2 heads or more in 3 coin toss formula what is the formula to calculate the probabilities of getting 2 heads or more in 3 coin toss ?
i've seen a lot of solution but almost all of them were using method of listing all of possible combination like HHT,HTH,etc

what i am trying to ask is the formula and/or method to calculate this using a formula and
no need to list all of the possible combination.

listing 3 coin toss combination is easy(8 possible combination),but suppose i change the coins to dice or say 20-side dice. that would take a long time to list all the possible combination.
The probability of exactly $k$ heads in $n$ tosses is
$$
p_{n,k} = \left(\frac{1}{2}\right)^n \binom{n}{k}
$$
so, the probability of getting at least $k$ heads is
$$
\sum_{i=k}^n p_{n,i} = \frac{1}{2^n}\sum_{i=k}^n \binom{n}{i}
$$
or
$$
1 - \sum_{i=0}^{k-1} p_{n,i} = 1-\frac{1}{2^n}\sum_{i=0}^{k-1} \binom{n}{i}
$$
For $n=3$ and $k=2$ you get $\frac{1}{8}\left(\binom{3}{2}+\binom{3}{3}\right)=\frac 12$.
The question is about the chances of $~2~$ or more success occurring so. $~x\ge 2~$.

Not $~x=2~$  so the $~1$-binomial$(n,p,x-1)~$ function is supposed to give $~.5~$ but it doesn't, because Aliens wear pure hats.