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• NCERT Solutions for Class 10 Maths Chapter 15 - Probability
• NCERT Solutions

NCERT Class 10 Maths Chapter 15: Complete Resource for Probability

The branch of Mathematics involving the numerical descriptions of what will be the outcome of an event, or whether it is true or not is termed as Probability. This is a scoring yet tricky Chapter of Class 10 Maths. Hence, you need to be familiar with the tips and tricks necessary to solve the numerical problems quickly. In this regard, Vedantu provides precise  NCERT Solutions for Class 10 Maths probability that contains different kinds of sums you can expect in exams.

Chapter 15 Maths Class 10  NCERT Solutions guide is curated following the latest rules of the CBSE board. Sums from every topic are precisely solved to clear your concepts of the same. Additionally, it is available in a PDF format, and you can study the same from Vedantu’s website. Alternately, you can even download it that too at free of cost. You can also download NCERT Solutions for Class 10 Science and make use of it in your preparation.

Important Topics under NCERT Solutions for Class 10 Maths Chapter 15 Probability

Chapter 15 of the class 10 Maths syllabus covers the concept of Probability. It is one of the most significant chapters covered in class 10 Maths and includes 4 important sections that practically cover all solutions and ideas that are related to Probability. The following table includes these 4 topics under the chapter and it is highly recommended that students pay close attention to these sections to gather and retain all the information required to master this chapter on Probability.

Importance of Probability

The probability theory is an important concept in maths that is used to make predictions and estimates. Estimates and predictions are very much used in research and gathering information and data from a source. Therefore, it is highly in use in statistics and we recommend that students go through the concepts of probability and practise the solved questions thoroughly to be able to score well in their exams.

Related Chapters

Exercises under NCERT Solutions for Class 10 Maths Chapter 15 – Probability

Exercise 15.1 : This exercise consists of questions based on the theoretical probability of an event, such as finding the probability of an event using the classical definition of probability, understanding the addition rule of probability and its properties, calculating the probability of complementary events, and finding the probability of the intersection and union of events.

Exercise 15.2 : This exercise involves questions based on the experimental probability of an event, such as finding the probability of an event using experimental data, understanding the difference between theoretical and experimental probability, and calculating the mean and variance of a given set of data.

Overall, the exercises in NCERT Solutions for Class 10 Maths Chapter 15 "Probability" are designed to help students develop their understanding of the theoretical and experimental probability of an event, and to enhance their problem-solving skills related to probability. The solutions to each exercise provide step-by-step instructions and explanations to help students understand the concepts better.

Access NCERT Solutions for Class 10 Maths Chapter 15 – Probability

1. Complete the following statements:

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

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 _____.

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 _____.

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 _____.

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 _____.

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.

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. 

The outcomes are not equally likely outcome.

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

The outcomes are equally likely outcome.

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

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?

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) $\frac{2}{3}$

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.

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.

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) $0.7$ 

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 $E$’? 

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?

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? 

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?

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 ?

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? 

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 ?

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 ? 

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? 

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 ?

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?

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, $1-\frac{1}{18}=\frac{17}{18}$ .

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? 

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)$8?$ 

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) $\text{an odd number?}$

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) $\text{a number greater than 2?}$ 

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}$$?$ 

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; 

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 $6$ 

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. 

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

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 

There are 12 face cards in the deck.

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

(iii) a red face cards 

There are 6 red face cards in the deck.

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

(iv) the jack of hearts

There are 1 jack of hearts card in the deck.

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

(v) a spade

There are 13 spade cards in the deck.

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

(vi) the queen of diamonds 

There are 1 queen of diamonds card in the deck.

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?

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? 

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.

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?

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 ?

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. 

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 

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 $5$. 

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? 

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? 

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

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

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?

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 ? 

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 ? 

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:

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, $=\frac{1}{36}$ 

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

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

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

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

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

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

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

Probability of getting a sum of 10, \[=\frac{3}{36}=\frac{1}{12}\] 

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

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

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.

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 $\frac{1}{11}$. 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.

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, 

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

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)

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?

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 $\frac{1}{3}$. 

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 $\frac{1}{2}$. 

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.

Exercise 15.2

1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day?

Two customers are visiting the shop from Tuesday to Saturday (5 days).

So, the total number of ways they can visit the shop will be, $=5\times 5=25$ .

Again, the number of ways they can shop the same day is, $5$ .

So, the probability of both reaching the shop the same day,

$=\frac{shop\,on\,same\,day}{total\,number\,of\,ways}=\frac{5}{25}=\frac{1}{5}$.  

(ii) consecutive days? 

And, they reaching the shop in consecutive days in 8 number of ways.

So, the probability of both reaching the shop on consecutive days,

$=\frac{shop\,on\,consecutive\,days}{total\,number\,of\,ways}=\frac{8}{25}$ .

(iii) different days? 

We also have, 

$P(both\,of\,them\,reaching\,on\,same\,days)+P(both\,of\,them\,reaching\,not\,on\,same\,days)=1$ 

Putting the values,

$\frac{1}{5}+P($ both of them reaching not on same days $)=1$

$\Rightarrow P($ both of them reaching not on same days $)=\frac{4}{5}$

So, the probability of not reaching on the same day$=\frac{4}{5}$.  

2. A die is numbered in such a way that its faces show the numbers $1,2,2,3,3,6$ . It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws: 

The total number of ways will be, $=6\times 6=36$.

i. The probability of the total scores as even numbers,

$=\frac{the\,ways\,of\,getting\,even\,sum}{total\,no\,of\,ways}=\frac{18}{36}=\frac{1}{2}$ 

ii. The probability of getting the total score 6,

$=\frac{the\,ways\,of\,getting\,even\,sum}{total\,no\,of\,ways}=\frac{4}{36}=\frac{1}{9}$

iii. The probability of the total sum as 6 or less than 6,

$=\frac{the\,ways\,of\,getting\,even\,sum}{total\,no\,of\,ways}=\frac{15}{36}=\frac{5}{12}$

3. A bag contains $5$  red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag. 

Let us consider, the number of blue balls be,$x$ blue balls.

The total number of balls, $5+x$ balls.

