NCERT Solutions for Class 12 Chapter 13 Probability Exercise 13.1

Collegedunia Team logo

Collegedunia Team

Content Curator

Class 12 Maths NCERT Solutions Chapter 13 Probability Exercise 13.1 is based on following concepts: Conditional Probability and Properties of conditional probability.

Download PDF NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.1

Check out the solutions of Class 12 Maths NCERT solutions Chapter 13 Probability Exercise 13.1

Read More: NCERT Solutions For Class 12 Mathematics Chapter 13 Probability

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

Also Read:

Also Read:

CBSE CLASS XII Related Questions

  • 1.
    Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \).
    Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B)}{P(B)} \).

      • Both Assertion (A) and Reason (R) are true and the 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 and Reason (R) is false.
      • Assertion (A) is false and Reason (R) is true.

    • 2.

      Find the domain of \[ q(x)=\cos^{-1}(4x^2-3). \] Hence, find the value of \(x\) for which \[ q(x)=0. \] Also, write the range of \[ 3q(x)-\pi. \] 


        • 3.
          Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]


            • 4.

              Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not. 


                • 5.
                  For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

                    • local maximum value is 2
                    • local minimum value is \( -2 \)
                    • local maximum value is \( -2 \)
                    • local minimum value \( < \) local maximum value

                  • 6.

                    If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is: 

                      • \(\frac{1}{3}\)
                      • \(\frac{1}{9}\)
                      • \(3\)
                      • \(9\)
                    CBSE CLASS XII Previous Year Papers

                    Comments


                    No Comments To Show