards}=\frac{13}{52}=\frac{1}{4}$

(vi) the queen of diamonds

Ans:

There are 1 queen of diamonds card in the deck.

Thus, the probability,

$=\frac{total\,number\,of\,red\,face\,cards}{total\,number\,of\,cards}=\frac{1}{52}$

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.

(i) What is the probability that the card is the queen?

Ans:

There are total 5 cards given in our deck.

Thus, the probability of getting a queen card among the 5 cards,

$=\frac{number\,of\,queen}{number\,of\,total\,cards}=\frac{1}{5}$

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Ans:

Now, the queen is put aside, so there will be 4 cards left.

a. Thus, the probability of getting a queen card among the 5 cards,

$=\frac{number\,of\,ace}{number\,of\,total\,cards}=\frac{1}{4}$

b. There are no queen cards left in the deck.

Thus, the probability of getting a queen card among the 4 cards,

$=\frac{number\,of\,queen}{number\,of\,total\,cards}=\frac{0}{4}=0$

16. 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 taken out is a good one.

Ans:

There are total (132+12)=144 number of pens in the lot.

And also there are 132 good pens in the given collection.

Thus, the probability of getting a good pen,

$=\frac{number\,of\,good\,pens}{number\,of\,total\,pens}=\frac{132}{144}=\frac{11}{12}$

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

Ans:

There are 4 defective bulbs among 20 bulbs.

Thus, the probability of getting a defective bulb,

$=\frac{number\,of\,defective\,bulb}{number\,of\,total\,bulb}=\frac{4}{20}=\frac{1}{5}$

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 ?

Ans:

After the first draw, there are 19 bulbs left in the lot. Again, as the bulb was a non-defective bulb, the total non-defective bulbs are 15.

Therefore, the probability of not getting a defective bulb this time,

$=\frac{15}{19}$ .

18. 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.

Ans:

There are 81 two digit numbers between 1 to 90.

Thus, the probability of getting a two digit number in the draw,

$=\frac{the\,total\,number\,of\,two\,digit\,numbers}{total\,numbers}=\frac{81}{90}=\frac{9}{10}$

(ii) a perfect square number

Ans:

The number of perfect square numbers between 1 to 90.

Thus, the probability of getting a perfect number in the draw,

$=\frac{the\,total\,number\,of\,perfect\,number}{total\,numbers}=\frac{9}{90}=\frac{1}{10}$

(iii) a number divisible by $5$.

Ans:

The number of numbers divisible by 5, 18.

Thus, the probability of getting a number divisible by 5 in the draw,

$=\frac{the\,total\,number\,of\,number\,divisible\,by\,5}{total\,numbers}=\frac{18}{90}=\frac{1}{5}$

19. A child has a die whose six faces show the letters as given below: A, A, B, C, D, E. The die is thrown once. What is the probability of getting

(i) A?

Ans:

There are two A’s in the six faces, so, the probability of getting an A,

$=\frac{total\,number\,of\,A's}{total\,number\,of\,sides}=\frac{2}{6}=\frac{1}{3}$

(ii)  D?

Ans:

There are one D in the six faces, so, the probability of getting an D,

$=\dfrac{total\,number\,of\,D's}{total\,number\,of\,sides}=\dfrac{1}{6}$

20. 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 diameter 1m?

Ans:

It is given that it is a rectangle with sides 3 m and 2 m.

Thus, the area of the rectangle,

$=$ length $\times$ breadth

$=3 \times 2=6 \mathrm{~m}^{2}$

The radius of the circle, half of diameter $=\frac{1}{2}\,m$ .

The area of the circle, $=\pi .{{\left( \frac{1}{2} \right)}^{2}}=\frac{\pi }{4}\,{{m}^{2}}$

Thus, the probability of the die landing inside the circle is,

$=\frac{area\,of\,the\,circle}{area\,of\,the\,rectangle}=\frac{\frac{\pi }{4}}{6}=\frac{\pi }{24}$

21. 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 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 ?

