Median of discrete and continuous random variables. If case of continuous random variables, we define median 'm' as a value such that the $P(X\ge m)=0.5$ and $P(X\le m)=0.5$.
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For discrete random variables, $P(X\ge m)\ge0.5$ and $P(X\le m)\ge0.5$
Why is it greater than or equal to in case of discrete random variables and only equals to in case of continuous random variable?
 A: For strictly continuous random variables the value $m$ is unique and always in the support (the set of possible values) of $X$, thus there is no problem to define it as $P(X\ge m) = 0.5$. In discrete random variable whether with finite or infinite countable support, it is possible that no possible value of $X$ will satisfy $P(X\ge m )=0.5$. Let us consider a simple example. Let $X \sim \mathcal{B}in (3, 1/2)$, hence we have to find $m$ such that
$
P(X \le m ) =0.5
$.
Let us check possible value of $m$
$$
P(X\le 1) = \frac{1}{2^3} + \frac{3}{2^3} = 1/2.
$$
Bingo! $1$  is the median. Now, take $p=1/4$ instead of $1/2$,
$$
P(X\le 1) = \frac{3^3}{4^3} + \frac{3}{2^3} = 0.74,
$$
that is way too high probability, however
$$
P(X\le 0) = \frac{3^3}{4^3} = 0.42.
$$
That is too low. I.e., there is no value that $X$ can get and it will satisfy 
$
P(X\le m) = 0.5,
$
hence you have to chose $0$ or $1$ as the median. Any value in $(0,1)$ will not change a thing as $X$ cannot receive these values.
A: It is possible in the discrete case for the X=m not be possible.  Simple example: toss one die, possible values of X 1,2,3,4,5,6.  Median is 3.5.
