NCERT Solutions for Class 12 Chapter 13 Probability Exercise 13.5

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Class 12 Maths NCERT Solutions Chapter 13 Probability Exercise 13.5 is based on Bernoulli Trials and Binomial Distribution.

Bernoulli Trials must satisfy the following conditions:

There should be a finite number of trials.
Trials should be conducted independently.
Each trial has exactly two outcomes: success or failure.
Probability of success remains the same in each trial.

Download PDF NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.5

Check out the solutions of Class 12 Maths NCERT solutions Chapter 13 Probability Exercise 13.5

Also check other Exercise Solutions of Class 12 Maths Chapter 13 Probability

Also Read:

Chapter 13 Probability Important Topics
Conditional Probability	Conditional Probability Formula	Multiplication Theorem on Probability
Bias	Independent Events	Mean and variance of random variable
Difference between mutually exclusive and independent events	Uniform distribution formula	T distribution formula
Coin toss probability formula	T distribution	Uniform distribution
Probability and statistics symbols	Probability density function	Geometric distribution
Exponential distribution	Beta distribution	NCERT Solutions for Class 12 Mathematics Chapter 13 Probability

Also Read:

Class 12 Mathematics Important Guides
Class 12 Mathematics Notes	Math MCQs	Trigonometry
Number Systems	Probability and Statistics	Arithmetic
Algebra	Difference Between Topics in Maths	Math Study Guides
Math Formula	NCERT Solutions for Class 12 Maths	Mensuration

CBSE CLASS XII Related Questions

1.
Evaluate :
\[ I = \int_{-\pi}^{\pi} \frac{2x\,(1+\sin x)}{1+\cos^2 x}\,dx \]
View Solution
2.
If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that
\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]
View Solution
3.
Find the value of \(x\) if \(A = B\), where
\[ A = \begin{bmatrix} 6 & 3x \\ 8 & -6 \end{bmatrix} \text{ and } B = \begin{bmatrix} 6 & x^2+2 \\ 8 & x^2-5x \end{bmatrix}. \]
View Solution
4.
Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).
View Solution
5.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :
- \( -6 \) unit/sec
- \( 0 \) unit/sec
- \( 18 \) unit/sec
- \( 6 \) unit/sec
View Solution
6.
The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :
- \( \frac{7\pi}{12} \)
- \( \frac{5\pi}{12} \)
- \( -\frac{5\pi}{12} \)
- \( \frac{\pi}{3} \)
View Solution

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NCERT Solutions For Class 12 Mathematics Chapter 13: Probability

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You can also check

NCERT Solutions for Class 12 Chapter 13 Probability Exercise 13.5

CBSE CLASS XII Related Questions

1.
Evaluate :
\[ I = \int_{-\pi}^{\pi} \frac{2x\,(1+\sin x)}{1+\cos^2 x}\,dx \]

2.
If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that
\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]

3.
Find the value of \(x\) if \(A = B\), where
\[ A = \begin{bmatrix} 6 & 3x \\ 8 & -6 \end{bmatrix} \text{ and } B = \begin{bmatrix} 6 & x^2+2 \\ 8 & x^2-5x \end{bmatrix}. \]

4.
Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

5.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

6.
The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :

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NCERT Solutions for Class 12 Chapter 13 Probability Exercise 13.5

CBSE CLASS XII Related Questions

1.Evaluate :\[ I = \int_{-\pi}^{\pi} \frac{2x\,(1+\sin x)}{1+\cos^2 x}\,dx \]

2.If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]

3.Find the value of \(x\) if \(A = B\), where\[ A = \begin{bmatrix} 6 & 3x \\ 8 & -6 \end{bmatrix} \text{ and } B = \begin{bmatrix} 6 & x^2+2 \\ 8 & x^2-5x \end{bmatrix}. \]

4.Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

5.The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

6.The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Evaluate :
\[ I = \int_{-\pi}^{\pi} \frac{2x\,(1+\sin x)}{1+\cos^2 x}\,dx \]

2.
If \(y = \sqrt{\log \left\{ \sin \left( \dfrac{x^2}{3} + 1 \right) \right\}}\), then show that
\[ \frac{dy}{dx} = \frac{x \cot \left( \dfrac{x^2}{3} + 1 \right)}{3y}. \]

3.
Find the value of \(x\) if \(A = B\), where
\[ A = \begin{bmatrix} 6 & 3x \\ 8 & -6 \end{bmatrix} \text{ and } B = \begin{bmatrix} 6 & x^2+2 \\ 8 & x^2-5x \end{bmatrix}. \]

4.
Solve the differential equation \( (x+y)\,dy + (x-y)\,dx = 0 \), given that \( y(1) = 1 \).

5.
The motion of a particle is described as \( S = 2t^{3} - 6t^{2} + 8 \). Its velocity \( \frac{dS}{dt} \), when the acceleration \( \frac{d^{2}S}{dt^{2}} \) vanishes is :

6.
The value of \( \left[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right) \right] \) is :