If a random variable X has the following probability distribution of X

$$ \begin{array}{|l|c|c|c|c|c|c|c|c|} \hline \mathrm{X}=x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathrm{P}(\mathrm{X}=x) & 0 & \mathrm{k} & 2 \mathrm{k} & 2 \mathrm{k} & 3 \mathrm{k} & \mathrm{k}^2 & 2 \mathrm{k}^2 & 7 \mathrm{k}^2+\mathrm{k} \\ \hline \end{array} $$

Then $P(x \geq 6)=$

If $X \sim B\left(6, \frac{1}{2}\right)$, then $P(|X-2| \leqslant 1)=$

A random variable X has following p.d.f. $\mathrm{f}(x)=\mathrm{kx}(1-x), 0 \leqslant x \leqslant 1 \quad$ and $\quad \mathrm{P}(x>\mathrm{a})=\frac{20}{27}$, then $\mathrm{a}=$

If A and B are independent events such that $\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)=\frac{3}{25}$ and $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{8}{25}$, then $P(A)=$