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

The constant $k$ is determined by the normalization condition $\int_{0}^{1} kx(1-x) dx = 1$.

$k \int_{0}^{1} (x - x^2) dx = k \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = k \left( \frac{1}{6} \right) = 1$

Thus $k = 6$ and $f(x) = 6x(1-x)$.

Given $P(x > a) = \frac{20}{27}$, we require $\int_{a}^{1} 6x(1-x) dx = \frac{20}{27}$.

$6 \int_{a}^{1} (x - x^2) dx = 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{a}^{1} = 1 - 3a^2 + 2a^3 = \frac{20}{27}$

Multiplying through by 27 yields $54a^3 - 81a^2 + 7 = 0$.

Testing $a = \frac{1}{3}$ gives $54\left(\frac{1}{27}\right) - 81\left(\frac{1}{9}\right) + 7 = 2 - 9 + 7 = 0$.

The correct choice is A.