Now, it is also said, the probability of drawing a blue ball is double that of a red ball.

2 (probability of getting a blue ball)=probability of getting a redball

$\Rightarrow 2\left(\frac{\text { number of blue balls }}{\text { totalballs }}\right)=\frac{\text { number of redballs }}{\text { total balls }}$

$2\left(\frac{5}{5+x}\right)=\frac{x}{5+x}$

$\Rightarrow x=10$ 

Thus, the number of blue balls is, $x=10$.

4. A box contains $12$  balls out of which $x$  are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? 

There are $x$ balls among 12 balls which are black. Thus, the number of other balls, $12-x$ . 

The probability of getting a black ball, $=\frac{no\,of\,black\,balls}{total\,no\,of\,balls}=\frac{x}{12}$. 

If $6$  more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find $x$.

Now, 6 more black balls are added in the box. So, the total number of balls, $12-x+6=18-x$ .

And total no of black balls, $12+6=18$ .

Now, the probability of getting a black ball, $=\frac{no\,of\,black\,balls}{total\,no\,of\,balls}=\frac{x+6}{18}$.

So, we get,

$2\left(\frac{x}{12}\right)=\frac{x+6}{18}$

$\Rightarrow \frac{x}{6}=\frac{x+6}{18}$

$\Rightarrow x=\frac{x+6}{3}$

$\Rightarrow 2 x=6$

$\Rightarrow x=3$ 

Before the total numbers of black balls were, $3$.

5. A jar contains $24$ marbles, some are green, and others are blue. If a marble is drawn at random from the jar, the probability that it is green is $\frac{2}{3}$ . Find the number of blue balls in the jar. 

Let us consider, the number of green marbles be, $x$ .

The total number of marbles in the jar is 24.

And the total number of blue marbles, $24-x$ .

The probability of getting green marbles in the draw be, $\frac{total\,no\,of\,green\,marbles}{total\,marbles}=\frac{x}{24}$ .

And, from the question, The probability of getting green marbles in the draw, $\frac{2}{3}$ .

$\frac{x}{24}=\frac{2}{3}$

$\Rightarrow x=\frac{24 \times 2}{3}=16$

So, number of blue marbles is $24-16=8$.

Chapter 15 Class 10 Maths NCERT Solutions Overview

The entire chapter 15 Class 10 Maths Solution material is divided into two exercises. Please go through the following section to get familiar with the probability problems.

You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 10 Maths

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Pair of Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometry

Chapter 8 - Introduction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Constructions

Chapter 12 - Areas Related to Circles

Chapter 13 - Surface Areas and Volumes

Chapter 14 - Statistics

Chapter 15 - Probability

NCERT Solutions for Class 10 Maths Chapter 15 Exercises

Ncert solutions for class 10 maths probability - exercise 15.1.

Q1: The first question contains five fill in the blanks, and you need to have a proper understanding of the text to answer the same. However, the answers are also provided in Class 10 Maths Ch 15 Solutions.

Q2: This question requires you to answer whether the events mentioned have equally likely outcomes or not.

Q3: Question 3 is an interesting one. It asks the students why tossing a coin is considered an appropriate method of deciding which team will do the batting first.

Q4 and Q5: Question 4 asks you to figure out which of the options given is not a probability of an event. On the other hand, question 5 deals with what will be the probability of ‘not E’ when P(E) = 0.05.

Q6, 7, 8, 9, 10 and 11: All of these six questions deal with finding the probability of the given events. For instance, question 6 asks the probability of taking an orange coloured candy out of a bag full of lemon candies. Similarly, by solving problem 7 you will know how to find the probability of two students having a birthday on the same date.

Q12, 13, 14, 15 and 16: These questions also deal with finding the probable outcomes but are mainly related to numbers and pack of cards.

Q17, 18, 19, 20, 21 and 22: These problems are a bit more complicated, but require you to approach with similar techniques used in the previous sums. 

Q23, 24 and 25: The last three questions in this exercise deals with the tossing of coins and throwing dice.

NCERT Solutions for Class 10 Maths Probability - Exercise 15.2

Q1: In this exercise, the first question is about two customers Shyam and Ekta. You need to calculate the probability of them both visiting a shop on the same day, consecutive days or different days.

Q2: This question is again about a numbered die which is thrown two times, and you need to find the value of the total scores.

Q3 and 4: Both these problems are related to picking a coloured ball out from a collection of differently coloured balls.

Q5: The final question is related to drawing marbles out of a jar. The data provided is the probability of picking green colour marbles, and you need to evaluate the same for blue marbles.

So, make sure to download NCERT Solutions for Class 10 Maths Chapter 15 PDF prepared by Vedantu and gain an extra edge over your fellow competitors.

Vedantu's NCERT Solutions for Class 10 Maths Chapter 15 - Probability offer an exceptional resource for students seeking to grasp the complexities of probability theory. With comprehensive and well-structured explanations, the platform empowers learners to tackle real-world challenges with confidence. The adept faculty ensures that each concept is elucidated with clarity, making learning an engaging and rewarding experience. Vedantu's commitment to providing accurate solutions and a wide array of practice questions aids students in honing their problem-solving skills. By utilising this invaluable tool, students can enhance their understanding of probability, paving the way for academic success and a deeper appreciation of its practical applications in various fields.

FAQs on NCERT Solutions for Class 10 Maths Chapter 15 - 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 15 Maths Class 10?

Subtopics of Chapter 15 of the Class 10 Maths NCERT textbook:

Introduction

Experiment and Outcome

Sample Space

Elementary Events

Favourable Outcomes

Probability of an Event

Theoretical Probability

Experimental Probability

Addition Theorem of 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 15 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.

NCERT Solutions for Class 10

AssignmentsBag.com

Assignments For Class 10 Mathematics Probability

Assignments for Class 10 Mathematics Probability have been developed for Standard 10 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 10 Mathematics Probability from our website as we have provided all topic wise assignments free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 10 Mathematics Probability. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Mathematics Probability book and get good marks in class 10 exams.

