Let X be a continuous random variable with PDF

$f_{x} (x) = {\begin{matrix} 4 x^{3} 0 < x \leq 1 \\ 0 o t h e r w i s e \end{matrix}$

The value of $P (X \leq \frac{2}{3} | X > \frac{1}{3})$ is

Question

Question

Answer 1

$P (X \leq \frac{2}{3} | X > \frac{1}{3}) = \frac{P (X \leq \frac{2}{3} \cap X > \frac{1}{3})}{P (X > \frac{1}{3})}$

$= \frac{P (\frac{1}{3} < X \leq \frac{2}{3})}{P (X > \frac{1}{3})}$

$P (\frac{1}{3} < X \leq \frac{2}{3}) = \int_{1 / 3}^{2 / 3} 4 x^{3} d x = {[x^{4}]}_{\frac{1}{3}}^{\frac{2}{3}} = \frac{15}{81} = \frac{5}{27}$

$P (X > \frac{1}{3}) = \int_{1 / 3}^{1} 4 x^{3} d x$

$= {[x^{4}]}_{\frac{1}{3}}^{1}$

$= 1 - \frac{1}{81} = \frac{80}{81}$

$P (X \leq \frac{2}{3} | X > \frac{1}{3}) = \frac{\frac{5}{27}}{\frac{80}{81}} = \frac{5}{27} \times \frac{81}{80} = \frac{3}{16}$