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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Check out NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1 Solutions

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.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.
View Solution
3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
View Solution
4.
If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is
- 3
- –3
- –4
- \(\pm 3\)
View Solution
5.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
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.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

4.
If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is

5.
The HCF of 960 and 432 is :

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

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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.In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

3.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

4.If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is

5.The HCF of 960 and 432 is :

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.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

4.
If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is

5.
The HCF of 960 and 432 is :

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.