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’ = _____. 

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? 

Ans:

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

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

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

(i)$8?$ 

Ans:

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

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

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

Ans:

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

Thus, the probability of getting an odd number is,  

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

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

Ans:

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

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

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

Ans:

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

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

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

Ans:

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

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

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

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

Ans:

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

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

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

(iii) an odd number. 

Ans:

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

Thus, the probability of getting a odd number,

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

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

(i) a king of red color

Ans:

We know there are 52 numbers in the deck.

There are 2 kings of red color in the deck.

Thus, the probability,

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

(ii) a face cards 

Ans:

There are 12 face cards in the deck.

Thus, the probability,

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

(iii) a red face cards 

Ans:

There are 6 red face cards in the deck.

Thus, the probability,

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

(iv) the jack of hearts

Ans:

There are 1 jack of hearts card in the deck.

Thus, the probability,

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

(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 (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,

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

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?

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:

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, $=\frac{1}{36}$ 

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.

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 $\frac{1}{11}$.

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.

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, 

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

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)

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.

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?

Ans:

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? 

Ans:

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? 

Ans:

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: 

Ans:

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. 

Ans:

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

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

Putting the values,

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

$\Rightarrow x=10$ 

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? 

Ans:

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 $x$.

Ans:

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,

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

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

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

Simplifying further,

$3 x=x+6$

$\Rightarrow 2 x=6$

$\Rightarrow x=3$ 

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. 

Ans:

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, $\frac{2}{3}$ .

Now, 

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

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

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.

Conclusion

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.