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NCERT Solutions for Class 10 Maths Chapter 15: Probability - Exercise 15.1

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NCERT Solutions for Class 10 Maths Chapter 15 Probability

If the students are searching for proper guidelines and material for the preparation of final examinations, they have reached out a perfect platform to clear all these doubts. The Class 10 Maths NCERT Solutions Chapter 15 Exercise 15.1 help students to learn the subject, which automatically gives a better score. As the materials of NCERT Solutions are prepared by well-experienced mathematicians at Vedantu, even the slow learners can learn and understand easily. The solved and unsolved questions available in the PDF help their students to get practice with every concept and every exercise in the chapter.  You can also download the NCERT Solution for Class 10 Science to score more marks in the examinations.


Class:

NCERT Solutions for Class 10

Subject:

Class 10 Maths

Chapter Name:

Chapter 15 - Probability

Exercise:

Exercise - 15.1

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

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NCERT Solutions for Class 10 Maths Chapter 15: Probability - Exercise 15.1
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Access NCERT Solutions for Class 10 Maths Chapter 15 – Probability

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: If Shyam and Ekta are visiting a particular shop on the same week which starts from Tuesday and ends on Saturday then the total days in the week $=5$

Now, if each of them visiting the shop is equally likely then,

(The outcome of the event of Shyam visiting the shop) $=5$

(The outcome of the event of Ekta visiting the shop) $=5$

Thus, the total outcome of events

$\begin{aligned} \Rightarrow \mathrm{P}(\text { Total outcome }) &=5 \times 5 \\ &=25 \end{aligned}$


(ii) Consecutive Days?

Ans: Probability of both of them reaching the shop on the same day:

$\mathrm{P}($ Same day $)=(\mathrm{T}, \mathrm{T}),(\mathrm{W}, \mathrm{W}),(\mathrm{Th}, \mathrm{Th}),(\mathrm{F}, \mathrm{F}),(\mathrm{S}, \mathrm{S})$.

$=5$

$P$ (reaching on the same day) $=\frac{P(\text { sameday })}{P(\text { Total outcomes })}$

$ \begin{aligned} &=\frac{5}{25} \\ &=\frac{1}{5} \end{aligned} $

Answer $=\frac{1}{5}$ 


(ii) Consecutive days?

Ans: The probability of both of them reaching on consecutive days:

$\mathrm{P}($ Consecutive days $)=(\mathrm{T}, \mathrm{W}),(\mathrm{W}, \mathrm{Th}),(\mathrm{Th}, \mathrm{F}),(\mathrm{F}, \mathrm{S}),(\mathrm{W}, \mathrm{T}),(\mathrm{Th}, \mathrm{W}),(\mathrm{F}, \mathrm{Th})$, $(\mathrm{S}, \mathrm{F})$

$=8$

$\mathrm{P}$ (reaching on consecutive days) $=\frac{\mathrm{P}(\text { consecutive days })}{\mathrm{P}(\text { Total outcomes })}$

$ =\frac{8}{25} $

Answer $=\frac{8}{25}$


(iii) Different days?

Ans: Let $P$ (both of them reaching on same days) $+P$ (both of them not reaching on the same days) $=1$

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

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

$ =\frac{4}{5} $

Answer $=\frac{4}{5}$


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:

i) $\mathrm{P}$ (getting the total score as even) $=\frac{\text { ways of getting even sum }}{\text { total ways }}$ 

$ =\frac{18}{36} $


ii) $P$ ( getting total as 6 ) $=\frac{\text { ways of getting total as } 6}{\text { total ways }}$


$ =\frac{5}{36} $


iii) $\mathrm{P}$ ( getting total sum as 6 or less than 6 ) $=$ ways of getting sum equal orless than 6


$ \begin{aligned} \text { total ways } &=\frac{15}{36} \end{aligned} $


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

Ans: A bag of 5 red balls and x blue balls (assume) will be having total $5+x$ balls.

$P($ getting a blue ball $)=\frac{x}{5+x}$

$P($ getting a red ball $)=\frac{5}{5+x}$

If the probability of getting a blue ball is double the probability of getting a red ball then

$2 \mathrm{P}$ (getting a blue ball) $=\mathrm{P}$ (getting a red ball)

$ \begin{aligned} 2\left(\frac{x}{5+x}\right) &=\frac{5}{5+x} \\ 2 x^{2}+5 x-25 &=0 \end{aligned} $

On solving the following equation

$ \mathrm{x}=10,-5 $

Number of blue balls cannot be negative thus $x=10$.

