We define p-coin as having a $p$ probability to land on Tails and $1-p$ to land on Heads.
$X$ is a random variables that given $n$-flips results, gives the number of tails ($T$) that we got:
$$X(x_1x_2 \ldots x_n) = \bigl| \{i: x_i = T \} \bigr| ~.$$
What is the probability of event $ A = ( X \geq {n}/{2}) $? Namely, the probability of the set of $ \{ x_1x_2 \ldots x_n \}$ such that $X(x_1x_2 \ldots x_n) \geq n/2$.
How does one go calculating this? Should I try to calculate the opposing case and reduce it from $1$? Whats the procedure here?
It is binomial distribution: $n$ independent trials, with only two outcomes.
We know the probability of head is $1-p$ and that of a tail is $p$. So $$\Pr(X=k)=C^n_k p^k (1-p)^{n-k}$$ Hence the required probability is $$\Pr(X=n/2) + \Pr(X =n/2+1)+ \dotsb + \Pr(X=n) \\ =\frac12 \left( 1+C^n_{n/2}p^{n/2}(1-p)^{n/2} \right)$$