Question. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?  (A) 1/2 (B) 2/5 (C) 8/15 (D) 9/20

Question. Three coins are tossed, the probability of getting at most 2 heads is    (A) 3/8 (B) 1/2 (C) 7/8 (D) 1/8

Question. What is the probability of getting a sum 9 from two throws of a dice?    (A) 1/6 (B) 1/8 (C) 1/9 (D) 1/12

Question. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?    (A) 1/2 (B) 3/4 (C) 3/8 (D) 5/16

Question. Two dice are tossed. The probability that the total sum is a prime number is:  (A) 1/6 (B) 5/12 (C) 1/2 (D) 7/9

Question. A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:  (A) 1/13 (B) 2/13 (C) 1/26 (D) 1/52

Question. Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and other is a heart, is :    (A) 3/20 (B) 29/34 (C) 47/100 (D) 13/102

Question. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?  (A) 1/13 (B) 3/13 (C) 1/4 (D) 9/52

Question. There are 30 cards of the same size in a bag on which natural numbers 1 to 30 are written. One card is taken out of the bag at random. Then the probability that the number on the selected card is not divisible by 3 is    (A) 1/3 (B) 3/4 (C) 2/3 (D) 1/4

Question. If all the face cards are removed from a pack of 52 cards and then a card is randomly drawn then the probability of getting a ‘10 of heart’ will be    (A) 1/40 (B) 2/49 (C) 3/40 (D) 3/17

Question. The probability of guessing the correct answer to a certain question is p/12. If the probability of not guessing the correct answer to the same question is 3/4, the value of p is    (A) 3 (B) 4 (C) 2 (D) 1

Question. From a well shuffled pack of 52 cards, black aces and black queens are removed and from the remaining cards, a card is drawn at random. Then the probability of drawing a king or a queen is    (A) 7/8 (B) 3/4 (C) 1/8 (D) 1/2.

Question. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:    (A) 21/46 (B) 25/117 (C) 1/50 (D) 3/25

Question. A dice is tossed 100 times and the data is recorded as below  The probability that we get at even number in a trial is (A) 2/5 (B) 3/5 (C) 1/5 (D) 4/5

Question. From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?  (A) 1/15 (B) 25/57 (C) 35/256 (D) 1/221

Question. The king, queen and jack of hearts are removed from a deck of 52 playing cards and then well shuffled. One card is selected from the remaining cards. Then the probability of getting a king is    (A) 1/49 (B) 2/49 (C) 3/49 (D) 1

Question. Sum of probabilities of all the events in a sample space related to any event is    (A) 1 (B) 0 (C) –1 (D) not defined

VERY SHORT ANSWER TYPE QUESTIONS

Question. A card is drawn at random from a pack of 52 playing cards. What is the probability that the card drawn is neither a red card nor a black king?

Question. A child has a die whose six faces show the letters as given below : 

The die is throw once. What is the probability of getting ‘B’?

Question. A die is thrown once. What is the probability of getting a prime number?

Question. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is four times that of a red ball, how many blue balls are in the bag?

Question. A letter is chosen from the word ‘RANDOM’. What is the probability that it is a vowel?

Question. Out of 200 bulbs in a box, 20 bulbs are defective. One bulb is taken out at random from the box. What is the probability that the drawn bulb is not defective?

Question. Three coins are tossed simultaneously. What is the probability of getting no head?

Question. A coin is tossed twice. Write the possible outcomes.

{HH, HT, TH, TT}

Question. Two dice are thrown simultaneously. What is the probability of getting a doublet?

Question. A bag contains 6 red and 4 blue balls. One ball is drawn at random. What is the probability that the ball drawn is not red?

Question. A die is rolled twice. Write the possible outcomes.

{(1, 1), (1, 2),…., (1, 6), (2, 1),…., (2, 6),….., (6, 1),……, (6, 6)}; Total 36 cases.

Question. If the probability of winning a game is 0.4. What is the probability of losing it?

Question. Two dice are thrown simultaneously. What is the probability of getting a multiple of 3 as the sum?

Question. Two coins are tossed simultaneously. What is the probability of getting atleast one head?

Question. A box contains 30 cards, numbered from 1 to 30. One card is drawn at random from the box. What is the probability that the number on the drawn card is a multiple of 2 or 3?

Probability

Different types of events and their probability representation

• Sure Events (PE) =1
• Impossible Event ,(PE) =0
• Certain Events,0 ≤ (PE) ≤ 1
• Elementary Events, (An event having only one outcome)
• Complementary Event,

Relationship between events

Question. Three unbiased coins are tossed together. Find the probability of getting : (i) all heads (ii) two heads (iii) one head (iv) at least two heads Solution.  Here, possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. So, total no. of possible outcomes = 8 (i) favourable outcome = HHH So, No. of favourable outcome = 1 ∴ P (all heads) = no. of favourable outcomes / Total no. possible outcome = 1/8 (ii) favourable outcomes are HHT, THH and HTH. So, no. of favourable outcomes = 3 ∴ P (two heads) = 3/8 (iii) favourable outcomes are HTT, THT and TTH. So, no. of favourable outcomes = 3 ∴ P (one head) = 3/8 (iv) favourable outcomes are HHH, HHT, HTH and THH. So, no. of favourable outcomes = 4 ∴ P (at least two heads) = 4/8 = 1/2

Question. Find the probability that a leap year selected at random will contain 53 sundays. Solution.  In a leap year, there are 366 days. We have, 366 days = 52 weeks + 2 days. Thus, a leap year has always 52 sundays. The remaining 2 days can be : (i) Sunday and Monday (ii) Monday and Tuesday (iii) Tuesday and Wednesday (iv) Wednesday and Thursday (v) Thursday and Friday (vi) Friday and Saturday (vii) Saturday and Sunday Clearly, there are seven elementary events associated with this random experiment. Let E be the event that a leap year has 53 sundays. Clearly, the event E will happer if the last two days of the leap year are either Sunday and Monday or Saturday and Sunday. ∴ Favourable no. of elementary events = 2 Hence, required probability = 2/7 Ans.

