NCERT Solutions for Class 12 Chapter 13 Probability Exercise 13.4

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Class 12 Maths NCERT Solutions Chapter 13 Probability Exercise 13.4 is based on Random Variables and its Probability Distributions. Key topics covered under these concepts are:

Probability distribution of a random variable
Mean and variance of a random variable

Download PDF NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.4

Check out the solutions of Class 12 Maths NCERT solutions Chapter 13 Probability Exercise 13.4

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.
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
2.
If \(P(A) = 0.3\), \(P(B) = 0.7\) and \(P(B/A) = 0.5\), then \(P(A \cup B)\) is :
- \(0.35\)
- \(0.85\)
- \(0.5\)
- \(0.21\)
View Solution
3.
For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :
Minimize \(Z = 18x + 10y\)
subject to constraints
\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]
View Solution
4.
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
5.
For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :
- \(\frac{1}{2}\)
- \(\frac{3}{5}\)
- \(\frac{1}{5}\)
- \(\frac{3}{10}\)
View Solution
6.
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

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NCERT Solutions For Class 12 Mathematics Chapter 13: Probability

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NCERT Solutions For Class 12 Mathematics Chapter 12: Linear Programming

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You can also check

NCERT Solutions for Class 12 Chapter 13 Probability Exercise 13.4

CBSE CLASS XII Related Questions

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

2.
If \(P(A) = 0.3\), \(P(B) = 0.7\) and \(P(B/A) = 0.5\), then \(P(A \cup B)\) is :

3.
For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :
Minimize \(Z = 18x + 10y\)
subject to constraints
\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]

4.
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 :

5.
For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :

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

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 12 Chapter 13 Probability Exercise 13.4

CBSE CLASS XII Related Questions

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

2.If \(P(A) = 0.3\), \(P(B) = 0.7\) and \(P(B/A) = 0.5\), then \(P(A \cup B)\) is :

3.For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :Minimize \(Z = 18x + 10y\)subject to constraints\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]

4.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 :

5.For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :

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

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

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

2.
If \(P(A) = 0.3\), \(P(B) = 0.7\) and \(P(B/A) = 0.5\), then \(P(A \cup B)\) is :

3.
For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :
Minimize \(Z = 18x + 10y\)
subject to constraints
\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]

4.
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 :

5.
For events A and B, if \(P(B)=\frac{3}{5}\), \(P(A/B)=\frac{1}{2}\) and \(P(A\cup B)=\frac{4}{5}\), then \(P(A)\) is :

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