A random variable $X$ has the following distribution \begin{array}{lllllllll} \hline \begin{array}{l} Values of \ X(x) \end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \ \hlineP(X=x) \quad 0 & k & 2 k & 2 k & 3 k & k^2 & 2 k^2 & 7 k^2+k \ \hline \end{array}

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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { Since, } \Sigma P\left(X=x_iight)=1 \ & \Rightarrow 0+K+2 K+2 K+3 K+K^2+2 K^2 \ & +7 K^2+K=1 \ & \Rightarrow \quad 9 K+10 K^2=1 \ & \Rightarrow \quad K=\frac{1}{10}, \quad[\because K>0] \ & \end{aligned} $ So, $ \begin{aligned} & P(0 < K < 6)=K+2 K+2 K+3 K+K^2 \ & =8 K+K^2=\frac{8}{10}+\frac{1}{100}=\frac{81}{100}=\left(\frac{9}{10}ight)^2 \end{aligned} $

A random variable $X$ has the following distribution \begin{array}{lllllllll} \hline \begin{array}{l} Values…

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