Q. 8. A random variable $X$ has the following probability distribution:
iii) $P(X>6)$
The sum of probabilities of the probability distribution of the random variables is 1.
$\therefore 0+k+2 k+2 k+3 k+k^2+2 k^2+7 k^2+k=1$
$10 k^2+9 k-1=0$
$(10 k-1)(k+1)=0$
$k= \frac{1}{10}$ and $k=-1$
$k=-1$ is not possible.
So, $k=\frac{1}{10}$
$P(X>6) & =P(X=7)$
$=7 K^2+K$
$=7 \times\left(\frac{1}{10}\right)^2+\frac{1}{10}$
$=\frac{7}{100}+\frac{1}{10}$
$=\frac{17}{100}$