PRACTICE EXERCISE

Question. A girl calculates that the probability of her winning the third prize in a lottery is 0.08. If 6000 tickets are sold, how many ticket has she bought. Solution. 480

Question. What is probability that a non-leap year selected at random will contain 53 Sundays. Solution. 1/7

Question. In a lottery, there are10 prizes and 25 blanks. Find the probability of getting a prize. Solution. 2/7

Question. Find the probability of prime numbers selected at random from the numbers 3,4,5,6…25 . Solution. 8/23

Question. A bag contains 5 red, 4 blue and 3 green balls. A ball is taken out from the bag at random. Find the probability that the selected ball is (a) of red colour (b) not of green colour. Solution. A. 5/12 B. 3/4

Question. A jar contains 54 marbles each of which is blue ,green or white . The probability of selecting a blue marble at random from the jar is 1/3, and the probability of selecting a green marble at random is 4/9 . How many white marbles does the jar contain. Solution. 12

Question. two coins are tossed simultaneously. Find the probability of getting exactly one head. Solution. 1/2

Question. A dice is thrown once. What is the probability of getting a number greater than 4? Solution. 1/3

Question. The probability of getting bad egg in a lot of 400 is 0.035. Then find the number of bad eggs in the lot. Solution. 14

Question. Write the probability of a sure event. Solution. 1

Question. Cards with numbers 2 to 101 are placed in a box. A card selected at random from the box. Find the probability that the card which is selected has a number which is a perfect square. Solution. 9/100

Question. Two dice are thrown at the same time. Find the probability that the sum of two numbers appearing on the top of the dice is more than 9. Solution. 1/6

Question. Two dice are thrown at the same time. Find the probability of getting different numbers on both dice. Solution. 5/6

Question. Cards marked with the number 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from the box. Find the probability that the number on the card is: (i) An even number (ii) A number less than14 (iii) A number is perfect square (iv) A prime number less than 20 Solution. A. 1/2 B. 3/25 C. 9/100, D. 2/25

Question. A coin is tossed two times. Find the probability of getting almost one head. Solution. 3/4

Question. What is the probability of an impossible event? Solution. 0

Question. When a dice is thrown, and then find the probability of getting an odd number less than 3. Solution. 1/6

Question. A bag contains card which are numbered from 2 to 90. A card is drawn at random from the bag. Find the probability that it bears. a.) A Two digit number b.) A number which is perfect square. Solution. 81/89, 8/89

Question. Two dice are thrown at the same time. Find the probability of getting (a) same no. on the both side(b) different no. on both dices. Solution. A. 1/6 B. 5/6

Question. Find the probability of getting the letter M in the word “MATHEMATICS”. Solution. 2/11

Question. Two dice are thrown simultaneously. Find the probability of getting : (i) a doublet i.e. same number on both dice. (ii) the sum as a prime number. Solution.  Possible outcomes associated to the random experiment of throwing two dice are : (1, 1), (1, 2), ……., (1, 6) (2, 1), (2, 2), ……., (2, 6) ………………………………… ………………………………… (6,1) , (6, 2), ……., (6, 6) ∴ Total number of possible outcomes = 6 × 6 = 36 (i) The favourable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6). ∴ Total no. of favourable outcomes = 6 So, P(a doublet) = no. of favourable outcomes / Total no. of possible outcomes = 6/36 = 1/6  (ii) Here, favourable sum (as a prime number) are 2, 3, 5, 7 and 11. So, favourable outcomes are (1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4,3), (6, 5) and (5, 6). ∴ no. of favourable outcomes = 15 ∴ P (the sum as a prime number) = 15/36 = 5/12 Ans. 

Question. One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is : (i) an ace (ii) either red or king (iii) a face card (iv) a red face card Solution.  here, total no. of possible outcomes = 52. (i) There are 4 ace cards in a pack of 52 cards. One ace can be chosen in 4 ways. So, favourable no. of outcomes = 4 ∴ P (an ace) = no. of favourable outcomes / Total no. of possible outcomes = 4/52 = 1/13 (ii) There are 26 red cards, including 2 red kings. Also, there are 4 kings, two red and two black. ∴ card drawn will be a red card or a king if it is any one of 28 cards (26 red cards and 2 black kings) So, favourable no. of outcomes = 28 ∴ P(either red or king) = 28/52 = 7/13 (iii) Kings, queens and jacks are the face cards. So, favourable no. of outcomes = 3 × 4 = 12 ∴ P(a face card) = 12/52 = 7/13 (iv) There are 6 red face cards, 3 each from diamonds and hearts. So, favourable no. of outcomes = 6 ∴ P(a red face card) = 6/52 = 3/26

Question. An unbiased die is thrown. What is the probability of getting : (i) an odd number (ii) a multiple of 3 (iii) a perfect square number (iv) a number less than 4. Solution.  Here, total number of all possible outcomes= 6 (i) favourable outcomes are 1, 3, 5. So, no. of favourable outcomes = 3 ∴ P (an odd number) =   No. of favourable outcomes / Total no. of possible outcome = 3/6 = 1/2 (ii) favourable outcomes are 3 and 6. So, no. of favourable outcomes = 2 ∴ P (a multiple of 3) = 2/6 = 1/3 (iii) favourable outcomes are 1 and 4. So, no. of favourable outcomes = 2 ∴ P (a perfect square number) = 2/6 = 1/3 (iv) favourable outcomes are 1, 2 and 3. So, no. of favourable outcomes = 3 ∴ P (a number less than 4) = 3/6 = 1/2 

Question. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is  drawn from this box. Find the probability that the number of the card is : (i) an even number (ii) a number less than 14 (iii) a number which is a perfect square (iv) a prime number less than 20. Solution.  From 2 to 101, these are (101–2) + 1 = 100 numbers. So, total no. of possible outcomes = 100. (i) From 2 to 101, the even numbers are 2, 4, 6, …., 100 which are 50 in number. So, number of favourable outcomes = 50 ∴ P(an even number) = no. of favourable outcomes / Total no. of possible outcomes = 50/100 = 1/2 (ii) From 2 to 101, the numbers less than 14 are 2, 3, …., 13 which are 12 in number. So, no. of favourable outcomes = 12 ∴ P(a number less than 14) = 12/100 = 3/25 (iii) From 2 to 101, the perfect squares are 4, 9, 16, ….. 100, which are 9 in number. So, no. of favourable outcomes = 9 ∴ P (a number which is a perfect square) = 9/100 (iv) From 2 to 101, the prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19 which are 8 in number. So, no. of favourable outcomes = 8 ∴ P (a prime no. less than 20) = 8/100 = 2/25

Question. A bag contains 3 red balls and 5 black balls.A ball is drawn at random from a bag. What is the probability that the ball drawn is : (i) red (ii) not red  Solution.  Total number of balls = 3 + 5 = 8 (i) P (red ball) = no. of red balls / Total no.of balls = 3/8 (ii) P (not red ball) = 1 – P(red ball) = 1 – 3/8 = 5/8

Question. Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 1 m?   

