Let X be a continuous random variable with PDF

\({f_x}\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {4{x^3}\;\;\;\;0 < x \le 1}\\ {0\;\;\;\;\;otherwise} \end{array}} \right.\)

The value of \(P\left( {X \le \frac{2}{3}{\rm{|}}X > \frac{1}{3}} \right)\;\)is

Option 4 : \(\frac{3}{{16}}\)

CT 1: Ratio and Proportion

2846

10 Questions
16 Marks
30 Mins

\(P\left( {X \le \frac{2}{3}{\rm{|}}X > \frac{1}{3}} \right) = \frac{{P\left( {X \le \frac{2}{3} \cap X > \frac{1}{3}} \right)}}{{P\left( {X > \frac{1}{3}} \right)}}\)

\( = \frac{{P\left( {\frac{1}{3} < X \le \frac{2}{3}} \right)}}{{P\left( {X > \frac{1}{3}} \right)}}\)

\(P\left( {\frac{1}{3} < X \le \frac{2}{3}} \right) = \mathop \smallint \limits_{1/3}^{2/3} 4{x^3}dx = \left[ {{x^4}} \right]_{\frac{1}{3}}^{\frac{2}{3}} = \frac{{15}}{{81}} = \frac{5}{{27}}\)

\(P\left( {X > \frac{1}{3}} \right) = \mathop \smallint \limits_{1/3}^1 4{x^3}dx\;\)

\( = \left[ {{x^4}} \right]_{\frac{1}{3}}^1\)

\( = 1 - \frac{1}{{81}} = \frac{{80}}{{81}}\)

\(P\left( {X \le \frac{2}{3}{\rm{|}}X > \frac{1}{3}} \right) = \frac{{\frac{5}{{27}}}}{{\frac{{80}}{{81}}}} = \frac{5}{{27}} \times \frac{{81}}{{80}} = \frac{3}{{16}}\)