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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 31 Mean and Variance of a random Exercise Fill in the Blanks Question 4 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 31 Mean and Variance of a random Exercise Fill in the Blanks  Question 4 Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{3}

Hint: - You must know the rules for finding values of random variables.

Given: -

              \begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline X & \frac{1}{2} & 1 & \frac{3}{2} & 2 \\ \hline P(X) & c & c^{2} & 2 c^{2} & c \\ \hline \end{array}\\ \end{aligned}

Solution:

               \begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline X & \frac{1}{2} & 1 & \frac{3}{2} & 2 \\ \hline P(X) & c & c^{2} & 2 c^{2} & c \\ \hline \end{array} \end{aligned}

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

\begin{aligned} &P\left(X=\frac{1}{2}\right)+P(X=1)+P\left(X=\frac{3}{2}\right)+P(2)=1 \\ \end{aligned}

\begin{aligned} &c+c^{2}+2 c^{2}+c=1 \\\\ &3 c^{2}+2 c-1=0 \\\\ &3 c^{2}+3 c-c-1=0 \\ \end{aligned}

\begin{aligned} &3 c(c+1)-1(c+1)=0 \\\\ &(3 c-1)(c+1)=0 \\\\ &3 c-1=0 \\\\ &c=\frac{1}{3} \end{aligned}

Or c+1=0

c=-1

But value of (X) does not support the negative value.

c=\frac{1}{3}

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