Ans:

There are total 144 ball pens in the lot and 20 of them are defective.

Thus, the total number of non-defective pens, $(144-20)=124$ .

Nuri will not buy the pen if it is defective, thus,

The probability of getting a good pen is,

$=\frac{total\,no\,of\,good\,pens}{total\,no\,of\,pens}=\frac{124}{144}=\frac{31}{36}$

(ii) She will not buy it ?

Ans:

Now, the probability of Nuri not buying the pen is,

$=1-\frac{31}{36}=\frac{36-31}{36}=\frac{5}{36}$

22. Refer to example 13: (i) Complete the following table:

 Event: ‘Sum on 2 dices’ 2 3 4 5 6 7 8 9 10 11 12 Probability $\frac{1}{36}$ $\frac{5}{36}$ $\frac{1}{36}$

Ans:

If there are two dices thrown simultaneously, then we can get the following results,

$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Thus, the total number of results is 36.

Probability of getting a sum of 2, $=\dfrac{1}{36}$

Probability of getting a sum of 3, $=\dfrac{2}{36}=\dfrac{1}{18}$

Probability of getting a sum of 4, $=\dfrac{3}{36}=\dfrac{1}{12}$

Probability of getting a sum of 5, $=\dfrac{4}{36}=\dfrac{1}{9}$

Probability of getting a sum of 6, $=\dfrac{5}{36}$

Probability of getting a sum of 7, $=\dfrac{6}{36}=\dfrac{1}{6}$

Probability of getting a sum of 8, $=\dfrac{5}{36}$

Probability of getting a sum of 9, $=\dfrac{4}{36}=\dfrac{1}{9}$

Probability of getting a sum of 10, $=\dfrac{3}{36}=\dfrac{1}{12}$

Probability of getting a sum of 11, $=\dfrac{2}{36}=\dfrac{1}{18}$

Probability of getting a sum of 12, $=\dfrac{1}{36}$

Thus, we get the values of our table.

(ii) A student argues that there are $11$ possible outcomes$2,3,4,5,6,7,8,9,10,11\text{ and 12}$ .Therefore, each of them has a probability $\frac{1}{11}$ . Do you agree with this argument?

Ans:

As we can see different values all over the table, we can conclude that, the given statement is wrong. The probability of each of them can never be $\dfrac{1}{11}$.

23. 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.

Ans:

Hanif will win if he gets 3 heads and 3 tails consecutively.

The probability of Hanif losing the game, = The probability of not getting 3 heads and 3 tails.

The possible outcomes of the tosses, $(HHH,HHT,HTH,HTT,THH,THT,TTH,TTT)$

The total number of outcomes is, 8.

Thus, the probability of not getting 3 heads and 3 tails,

$=1-\frac{2}{8}=\dfrac{6}{8}=\dfrac{3}{4}$.

24. A die is thrown twice. What is the probability that

(i) $5$  will not come up either time?

(Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment)

Ans:

We can see that two dices are thrown altogether, thus the total number of outcomes =36.

Now, the total cases where atleast 5 occurs is, 11, i.e,$(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6)$

So, the probability of not getting 5 either time is,$=1-\frac{11}{36}=\frac{25}{36}$.

(ii) $5$  will come up at least once?

Ans:

And, probability of getting 5 atleast once is,$\frac{11}{36}$.

25. 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 $\frac{1}{3}$.

Ans:

In this problem, two coins are tossed simultaneously, thus we get 4 outcomes, i.e, $(HH,HT,TH,TT)$ .

So, the probability of getting both heads and tails, $=\frac{2}{4}=\frac{1}{2}$ .

Thus, the statement is wrong.

(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 $\frac{1}{2}$.

Ans:

By throwing a die, we get 6 possible outcomes.

The odd numbers are $1,3,5$ .

Thus the probability of getting a odd number, $=\frac{3}{6}=\frac{1}{2}$ .

So, the statement is true.