$ \text { Answer }=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? 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 $\mathbf{x}$.

Ans:

i) If a box of 12 balls has $x$ black balls and $12-\mathbf{x}$ 0ther balls then

$\mathrm{P}$ (getting a black ball) $=\frac{\text { numberof black balls }}{\text { total balls }}$

$ =\frac{\mathrm{x}}{12} $


ii) If 6 more balls are added then the total number of black balls $=x+6$

And the total number of balls $=12+6=18$

So, $\mathrm{P}$ (getting a black ball) $=\frac{\text { number of black balls }}{\text { total balls }}$

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

Now, if the probability of getting a black ball is double of before then, $2 \mathrm{P}$ ( getting a black ball before) $=\mathrm{P}$ (getting a black ball)

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

$ 36 x-12 x=72 $

$ \begin{aligned} & 24 x=72 \\ & \mathrm{x}=3 \end{aligned} $

Thus, the number of black balls before was 3 .

Answer $=3$


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 marbles in the jar. Ans: The total number of marbles $=24$

Let the total number of green marbles be $x$ then

The total number of blue marbles $=24-\mathrm{X}$

$\mathrm{P}$ (getting a green marble) $=\frac{\text { numberof green marbles }}{\text { total number of marbles }}$

$ =\frac{x}{24} $

Given, $P$ (getting a green marble) $=\frac{2}{3}$

Putting the value, $\frac{x}{24}=\frac{2}{3}$

$ \begin{aligned} 3 x &=48 \\ x &=16 \end{aligned} $

Therefore, if the number of green marbles is 16 then, the number of blue marbles is, $24-16=8$

Answer $=8$


Exercise 15.1 Class 10 NCERT Solutions Free PDF

The students can avail the opportunity of downloading NCERT Solutions Class 10 Maths Chapter 15 Exercise 15 PDF from the official website of Vedantu for free. Taking a soft copy or hard copy is up to the student. Students may not need to bother about the internet connection. They can clarify their doubts through live chat from the official website. The downloading process is simple and strikes forward, so that either the students or parents can take these PDFs easily.


Chapter 15 Maths Class 10 Exercise 15.1 Question 1 

Question - 1 of this exercise is a set of filling the blanks to be answered by the students after understanding the concept of probability.


Exercise 15.1 Maths Class 10 PDF Question 2 

The next question is also about a set of four very short answer type questions. These can be answered with one or two words.


Exercise 15.1 Class 10 Maths NCERT Solutions Question 3

The third question is about the football game, and the students need to understand the question properly and give the reason for tossing a coin.


Maths Class 10 Chapter 15 Exercise 15.1 Question 4

The fourth question of this exercise is like choosing the correct option from the given options.


Class 10 Maths Chapter 15 Exercise 15.1 Solutions Question 5

The fifth question is tricky but very easy after understanding the logic. It is like the inverse property. After verifying the NCERT PDF thoroughly, students can answer this question very easily.


Ex 15.1 Class 10 Question 6

The next question is a direct question that can be answered by selecting one of the two options.


Exercise 15.1 Class 10 Question 7

The 7th question is about identifying the probability of having the same birthday. It is easy to answer but reading questions twice or thrice is always suggestible.


15.1 Class 10 Question 8

The other question is about taking out a ball and finding out the probability of the color of the ball. Students need to answer both the questions involved in it.


Ex 15.1 Class 10 NCERT Maths Question 9

This question is similar to that of the previous question here also students need to answer all the three probabilities asked in the question.


Class 10 Maths Chapter 15 Exercise 15.1 Question 10

The next equation is a shorter answer type question which is about the piggy bank. The coin has several probabilities, but the students need to answer to probability conditions asked at the end of the question.


Exercise 15.1 Maths Class 10 PDF Question 11

The 11th question is a direct question that can be answered easily by the students. It is about the probability of getting a male fish or a female fish.


Exercise 15.1 Class 10 NCERT Solutions Question 12

The next equation is about a game board. It is a long answer type of question that has four questions in it.


Exercise 15.1 Class 10 Maths NCERT Solutions Question 13

The 13th question is about throwing a dice. Dice is the best example of probability concepts. So students should concentrate on this kind of question and understand the logic and answer correctly.


Maths Class 10 Chapter 15 Exercise 15.1 Question 14

The next question is a long answer type of question. It has six sub-questions in it. The question is about picking up a card from the pack.


