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A random variable $X$ has the following p.d.f\[\]$ \begin{array} {llllllll} \textbf{X:}& 0& 1& 2& 3& 4& 5&6&7 \\ \textbf{P(X):}& 0& k &2k& 2k& 3k& k^{2}&2k^{2}&7k^{2}+k \end{array} $\[\]The value of $k$ is

\[\begin{array}{1 1}(1)\frac{1}{8}&(2)\frac{1}{10}\\(3)0&(4)-1 \;or\; \frac{1}{10}\end{array}\]

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$P(X=0)+P(X=1) +.....+P(X=7)=1$
$10k (k+1) -1(k+1) =0$
$(10 k -1)(k+1)=0$
$k=\large\frac{1}{10} $ or $k=-1$
$k=-1 $ is not possible
Since $P(X=1),P(X=2),P(X=3),P(X=4)$ are negative which is impossible.
So, $k= \large\frac{1}{10}$
Hence 2 is the correct answer.
answered May 23, 2014 by meena.p

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