## Overview of Deleted Syllabus for CBSE Class 10 Maths Probability

 Chapter Dropped Topics Probability 15.1 Introduction Exercise 15.2 (Optional)

## Conclusion

Vedantu's NCERT Solutions for Class 10 Maths Chapter 14 - Probability offers an exceptional resource for students seeking to grasp the complexities of probability theory.

You will be able to solve problems involving coin tosses, dice rolls, card games, and other scenarios where chance plays a role.

With comprehensive and well-structured explanations, the platform empowers learners to tackle real-world challenges with confidence.

## Other Study Material for CBSE Class 10 Maths Chapter 14

 S. No Important Links for Chapter 14 Probability 1 Class 10 Probability Important Questions 2 Class 10 Probability Revision Notes 3 Class 10 Probability Important Formulas 4 Class 10 Probability RD Sharma Solutions

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 14 Probability

1. What is the curriculum of CBSE Board Class 10 Mathematics?

There are 15 units in Class 10 Mathematics. These include Arithmetic Progressions, Pair of Linear Equations in Two Variables, Real numbers, Polynomials, Quadratic Equations, Triangles, Some Applications of Trigonometry, Introduction to Trigonometry, Constructions, Circles, Coordinate Geometry, Surface Areas and Volumes, Areas Related to Circles, Probability and Statistics.

2. What are the applications of probability in day to day life?

Probability is a branch of mathematics that deals with the likelihood of events happening. It is used in a wide variety of applications in our daily lives, including:

• Weather forecasting: Weather forecasters use probability to estimate the likelihood of rain, snow, or other types of precipitation.

• Sports betting: Sports bettors use probability to assess the likelihood of a particular team or player winning a game.

• Insurance: Insurance companies use probability to calculate the risk of an event happening and to determine the premiums that they charge their customers.

• Medical diagnosis: Doctors use probability to assess the likelihood of a patient having a particular condition.

• Stock market investing: Stock market investors use probability to assess the risk of an investment and to determine the potential returns.

• Marketing: Marketers use probability to target their advertising campaigns to the most likely customers.

• Decision making: In many situations, we need to make decisions based on uncertain information. Probability can help us to make more informed decisions by providing us with a framework for assessing the likelihood of different outcomes.

Here are some other examples of how probability is used in our daily lives:

• When we decide what to wear, we are essentially using probability to assess the likelihood of different weather conditions.

• When we choose a route to drive to work, we are using probability to assess the likelihood of traffic congestion.

• When we decide what to eat, we are using probability to assess the likelihood of different foods being available at the grocery store.

• When we decide what to watch on TV, we are using probability to assess the likelihood of different shows being interesting to us.

3. What are the subtopics of Ch 14 Maths Class 10?

Subtopics of Chapter 14 of the Class 10 Maths NCERT textbook:

• Introduction

• Experiment and Outcome

• Sample Space

• Events

• Elementary Events

• Favourable Outcomes

• Probability of an Event

• Theoretical Probability

• Experimental Probability

• Complement of an Event

• Multiplication Theorem of Probability

This chapter introduces the concept of probability, which is the likelihood of an event occurring. It discusses the different types of events, such as elementary events, favourable outcomes, and sure events. The chapter also covers the two methods of calculating probability: theoretical probability and experimental probability. The addition and multiplication theorems of probability are also discussed in this chapter.

4. What can we learn in the chapter Probability?

In the chapter Probability, we can learn about the following:

• The definition of probability and how it is measured.

• The different types of events, such as elementary events, favourable outcomes, and sure events.

• The two methods of calculating probability: theoretical probability and experimental probability.

• The addition and multiplication theorems of probability.

• How to use probability to make predictions about the outcomes of random events.

Here are some specific things we can learn:

• The probability of an event can be anywhere from 0 to 1. A probability of 0 means that the event is impossible, while a probability of 1 means that the event is certain.

• The probability of an event can be calculated by dividing the number of favourable outcomes by the total number of possible outcomes.

• If we repeat an experiment many times, the experimental probability of an event will approach the theoretical probability of the event.