Chapter 15 Maths Class 10 Exercise 15.1 Question 15

This question is also similar to that of the previous question which related to the packing of cards.


Class 10 Maths Chapter 15 Exercise 15.1 Solutions Question 16

This question is about finding the probability of getting a defective pen and a good pen. Students can use formulae and answer the question directly.


Ex 15.1 Class 10 Question 17

It has two questions in it, and each is a different case. Students need to read properly and answer the questions by using formulae of probability.


Exercise 15.1 Class 10 Question 18

The next question is also given as a long answer type question as a whole or may split into three short answer questions. Students need to understand each case of probability separately and answer this kind of question.


15.1 Class 10 Question 19

The 19th question is about the probability of getting two faces of dice when a baby has thrown it.


Ex 15.1 Class 10 NCERT Maths Question 20

This question has a figure given in the pdf. Observe the figure and understand the question and then answer the probability conditions.


15.1 Class 10 Question 21

It is a question about the probability of a  scenario faced by a girl. The students need to find out the probability cases and answer this question.


Exercise 15.1 Class 10 Question 22

For answering this question, students need to refer to a PDF which contains examples. In that example the missing data is available. From the data, students need to answer another question.


Ex 15.1 Class 10 Question 23 

The 23rd question is about the tossing of a coin by two boys. We need to find out the probability.


15.1 Class 10 Question 24

Students need to answer the 24th question when a dice is thrown three times. The probability is also may triple this case.


Ex 15.1 Class 10 Question 25

The last question of this exercise is to give reasons for the arguments which have been made between people. 


Key Takeaways of NCERT Solutions

Students can get benefited with - 

  • A detailed explanation is available in the PDF.

  • It helps to store the material for future use.

  • It is a help-hand for students to revise the concepts of the chapter briefly.

FAQs on NCERT Solutions for Class 10 Maths Chapter 15: Probability - Exercise 15.1

1. If P(E) = 0.01, what is the probability of ‘not E’?

As we know 

P(E)+P(not E) = 1

It is given that, P(E) = 0.01

So, P(not E) = 1-P(E)

That is, P(not E) = 1-0.0

∴ P(not E) = 0.99

2. A box contains 5 red marbles, 8 white marbles, and 4 green marbles. One marble is removed from the box at random. What is the probability that the marble removed will be red?

The Total no. of balls = 5+8+4 = 17

P(E) = (Number of favorable outcomes/ Total number of outcomes)

So, Total number of red balls = 5

P (red ball) = 5/17 = 0.29

Thus, the probability of obtaining a red ball is 0.29. 

3. How many questions are there in Exercise 15.2 Probability?

CBSE Class 10 Maths Chapter 15 is on the topic of Probability. It has many exercises that include questions from various sub-topics, and among them, Exercise 15.2 has about five questions in total. These questions are less in number and hence very important to know all of them properly. You must make sure to refer to the solutions to these questions from the Vedantu website or from the Vedantu app so that you do not miss the opportunity to score full marks in your exams. These solutions are available free of cost for the students.

4. How many questions are there in Probability Exercise 15.1 Class 10 Maths?

Exercise 15.1 from the Probability Chapter 15 is a long enough exercise consisting of 25 questions in total where the questions range from various topics from the area of probability. You have to solve all the 25 questions because the questions are different from each other which will help you cover a whole range of topics and help you understand the pattern of solving each individual question.

5. What is the 3rd question in Exercise 15.1 Probability?

Exercise 15.1 has various questions. The third question is about a situation that describes a football game, and the students need to understand the question properly and give the reason for tossing a coin regarding the game. This is an easy question and if you face difficulty in solving this question, then you can refer to the solution given in Vedantu where you will have no problem understanding the method for solving it.

6. What is the range of probability?

Probability is a branch of Mathematics where the tendency or chance of an event to occur is measured. So there is a range fixed for this tendency. It is said that the probability of an event to take place will range between 0 and 1, that is the probability will be between greater than or equals to 0 and less than or equals to 1. It shows that it is not Mathematically possible to have probability values less than 0 or more than 1.

7. Can I score full marks in Probability?

Yes, you can score full marks in Probability. It is an entire practice based area of Maths and you need to understand the pattern for solving the questions in the easiest and best possible method. This will help you gain confidence in yourself so that you don’t back out from attempting any question in your exam and eventually solve them accurately and score full marks. In this process, Vedantu can prove to be your best guide and take you towards success.