Solution.  Total area of rectangular region = 3 m × 2 m = 6 m 2

Question. A bag contains 12 balls out of which x are white. (i) If one ball is drawn at random, what is the probability that it will be a white ball? (ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be double than that in (i). Find x. Solution.  (i) Total number elementary events = 12. There are x white balls out of which one can be chosen in x ways. So, favourable number of elementary events = x ∴ p 1  = P (white ball) = no. of favourable outcomes / Total no. of possible outcomes  = x / 12 (ii) If 6 more white balls are put in the bag, then total number of balls in the bag =12 + 6 = 18 and, no. of white balls in the bag = x + 6 ∴ p 2  = p (getting a white ball) = x + 6 / 18 It is given that, p 2  = 2p 1 ⇒ x + 6 /18 = 2x x / 12 ⇒ x + 6 /18 = x / 6 ⇒ 6( x+ 6) = 18 x ⇒ 6x + 36 = 18x ⇒ 12 x = 36 ⇒ x = 3 Ans. 

Question. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3 . Find the number of blue marbles in the jar. Solution.  Total number of elementary events = 24. Let there be x green marbles. ∴ P (green marbles is drawn) = x / 24 but, P(green marbles is drawn) = 2/3 (given) ⇒ x/24 = 2/3 ⇒ x = 2/3 x 24 ⇒ x = 16 ∴ Number of green marbles = 16 ⇒ Number of blue marbles = 24 – 16 = 8 Ans.

Assignments for Class 10 Mathematics Probability as per CBSE NCERT pattern

All students studying in Grade 10 Mathematics Probability should download the assignments provided here and use them for their daily routine practice. This will help them to get better grades in Mathematics Probability exam for standard 10. We have made sure that all topics given in your textbook for Mathematics Probability which is suggested in Class 10 have been covered ad we have made assignments and test papers for all topics which your teacher has been teaching in your class. All chapter wise assignments have been made by our teachers after full research of each important topic in the textbooks so that you have enough questions and their solutions to help them practice so that they are able to get full practice and understanding of all important topics. Our teachers at https://www.assignmentsbag.com have made sure that all test papers have been designed as per CBSE, NCERT and KVS syllabus and examination pattern. These question banks have been recommended in various schools and have supported many students to practice and further enhance their scores in school and have also assisted them to appear in other school level tests and examinations. Its easy to take print of thee assignments as all are available in PDF format.

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• Solving Assignments for Mathematics Probability Class 10 helps to further enhance understanding of the topics given in your text book which will help you to get better marks
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• You will be able to understand the type of questions which are expected in your Mathematics Probability class test
• You will be able to revise all topics given in the ebook for Class 10 Mathematics Probability as all questions have been provided in the question banks
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All free Printable practice assignments are in PDF single lick download format and have been prepared by Class 10 Mathematics Probability teachers after full study of all topics which have been given in each chapter so that the students are able to take complete benefit from the worksheets. The Chapter wise question bank and revision assignments can be accessed free and anywhere. Go ahead and click on the links above to download free CBSE Class 10 Mathematics Probability Assignments PDF.

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Class 10 Maths NCERT Solutions Chapter 15 Probability

• January 18, 2020 January 28, 2020

Class 10 Maths NCERT Solutions Chapter 15: All the Solutions pertaining to NCERT Class 10 Maths Chapter Probability acts as a Study Material to students preparing for their board exams. Prepare effectively for the exam by downloading the free PDF for NCERT Solutions Maths Class 10 Chapter 15. Subject experts prepared the NCERT Solutions for Class 10 Maths Probability as per the latest NCERT Textbooks to assist the students in scoring max. marks. Furthermore, Maths Class 10 Chapter 15 Probability Solutions are provided in Hindi and English Mediums for your comfort.

Solutions to all the questions are given in a detailed way so that it is easy for you to understand. Go through the Topics and Subtopics list existing here for the CBSE Class 10 Maths Chapter Probability so that you will have an idea of what to cover during your preparation. Revise the concepts thoroughly and excel in the board examination with the NCERT Class 10 Maths Solutions Chapter 15 Probability . Complete your assignments and homework in time with the NCERT Solutions provided for all the exercises(15.1, 15.2).

Topics and Subtopics listed for Class 10 Maths Chapter Probability

Class 10 Maths NCERT Solutions Chapter 15 Probability Ex 15.1

Class 10 Maths NCERT Solutions Chapter 15 Probability Ex 15.2

NCERT Solutions for Class 10 Maths Chapter 15 – Probability

We tried providing the Class 10 Maths NCERT Solutions Chapter Probability and they will be of extreme help during your preparation for the board exams. Step by step solutions provided for CBSE Class 10 Maths Probability will make complex fundamentals too simple. All the Important Questions, Multiple Choice Questions, Notes prepared here aids you in scoring you maximum marks. Get to know the methods used for solving the problems and apply them further in your board examination so that you won’t feel any difficulty.

NCERT Class 10 Maths Solutions Chapter 15 – Solved Exercises

Download the Class 10 Maths NCERT Solutions for Probability PDF and avail it offline so that it stays handy for you to practice. All the Questions given related to Probability are in a detailed manner and simple to understand. We gave the NCERT Solutions with an aim to help students understand difficult problems too with ease. You can practice the Maths Class 10 NCERT Solutions Chapter 15 Probability Exercises 15.1, 15.2 to know the various kind of questions being asked in exams.

Total Weightage from NCERT Class 10 Maths Chapter 15 Probability is 4 Marks. Out of which there will be 2 two mark questions. You need to concentrate on problems related to dice, cards, coin. All three topics are important and you will get questions mostly on this.

