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NCERT Solutions for Class 10 Maths Chapter 14 Probability

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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.

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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.

  • This article contains chapter notes, important questions, exemplar solutions, and exercise links for Chapter 14 - Probability, which you can download as PDFs.

  • 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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NCERT Solutions for Class 10 Maths Chapter 14 Probability
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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.

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

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?


Fig. 15.4 -The shopkeeper takes out one 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)$8?$


Fig. 15.5 spinning an arrow


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,

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

(v) a spade

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


A die of A, A, B, C, D, E


(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?


Fig. 15.6 drop a die at random on the rectangular region


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?

Justify your answer.

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

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.



Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

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FAQs on NCERT Solutions for Class 10 Maths Chapter 14 Probability

1. What is probability according to the Class 10 Maths NCERT Solutions for Chapter 14?

Probability is the measure of how likely an event is to occur, expressed as a number between 0 and 1. According to the NCERT Class 10 Maths syllabus, it is defined as the ratio of the number of favourable outcomes to the total number of possible outcomes, for equally likely events in an experiment.

2. How do you calculate the probability of an event in NCERT Class 10 Chapter 14?

The probability of an event (E) is calculated as:
Probability of E = (Number of favourable outcomes) / (Total number of possible outcomes)
This method is used for experiments where all outcomes are equally likely, as specified in the NCERT Solutions for Class 10 Maths Chapter 14.

3. What are elementary events and how are they addressed in the Class 10 Probability chapter?

Elementary events are outcomes that cannot be further broken down and have only one possible outcome. In Class 10 Probability, each single outcome of an experiment is an elementary event. For example, getting a head when tossing a coin is an elementary event.

4. What is a sure event and what is its probability as per the NCERT Solutions?

A sure event is one that is certain to happen. According to the Class 10 Maths Probability chapter, the probability of a sure event is always 1.

5. Can the probability of an event be negative or greater than one?

No. As explained in NCERT Class 10 Solutions for Probability, the probability of any event cannot be negative or greater than one. It must always satisfy 0 ≤ P(E) ≤ 1.

6. What are complementary events and how are they represented?

Complementary events refer to two outcomes that together cover all possible results of an experiment (e.g., E and not E). Their probabilities add up to 1: P(E) + P(not E) = 1 (as covered in Class 10 Chapter 14 Maths).

7. How does the addition theorem of probability apply in Class 10 NCERT Solutions?

The addition theorem states that the probability of either of two mutually exclusive events occurring is the sum of their individual probabilities. That is, P(A or B) = P(A) + P(B), provided A and B cannot happen together. This is a key focus in Chapter 14 Probability for Class 10.

8. What types of real-world problems does Chapter 14 Probability help solve?

The chapter helps solve problems involving:

  • Coin tosses
  • Dice rolls
  • Card draws
  • Random selection of objects (like marbles, bulbs, or pens)
  • Events in games of chance
It builds skills to model and quantify everyday uncertain situations using probability.

9. Why is tossing a coin considered a fair way to decide between two teams?

Tossing a fair coin has two equally likely outcomes (heads and tails), making it an unbiased method for decision-making. Both participants have the same chance of winning, aligning with the concept of equally likely outcomes covered in Class 10 NCERT Probability solutions.

10. What common misconceptions should students avoid in solving NCERT Probability problems?

  • Assuming outcomes are equally likely without verification. Carefully examine if each result truly has the same chance.
  • Forgetting the range of probability (0 to 1). Probabilities outside this range are not valid.
  • Confusing complementary events with mutually exclusive events. Complementary events cover all outcomes; mutually exclusive events do not overlap.

11. What is the significance of experimental probability in Class 10 Probability?

Experimental probability is based on observed outcomes from actual experiments. It is calculated as:
P(E) = (Number of times event E occurs) / (Total number of trials)
This approach is useful when theoretical calculation is difficult or not possible.

12. How is the concept of sample space used in Chapter 14 Probability?

The sample space is the set of all possible outcomes of an experiment. Accurate listing of the sample space is essential for calculating probabilities correctly, as emphasized in NCERT Class 10 Maths Probability exercises.

13. How does practicing NCERT Solutions for Probability benefit Class 10 students in exams?

Regular practice with NCERT Solutions for Probability Class 10 helps students:

  • Understand concepts deeply
  • Master stepwise solution techniques
  • Recognize and correct common mistakes
  • Boost confidence in applying probability to various scenarios
  • Score better in board exams as per CBSE guidelines

14. How can you differentiate between theoretical and experimental probability in Class 10?

  • Theoretical probability is calculated based on reasoning and the assumption of equally likely outcomes, without actual experiments.
  • Experimental probability is determined by performing the experiment and recording outcomes.
Both methods are discussed in NCERT Solutions for Class 10 Maths Chapter 14.

15. What is the role of probability in day-to-day decision making as taught in NCERT Class 10 Maths?

Probability provides a framework for making informed decisions when outcomes are uncertain, such as in weather forecasts, choosing lottery numbers, or assessing risks in real life. Class 10 Chapter 14 demonstrates these applications to show practical value beyond theory.