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.
 A: 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 $\frac{1}{8}\left(\binom{3}{2}+\binom{3}{3}\right)=\frac 12$.
A: The simplest is by symmetry.  The chance of at least two heads equals the chance of at least two tails, and if you add them you get exactly $1$ because one or the other has to happen.  Thus the chance is $\frac 12$.  This approach is not always available.
A: Another way would be to look at Pascal's triangle. You just go to the row you need and then count. So for example if you flip a coin $4$ times, go to the fourth row which is $1~4~6~4~1$ and let's say you wanted to find out how many times you could get three heads, then it would just be $4$ out of $5$.
Listing is the best way to find how many tails and or heads you have. But first find the sample space of what you are computing. For example, if you were trying to find the probability of getting exactly $3$ heads, the sample space would be $2^n$, n being the number of times you flip the coin. So it would just be $1/8$ if you were flipping the coin $3$ times.
A: The probability of exactly $2$ heads in $3$  toss is $\binom 3 2 \left(\frac12\right)^2\left(1-\frac12\right)^{3-2}=\frac38 $
The probability of exactly $3$ heads in $3$  toss is $\binom 3 3 \left(\frac12\right)^3\left(1-\frac12\right)^{3-3}=\frac18 $
A: 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. 