After going through the Class 10 Maths NCERT Solutions you will get in-depth knowledge on Chapter Probability. Clarify all your doubts by referring to the CBSE Maths Class 10 Solutions Chapter Probability. Revise the Complete Syllabus by making use of Multiple Choice Questions, Important Questions as well as Formulas prevailing on this page.

We believe the information shared related to Class 10 Maths NCERT Solutions Chapter 15 helps you stay updated and top in board exams. If you have any queries or doubts feel free to reach us via the comment section and we will be of your help at the soonest possible.

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Statistics and probability

Unit 1: analyzing categorical data, unit 2: displaying and comparing quantitative data, unit 3: summarizing quantitative data, unit 4: modeling data distributions, unit 5: exploring bivariate numerical data, unit 6: study design, unit 7: probability, unit 8: counting, permutations, and combinations, unit 9: random variables, unit 10: sampling distributions, unit 11: confidence intervals, unit 12: significance tests (hypothesis testing), unit 13: two-sample inference for the difference between groups, unit 14: inference for categorical data (chi-square tests), unit 15: advanced regression (inference and transforming), unit 16: analysis of variance (anova).

• Math Article

Probability

Probability  means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability has been introduced in Maths to predict how likely events are to happen. The meaning of probability is basically the extent to which something is likely to happen. This is the basic probability theory, which is also used in the  probability distribution , where you will learn the possibility of outcomes for a random experiment. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.

Learn More here: Study Mathematics

Probability Definition in Math

Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are going to happen, using it. Probability can range from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event. Probability for Class 10 is an important topic for the students which explains all the basic concepts of this topic. The probability of all the events in a sample space adds up to 1.

For example , when we toss a coin, either we get Head OR Tail, only two possible outcomes are possible (H, T). But when two coins are tossed then there will be four possible outcomes,  i.e {(H, H), (H, T), (T, H),  (T, T)}.

Download this lesson as PDF: – Download PDF Here

Formula for Probability

The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.

Sometimes students get mistaken for “favourable outcome” with “desirable outcome”. This is the basic formula. But there are some more formulas for different situations or events.

Solved Examples

1) There are 6 pillows in a bed, 3 are red, 2 are yellow and 1 is blue. What is the probability of picking a yellow pillow?

Ans: The probability is equal to the number of yellow pillows in the bed divided by the total number of pillows, i.e. 2/6 = 1/3.

2) There is a container full of coloured bottles, red, blue, green and orange. Some of the bottles are picked out and displaced. Sumit did this 1000 times and got the following results:

• No. of blue bottles picked out: 300
• No. of red bottles: 200
• No. of green bottles: 450
• No. of orange bottles: 50

a) What is the probability that Sumit will pick a green bottle?

Ans: For every 1000 bottles picked out, 450 are green.

Therefore, P(green) = 450/1000 = 0.45

b) If there are 100 bottles in the container, how many of them are likely to be green?

Ans: The experiment implies that 450 out of 1000 bottles are green.

Therefore, out of 100 bottles, 45 are green.

Probability Tree

The tree diagram helps to organize and visualize the different possible outcomes. Branches and ends of the tree are two main positions. Probability of each branch is written on the branch, whereas the ends are containing the final outcome. Tree diagrams are used to figure out when to multiply and when to add. You can see below a tree diagram for the coin:

Types of Probability

There are three major types of probabilities:

Theoretical Probability

Experimental probability, axiomatic probability.

It is based on the possible chances of something to happen. The theoretical probability is mainly based on the reasoning behind probability. For example, if a coin is tossed, the theoretical probability of getting a head will be ½.

It is based on the basis of the observations of an experiment. The experimental probability can be calculated based on the number of possible outcomes by the total number of trials. For example, if a coin is tossed 10 times and head is recorded 6 times then, the experimental probability for heads is 6/10 or, 3/5.

In axiomatic probability, a set of rules or axioms are set which applies to all types. These axioms are set by Kolmogorov and are known as Kolmogorov’s three axioms. With the axiomatic approach to probability, the chances of occurrence or non-occurrence of the events can be quantified. The axiomatic probability lesson covers this concept in detail with Kolmogorov’s three rules (axioms) along with various examples.

Conditional Probability is the likelihood of an event or outcome occurring based on the occurrence of a previous event or outcome.

Probability of an Event

Assume an event E can occur in r ways out of a sum of n probable or possible equally likely ways . Then the probability of happening of the event or its success is expressed as;

The probability that the event will not occur or known as its failure is expressed as:

P(E’) = (n-r)/n = 1-(r/n)

E’ represents that the event will not occur.

Therefore, now we can say;

P(E) + P(E’) = 1

This means that the total of all the probabilities in any random test or experiment is equal to 1.

What are Equally Likely Events?

When the events have the same theoretical probability of happening, then they are called equally likely events. The results of a sample space are called equally likely if all of them have the same probability of occurring. For example, if you throw a die, then the probability of getting 1 is 1/6. Similarly, the probability of getting all the numbers from 2,3,4,5 and 6, one at a time is 1/6. Hence, the following are some examples of equally likely events when throwing a die:

• Getting 3 and 5 on throwing a die
• Getting an even number and an odd number on a die
• Getting 1, 2 or 3 on rolling a die

are equally likely events, since the probabilities of each event are equal.

Complementary Events

The possibility that there will be only two outcomes which states that an event will occur or not. Like a person will come or not come to your house, getting a job or not getting a job, etc. are examples of complementary events. Basically, the complement of an event occurring in the exact opposite that the probability of it is not occurring. Some more examples are:

• It will rain or not rain today
• The student will pass the exam or not pass.
• You win the lottery or you don’t.

Also, read: 

• Independent Events
• Mutually Exclusive Events

Probability Theory

Probability theory had its root in the 16th century when J. Cardan, an Italian mathematician and physician, addressed the first work on the topic, The Book on Games of Chance. After its inception, the knowledge of probability has brought to the attention of great mathematicians. Thus, Probability theory is the branch of mathematics that deals with the possibility of the happening of events. Although there are many distinct probability interpretations, probability theory interprets the concept precisely by expressing it through a set of axioms or hypotheses. These hypotheses help form the probability in terms of a possibility space, which allows a measure holding values between 0 and 1. This is known as the probability measure, to a set of possible outcomes of the sample space.

