Probability that X is greater than the mode of X?

Question

How do I solve this problem:

Let X be a continuous random variable with density function \begin{equation} f(x) = \begin{cases} \hfill ax^2e^{-10x} \hfill & \text{for x $\geq$ 0} \\ \hfill 0 \hfill & \text{otherwise} \\ \end{cases} \end{equation} where a > 0. What is the probability of X greater than or equal to the mode of X?

I got stuck in integrating $$\int_{0}^{\infty} ax^2e^{-10x} dx = 1$$ and onwards. — Mestica, May 22 '16 at 07:47
@Ralf17 Good idea. Apply the method of integration by parts with $ u=x^2, v'=e^{-10x}$ — callculus42, May 22 '16 at 07:50

Olivier Oloa · Accepted Answer · 2016-05-22T08:46:14.743

What is the probability of $X$ greater than or equal to the mode of $X$?

Hint. One may find the maximum of $f$ over $\mathbb{R}$, using \begin{equation} f'(x) = \begin{cases} \hfill -2ax(5x-1)e^{-10x} \hfill & \text{for $x \geq$ 0} \\\\ \hfill 0. \hfill & \text{otherwise} \\ \end{cases} \end{equation} Then one may evaluate $$ P\left(X\geq \frac15\right)=\int_{1/5}^\infty-2ax(5x-1)e^{-10x}\:dx. $$ The latter integral may be computed by parts.

Probability that X is greater than the mode of X?

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