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  • Please Solve R.D.Sharma class 12 Chapter 31 Mean and Variance of a random Variable Exercise Fill in the Blanks Question 3 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 31 Mean and Variance of a random Variable  Exercise Fill in the Blanks Question 3 Maths Textbook Solution.

Answers (1)

Answer: \frac{3}{10}

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

Given: -

\begin{array}{|c|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(X) & c & 2 c & 2 c & 3 c & c^{2} & 2 c^{2} & 7 c^{2}+c \\ \hline \end{array}

Solution:P(X \leq 2)=P(X=1)+P(X=2)

Therefore, we know that the sum of probabilities in a probability distribution is always 1.

\begin{aligned} &\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=7)=1 \\ \end{aligned}

\begin{aligned} &c+2 c+2 c+3 c+c^{2}+2 c^{2}+7 c^{2}+c=1 \\\\ &10 c^{2}+9 c-1=0 \\ \end{aligned}

\begin{aligned} &10 c^{2}+10 c-c-1=0 \\\\ &10 c(c+1)-(c+1)=0 \\\\ &(10 c-1)(c+1)=0 \\\\ &c=\frac{1}{10}, c=-1 \end{aligned}

Negative value does not support, we use c=1/10

\begin{aligned} &P(X \leq 2)=P(X=1)+P(X=2) \\\\ &=\frac{1}{10} \times 1+\frac{1}{10} \times 2 \\\\ &=\frac{1}{10}+\frac{2}{10} \\\\ &=\frac{3}{10} \end{aligned}

 

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