Probability Density Function

The Probability Density Function (PDF) is the probability function which is represented for the density of a continuous random variable lying between a certain range of values. Probability Density Function explains the normal distribution and how mean and deviation exists. The standard normal distribution is used to create a database or statistics, which are often used in science to represent the real-valued variables, whose distribution is not known.

Probability Terms and Definition

Some of the important probability terms are discussed here:

Applications of Probability

Probability has a wide variety of applications in real life. Some of the common applications which we see in our everyday life while checking the results of the following events:

• Choosing a card from the deck of cards
• Flipping a coin
• Throwing a dice in the air
• Pulling a red ball out of a bucket of red and white balls
• Winning a lucky draw

Other Major Applications of Probability

• It is used for risk assessment and modelling in various industries
• Weather forecasting or prediction of weather changes
• Probability of a team winning in a sport based on players and strength of team
• In the share market, chances of getting the hike of share prices

Problems and Solutions on Probability

Question 1: Find the probability of ‘getting 3 on rolling a die’.

Sample Space = S = {1, 2, 3, 4, 5, 6}

Total number of outcomes = n(S) = 6

Let A be the event of getting 3.

Number of favourable outcomes = n(A) = 1

i.e. A  = {3}

Probability, P(A) = n(A)/n(S) = 1/6

Hence, P(getting 3 on rolling a die) = 1/6

Question 2: Draw a random card from a pack of cards. What is the probability that the card drawn is a face card?

A standard deck has 52 cards.

Total number of outcomes = n(S) = 52

Let E be the event of drawing a face card.

Number of favourable events = n(E) = 4 x 3 = 12 (considered Jack, Queen and King only)

Probability, P = Number of Favourable Outcomes/Total Number of Outcomes

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

P(the card drawn is a face card) = 3/13

Question 3: A vessel contains 4 blue balls, 5 red balls and 11 white balls. If three balls are drawn from the vessel at random, what is the probability that the first ball is red, the second ball is blue, and the third ball is white?

The probability to get the first ball is red or the first event is 5/20.

Since we have drawn a ball for the first event to occur, then the number of possibilities left for the second event to occur is 20 – 1 = 19.

Hence, the probability of getting the second ball as blue or the second event is 4/19.

Again with the first and second event occurring, the number of possibilities left for the third event to occur is 19 – 1 = 18.

And the probability of the third ball is white or the third event is 11/18.

Therefore, the probability is 5/20 x 4/19 x 11/18 = 44/1368 = 0.032.

Or we can express it as: P = 3.2%.

Question 4: Two dice are rolled, find the probability that the sum is:

• less than 13

Video Lectures

Introduction.

Solving Probability Questions

Probability Important Topics

Probability Important Questions

Probability Problems

• Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is: (i) 6 (ii) 12 (iii) 7
• A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being a (i) red ball (ii) green ball (iii) not a blue ball
• All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value (i) 7 (ii) greater than 7 (iii) less than 7
• A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded. (i) How many different scores are possible? (ii) What is the probability of getting a total of 7?

Frequently Asked Questions (FAQs) on Probability

What is probability give an example, what is the formula of probability, what are the different types of probability, what are the basic rules of probability, what is the complement rule in probability.

In probability, the complement rule states that “the sum of probabilities of an event and its complement should be equal to 1”. If A is an event, then the complement rule is given as: P(A) + P(A’) = 1.

What are the different ways to present the probability value?

The three ways to present the probability values are:

• Decimal or fraction

What does the probability of 0 represent?

The probability of 0 represents that the event will not happen or that it is an impossible event.

What is the sample space for tossing two coins?

The sample space for tossing two coins is: S = {HH, HT, TH, TT}

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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good basics concepts provide

Great video content. Perfectly explained for examinations. I really like to learn from BYJU’s

Thank you for your best information on probablity

Good explanation about probability and concept for simple understanding the overall chapter. I hope this is a good way to understand the CONCEPT

For learner of class X standard , it is providing all the relevant informations and approach towards the contenet is knitted in an elegant manner and students will have the opportunity to grasp the topic easily and will be immensely benefited.

this topic has been hard for me but now I know what it is all about and I have really enjoyed it thanks for your good explanation

Well explained

Excellent explanation of probability. One can easily understand about the probability.

I really appreciated your explanations because it’s well understandable Thanks

Love the way you teach

There are 3 boxes Box A contains 10 bulbs out of which 4 are dead box b contains 6 bulbs out of which 1 is dead box c contains 8 bulbs out of which 3 are dead. If a dead bulb is picked at random find the probability that it is from which box?

Probability of selecting a dead bulb from the first box = (1/3) x (4/10) = 4/30 Probability of selecting a dead bulb from the second box = (1/3) x (1/6) = 1/18 Probability of selecting a dead bulb from the third box = (1/3) x (3/8) = 3/24 = 1/8 Total probability = (4/30) + (1/18) + (1/8) = (48 + 20 + 45)360 =113/360

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Maths assignment class viii | quadrilateral, lesson plan math class x (ch-15) probability, e- lesson plan    subject mathematics     class 10.

• Classical definition of probability.
• Probability of sure event, impossible event and concept of equally likely events.
• Range of probability.
• Concept of probability with one coin, two coin three coins and method of writing their sample space.
• Concept of probability with one die, two dice and method of writing their sample space.
• Concept of probability with the playing cards.
• Concept of probability with the term at least (≥ ) , at most (≤ ) , "and "(common), "or" (all with common one time) etc.
• Simple problems on finding the probabilities of an event.
• Probability of sure event impossible event equally likely events. 
• Range of Probability.
• Method of writing sample space when different number of coins are tossed.
• Method of writing sample space when different number of  dice are thrown. 
• Students should know the terms associated with the playing cards and become able to tackle the problems with them.
• Able to find probability in different situations. 

Explain the terms associated with the playing cards i.e. possible outcomes, number of black(52) and red(52) cards, number ace(4) and face(12) cards etc and their probabilities.

