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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.

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In a number of important cases, there are relatively simple formulas that will do the counting for you, so you do not need to make a list and then count. However, it is a good idea to do the counting yourself in a number of "small" cases, so that you will know what is going on. –  André Nicolas Dec 29 '12 at 17:57
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3 Answers

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.

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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 $

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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 $\frac{1}{8}\left(\binom{3}{2}+\binom{3}{3}\right)=\frac 12$.

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