# NCERT Solutions for Class 10 Maths Chapter 15

NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1 and Exercise 15.2 in English and Hindi medium for CBSE, UP Board, MP Board free to download updated for new academic session 2020-21. These NCERT Solutions are applicable for UP Board also as UP Board has implemented NCERT Textbooks as UP Board Textbooks for the academic session 2020-2021. UP board Students can download UP Board solutions for Class 10 Maths Chapter 15 all exercises. These solutions are applicable for all the students who are following CBSE Board for academic session 2020-21. Not only NCERT Solutions but the NCERT Books also for all subjects along with Offline Apps of NCERT Textbook’s Solutions for class 10 all subjects.

Videos related to Ex 15.1 and Ex. 15.2 of class 10 Maths are given below explaining each and every questions. Explanation is in simple language, so that students can understand easily.## NCERT Solutions for Class 10 Maths Chapter 15

Class: 10 | Maths (English and Hindi Medium) |

Chapter 15: | Probability |

### 10th Maths Chapter 15 Solutions

NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1 and Exercise 15.2 of Probability is given below for academic session 2020-21. NCERT Solutions for class 10 is updated according to CBSE Curriculum 2020-2021. Join the discussion forum to ask your questions and answer the questions already asked by other users.

### 10th Maths Chapter 15 Exercise 15.1

### 10th Maths Chapter 15 Exercise 15.2

#### Class 10 Maths Chapter 15 Solutions in Videos

##### What do you understand by Random Experiment?

Random Experiment: A random experiment is one in which the exact outcome cannot be predicted.

##### What is meant by Trial in Probability?

Trial: Performing a random experiment is called a trial.

##### What is meant by Outcomes?

Outcomes: The result of a random experiment is called an outcome.

##### What is Sample space in terms of Probability?

Sample space: The collection of all possible outcomes of a random experiment is called a sample space.

##### What are Events?

Event: Any possible outcome or combination of outcomes of a random experiment is called an event.

##### Which events are called Equally likely events?

Equally likely events: Two or more events of a random experiment are said to be equally likely events if each one of them have an equal chance of occurrence.

##### What is meant by Probability of an event?

Probability of an event: The chance of occurrence of the event expressed quantitatively is known as the probability of an event and denoted by P(E).

#### Previous Years Questions

ONE MARK QUESTIONS

1. A number is chosen at random from the numbers -3, -2,-1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1? [CBSE 2017]

THREE MARKS QUESTIONS

1. Two different dice are thrown together. Find the probability that the numbers obtained

(i) have a sum less than7

(ii) have a product less than 16

(iii) is a doublet of odd numbers. [CBSE 2017]

FOUR MARKS QUESTIONS

1. Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25? [CBSE 2017]

##### About Probability

The measure of certainty of events in numerical values, under certain conditions, is provided by the branch of mathematics called ‘Theory of Probability. This theory has extensive use as one of the basic tools in statistics and wide range of applications in Science, engineering, biological science, medical, commerce, weather forecasting etc.

###### Historical facts!

- The concept of probability was developed in a very strange manner. In 1654, a gambler by name Chevalier de Mere approached the well-known 17th century French philosopher and mathematician Blaise Pascal regarding certain dice problems. Pascal discussed them with another French mathematician Pierre de Fermat and they found solution to dice problems. This work was the beginning of probability theory.
- Probability theory has its actual origin in the 16th century when an Italian physician and mathematician J. Cardan wrote the first book on the subject ‘The book on Games of Chance’. Since its inception, the study of Statistics and probability has attracted the attention of great mathematicians (James Bernoulli (1654 – 1705), A. de Moivre (1667 – 1754) and Pierre Simon Laplace (1749 – 1827)).
- In 1812, Pierre Simon Laplace or Pierre de Laplace (1749 – 1827, France) proposed a mathematical system of inductive reasoning based on probability. He introduced many principles of probability, one among them is, “Probability is the ratio of the favoured events to the total possible events”.
- Statistician Karl Pearson (1857 – 1936) had tossed the coin 24000 times and he got 12012 heads. Then calculated experimental probability 12012/2400 = 0.5005.

5. In the eighteenth century French De Buffon tossed a coin 4040 times and got 2048 heads. Then he calculated experimental probability 2048/4040 = 0.507.

### Important Questions on Class 10 Maths Chapter 15

इसलिए 0.05+P(नहीं E)=1

⇒ P(नहीं E)=1-0.05=0.95

P(संतरे की महक वाली गोली ) = 0

Probability that two students are having same birthday P (E)

= 1 − P (not E)

= 1 − 0.992

= 0.008

Number of possible outcomes of a dice = 6

Prime numbers on a dice are 2, 3, and 5.

Total prime numbers on a dice = 3

Probability of getting a prime number = 3/6 =1/2

₹ 1 के सिक्के = 50

₹ 2 के सिक्के = 20

₹ 5 के सिक्के = 10

कुल सिक्के = 100 + 50 + 20 + 10 = 180

P(50 पैसे के सिक्के ) = (50 पैसे के सिक्के )/(कुल सिक्के ) = 100/180 = 5/9

मादा मछली = 8

कुल मछली = 5 + 8 = 13

P(नर मछली ) = (नर मछली)/(कुल मछली) = 5/13

Total number of queens = 1

P(getting a queen)

= (Number of favourable outcomes)/(Number of total possible outcomes)

=1/5

Total number of defective bulbs = 4

P(getting a defective bulb)

= (Number of favourable outcomes)/(Number of total possible outcomes)

= 4/20

= 1/5

Total number of two-digit numbers between 1 and 90 = 81

P (getting a two-digit number)

= 91/90

= 9/10

कुल ख़राब पेन = 20

कुल अच्छे पेन = 144 – 20 = 122

P(आप वह पेन खरीदेंगे)

= ( कुल अच्छे पेन)/( कुल पेन)

= 122/144

= 61/72

लाल गेंदों की कुल संख्या = 5

कुल गेंदे = x + 5

P (लाल गेंद) = 5/(5 + x)

P (नीली गेंद) = x/(5 + x)

दिया है,

2(5/(5 + x)) = x/(5 + x)

⇒ 10(x + 5) = x^2 + 5x

⇒ x^2-5x-50=0

⇒ x^2 – 10x + 5x – 50 = 0

⇒ x(x – 10) + 5 (x – 10) = 0

⇒ (x – 10)(x + 5) = 0

⇒ x – 10 = 0 या x + 5 = 0

⇒ x = 10 या x = – 5

क्योंकि गेंदों की संख्या ऋणात्मक नहीं हो सकती, अतः नीली गेंदों की कुल संख्या 10 है ।