Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 31 Mean and Variance of a random Exercise Very Short Answer type Question 2 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 31 Mean and Variance of a random Exercise Very Short Answer type Question 2 Maths Textbook Solution.

Answers (1)

Answer: - k=\frac{1}{2}

Hint: - You must know the rules for solving the probability distribution problems.

Given: -

            \begin{array}{|c|c|c|c|c|} \hline X=x_{i} & 0 & 1 & 2 & 3 \\ \hline P\left(X=x_{i}\right) & 2 k^{4} & 3 k^{2}-5 k^{3} & 2 k-3 k^{2} & 3 k-1 \\ \hline \end{array}

Solution: - We know that the sum of probabilities in a probability distribution is always 1.

\sum_{i=1}^{n}p_{i}=1

\begin{aligned} &\therefore P(x=0)+P(x=1)+P(x=2)+P(x=3)=1 \\ \end{aligned}

\begin{aligned} &2 k^{4}+3 k^{2}-5 k^{3}+2 k-3 k^{2}+3 k-1=1 \\ \end{aligned}

\begin{aligned} &2 k^{4}-5 k^{3}+5 k=1+1 \\ \end{aligned}

\begin{aligned} &2 k^{4}-5 k^{3}+5 k-2=0 \\ \end{aligned}

\begin{aligned} &\left.(k-1)(k-2)\left(2 k^{2}+k-1\right)=0 \quad \text { \{using division }\right\} \\ \end{aligned}

\begin{aligned} &(k-1)(k-2)(2 k-1)(k+1)=0 \\ \end{aligned}

\begin{aligned} &k=-1, \frac{1}{2}, 1,2 \end{aligned}

Value of k cannot be 1 or greater than 1 because sum of probabilities should always be equal to 1

Hence,k=1/2

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Date Sheet 2025: CBSE, UPMSP, MPBSE, Class 10, 12 board exam dates out; BSEB, CGBSE schedules soon
anam-khan

CBSE Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4

Latest Question