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  • Explain solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise multiple choice questions question 4 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 31 Mean and Variance of a Random Variable exercise multiple choice questions question 4 maths textbook solution

Answers (1)

Answer:

              \frac{1}{10}

Hint:

              To solve this we add all the variables

Given:

              

X     0 1 2 3 4 5 6 7
P\left ( X \right )     0 2p 2p 3p p^{2} 2p^{2} 7p^{2} 2p

Solution:

\begin{aligned} &\text { As we know, } \sum_{i=0}^{n} P\left(X_{i}\right)=1 \\ &0+2 p+2 p+3 p+p^{2}+2 p^{2}+7 p^{2}+2 p=1 \\ &10 p^{2}+9 p-1=0 \\ &10 p^{2}+10 p-p-1=0 \\ &10 p(p+1)-1(p+1)=0 \\ &(10 p-1)(p+1)=0 \\ &p=\frac{1}{10} \text { or }-1 \end{aligned}

Probability cannot be negative. So value of p=\frac{1}{10}

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