Find the median of the following distribution
$p(x)=\frac{4!}{x!(4-x)!} {(\frac{1}{4})}^x {(\frac{3}{4})}^{4-x}$ , x=0,1,2,3,4, zero elsewhere.
In the question, the median is defined as the value of x such that $P(X<x)<\frac{1}{2}$ and $P(X\leq x)\geq\frac{1}{2}$.
I have calculated all values of P(X=x) but I am unable to understand how to find x from these values. These are the following values I have obtained: P(X=0) = 0.3164, P(X=1) = 0.4218, P(X=2) = 0.2109, P(X=3) = 0.0468, P(X=4) = 0.003906.
Also, from what I was taught in one of the class median is the value of x such that $P(X\leq x) = 0.5 = P(X\geq x)$. So, is the definition in the book incorrect?