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 Exercise 15.1 Solutions are based on the following concepts:

Occurrence of an event
Impossible Events
Sure Event

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Exercise Solutions of Class 10 Maths Chapter 15 Probability

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NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2 Solutions

Also Check:

Probability Chapter Related Topics
Theoretical Probability	Equally Likely Events	Empirical Probability
Events in Probability	Complimentary Events	A Theoretical Approach to Probability
P Value	Experimental Probability	Elementary Event
Sum of Probabilities	Geometric Probability	Chance and Probability
Empirical Probability Formula	Poisson Distribution Formula	Standard Normal Distribution

Also check:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Arithmetic
Class 10 Maths Notes	Permutations and Combinations	Mensuration
Trigonometry	Number Systems	Geometry
Math Formula	NCERT Solutions for Class 10 Maths	Algebra

CBSE X Related Questions

1.
$ABCD$ is a parallelogram such that $AF = 7 \text{ cm}$, $FB = 3 \text{ cm}$ and $EF = 4 \text{ cm}$, length $FD$ equals
- $\frac{21}{4} \text{ cm}$
- $\frac{28}{3} \text{ cm}$
- $\frac{12}{7} \text{ cm}$
- $5.5 \text{ cm}$
View Solution
2.
If $\alpha, \beta$ are the zeroes of the polynomial $p(x) = x^2 - 3x - 1$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
View Solution
3.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :
- $q - 9p$
- $p - 9q$
- $p + 9q$
- $2p + 9q$
View Solution
4.
A kite is flying at a height of $60 \text{ m}$ above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as $30^{\circ}$. From the bottom of the same building, the angle of elevation of kite is $45^{\circ}$. Find the length of the string and height of roof from the ground. (Use $\sqrt{3} = 1.73$)
View Solution
5.
Prove that :
$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$.
View Solution
6.
Assertion (A) : H.C.F. $(36 m^{2}, 18 m) = 18 m$, where $m$ is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution

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NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1

Exercise Solutions of Class 10 Maths Chapter 15 Probability

CBSE X Related Questions

1.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

2.
If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).

3.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

5.
Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

6.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

Similar Mathematics Concepts

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NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1

Exercise Solutions of Class 10 Maths Chapter 15 Probability

CBSE X Related Questions

1.\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

2.If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).

3.The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

5.Prove that : \(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

6.Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number. Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

2.
If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).

3.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

5.
Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

6.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.