• Review questions given by the teacher. 
• Students can prepare a presentation on the sample space of different number of coins and dice. 
• Solve NCERT problems with examples, 
• Solve the assignment given by the teacher.
• Assignment sheet will be given as home work at the end of the topic. 
• Separate sheets which will include questions of logical thinking and Higher order thinking skills will be given to the above average students.
• Class Test , Oral Test , worksheet and  Assignments.  can be made the part of assessment.
• Re-test(s) will be conducted on the basis of the performance of the students in the test.

INTEGRATION OF PROBABILITY WITH ARTIFICIAL INTELLIGENCE TOOL 

Very much useful.... Thank you...

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Probability Assignment 4 Worksheet Class 10 PDF with Answers

These Probability Assignment 4 worksheet PDF can be helpful for both teachers and students. Teachers can track their student’s performance in the chapter Probability Assignment 4. Students can easily identify their strong points and weak points by solving questions from the worksheet. Accordingly, students can work on both weak points and strong points. 

All students studying in CBSE class 10th, need to practise a lot of questions for the chapter Probability Assignment 4. Students can easily practise questions from the Probability Assignment 4 problems worksheet PDF. By practising a lot of questions, students can improve their confidence level. With the help of confidence level, students can easily cover all the concepts included in the chapter Probability Assignment 4. 

Probability Assignment 4 Worksheets with Solutions

Solutions is the written reply for all questions included in the worksheet. With the help of Probability Assignment 4 worksheets with solutions, students can solve all doubts regarding questions. Students can have deep learning in the chapter Probability Assignment 4 by solving all their doubts. By solving doubts, students can also score well in the chapter Probability Assignment 4. 

Probability Assignment 4 Worksheet PDF

Worksheet is a sheet which includes many questions to solve for class 10th students. The Probability Assignment 4 worksheet PDF provides an opportunity for students to enhance their learning skills. Through these skills, students can easily score well in the chapter Probability Assignment 4. Students can solve the portable document format (PDF) of the worksheet from their own comfort zone. 

How to Download the Probability Assignment 4 Worksheet PDF? 

To solve questions from the Probability Assignment 4 worksheet PDF, students can easily go through the given steps. Those steps are- 

• Open Selfstudys website.

• Bring the arrow towards CBSE which can be seen in the navigation bar. 

• Drop down menu will appear, select KVS NCERT CBSE Worksheet. 

• A new page will appear, select class 10th from the given list of classes. 
• Select Mathematics from the given list of subjects. Now click the chapter’s name that is Probability Assignment 4. 

Features of the Probability Assignment 4 Worksheet PDF

Before starting to solve questions from the Probability Assignment 4 problems worksheet PDF, students need to know everything about the worksheet. Those features are- 

• Variety of questions are included:  The Probability Assignment 4 Maths Worksheet for Class 10 includes varieties of questions. Those varieties of questions are- one mark questions, two mark questions, three mark questions, etc. 
• Solutions are provided:  Doubts regarding each question can be easily solved through the solutions given. Through solving questions, a student's comprehensive skill can be increased. 
• All concepts are covered:  By solving questions from the Probability Assignment 4 problems worksheet, students can easily cover all the concepts included in the chapter.  
• Created by Expert:  These worksheets are personally created by the subject experts. These Probability Assignment 4 worksheet pdf are created with proper research. 
• Provides plenty of questions:  The Probability Assignment 4 worksheet provides plenty of questions to practise. Through good practice, students can get engaged in the learning process. 

Benefits of the Probability Assignment 4 Worksheet PDF

With the help of Probability Assignment 4 problems worksheet PDF, students can easily track their performance. This is the most crucial benefit, other than this there are more benefits. Those benefits are- 

• Builds a strong foundation:  Regular solving questions from the worksheet can help students to build a strong foundation. Through the strong foundation, students can score well in the chapter Probability Assignment 4. 
• Improves speed and accuracy:  While solving questions from the chapter Probability Assignment 4, students need to maintain the speed and accuracy. Speed and accuracy can be easily maintained and improved by solving questions from the Probability Assignment 4 worksheet PDF. 
• Acts as a guide:  Probability Assignment 4 worksheets with solutions acts as guide for both the teachers and students. Through the worksheet, teachers can guide their students according to the answers given by them. Students can also analyse themselves with the help of answers and can improve accordingly. 
• Enhances the learning process:  Regular solving of questions from the worksheet can help students enhance their learning process. According to the learning skills, students can easily understand all topics and concepts included in the chapter Probability Assignment 4.  
• Improvisation of grades:  Regular solving of questions from the worksheet can help students to improve their marks and grades. With the help of good marks and good grades, students can select their desired field further. 

Tips to Score Good Marks in Probability Assignment 4 Worksheet

Students are requested to follow some tips to score good marks in the Probability Assignment 4 worksheet. Those tips are-

• Complete all the concepts:  First and the most crucial step is to understand all the concepts included in the chapter Probability Assignment 4.  
• Practise questions:  Next step is to practise questions from the Probability Assignment 4 problems worksheet. Through this students can identify all types of questions: easy, moderate, difficult, etc.  
• Note down the mistakes:  After practising questions, students need to note down the wrong sums that have been done earlier. 
• Rectify the mistakes:  After noting down the mistakes, students need to rectify all the mistakes made. 
• Maintain a positive attitude:  Students are requested to maintain a positive attitude while solving worksheets. By maintaining a positive attitude, students can improve speed and accuracy while solving the worksheets. 
• Remain focused:  Students need to remain focused while solving questions from the Probability Assignment 4 problems worksheet pdf. As it helps students to solve the questions as fast as possible. 

When should a student start solving the Probability Assignment 4 Worksheet PDF?

Students studying in class 10 should start solving worksheets after covering each and every concept included in the chapter. Regular solving questions from the Probability Assignment 4 worksheet PDF, can help students to have a better understanding of the chapter. Better understanding of the chapter Probability Assignment 4 can help students to score well in the class 10th board exam. 

Regular solving questions from the Probability Assignment 4 Worksheet PDF can help students to build a strong foundation for the chapter Probability Assignment 4. Strong foundation of the chapter Probability Assignment 4 can help students to understand further chapters. 

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