• The addition theorem of probability can be used to calculate the probability of two events occurring when the events are mutually exclusive.

• The multiplication theorem of probability can be used to calculate the probability of two events occurring when the events are not mutually exclusive.

5. State the basic law behind probability?

The law of probability informs us how likely particular occurrences are to occur. According to the rule of large numbers, the more trials you have in an experiment, the closer you come to a precise probability. The multiplication rule is used to calculate the likelihood of two occurrences occurring at the same time.

6. State an event where the probability is ½?

On tossing a random fair coin, there are two possible outcomes i.e. head and tail.

When a coin is tossed, the probability of getting ahead is ½ and the probability of getting a tail is also ½. This is based on the probability of the specific event occurring.

7. What is the chapter probability about?

Probability is a field of mathematics that deals with numerical descriptions of what will happen in the future, or whether something is true or not. This is a high-scoring yet challenging Chapter in Class 10 Maths. As a result, you must be conversant with the tips and methods required to swiftly answer numerical issues. In this regard, Vedantu offers exact NCERT Solutions for Class 10 Maths Probability, which include all types of sums that may be encountered in examinations.

8. Why choose Vedantu for Chapter 14 Class 10 Maths?

In a highly competitive world, students are thriving for the best educational services to score as much as they can, Vedantu offers the best solutions. NCERT Solutions for Class 10 Maths Probability is one of the best study guides for students, to help them achieve their desired scores. These solution PDFs are available at free of cost on the Vedantu app and the Vedantu website.

9. What is the main focus of Chapter 14 - Probability Class 10 Maths?

The main focus of Chapter 14 - Probability Class 10 is to introduce students to the basic concepts of probability, including the theoretical approach, experiments, outcomes, events, and the probability of an event occurring.

10. How does NCERT Solutions for Class 10 Probability Chapter 14 help in understanding probability?

The NCERT Solutions for class 10 probability provides detailed explanations and step-by-step solutions to textbook problems, helping students understand the fundamental principles of probability and how to apply them to solve various problems.

11. What key concepts are covered in the NCERT Solutions for Probability Class 10 solutions Chapter 14?

Key concepts covered in class 10 probability include:

• Probability of an event

• Complementary events

• Probability of simple and compound events

• Use of probability in real-life situations

12. Are the probability class 10 solutions aligned with the latest syllabus?

Yes, the NCERT probability class 10 solutions are aligned with the latest CBSE syllabus for the academic year 2024-25, ensuring that all topics and concepts are covered as per the curriculum.

13. How do probability class 10 solutions assist in exam preparation?

The probability class 10 solutions include a variety of problems and their detailed solutions, which help students practice extensively. This thorough practice enhances problem-solving skills and prepares students effectively for exams.

14. Can these solutions help in clearing doubts about class 10 probability concepts?

Absolutely. In probability class 10 NCERT solutions, detailed and clear explanations provided in the solutions help in resolving any doubts or misconceptions students might have about probability concepts.

15. What types of questions are included in the probability class 10 NCERT solutions?

The probability class 10 NCERT solutions include a wide range of questions such as multiple-choice questions, short answer questions, and long answer questions. These questions cover theoretical concepts, practical problems, and real-life applications of probability.

16. How can students benefit from practising probability class 10 NCERT solutions regularly?

Regular practice of chapter 14 class 10 maths helps students build a strong foundation in probability, enhances their problem-solving abilities, and boosts their confidence in tackling various types of questions in exams.

17. Are there any real-life examples provided in the solutions to explain chapter 14 class 10 maths concepts?

Yes, the chapter 14 class 10 maths solutions include real-life examples to illustrate how probability is used in everyday situations, making the concepts more relatable and easier to understand.

18. Where can students access the class 10 maths ch 14 for Class 10 Maths Chapter 14 - Probability?

Students can access the class 10 maths ch 14 solutions on educational platforms like Vedantu, where they are available for free download in PDF format, providing easy and convenient access for study and revision.