If X is a continuous random variable whose probability density function is given by \[f(x) = \left\{ \begin{array}{ll} k(5x - 2x^2) & \mbox{if } 0 \leq x \leq 2 \\ 0 & \mbox{otherwise,} \end{array} \right.\] then find $P(X > 1)$.

If X is a continuous random variable whose probability density function is given by \[f(x) = \left\{ \begin{array}{ll} k(5x - 2x^2) & \mbox{if } 0 \leq x \leq 2 \\ 0 & \mbox{otherwise,} \end{array} \right.\] then $P(X > 1)$ is 

(A) $\dfrac{3}{14}$ 

(B) $\dfrac{4}{5}$ 

(C) $\dfrac{14}{7}$ 

(D) $\dfrac{17}{28}$ 

Answer: (D) $\dfrac{17}{28}$ 

Explanation: 

$\int\limits_{-\infty}^{+\infty}f(x)dx = 1$ 

$\Rightarrow \int\limits_{0}^{2}k(5x-2x^2)dx = 1$ 

$\Rightarrow k = \dfrac{3}{14}$ 

Now, $P(X > 1)$ 

$= \int\limits_{1}^{\infty}f(x)dx$ 

$= \int\limits_{1}^{2} \dfrac{3}{14}(5x - 2x^2)dx$ 

$= \dfrac{17}{28}$