Generalising formula for probability: at least two heads in n throws How would one generalize the formula for the event of at least 2 heads in $n$ throws? 
I've tried by taking $1$ and subtracting what does not lead to at least 2 heads, i.e. either 1 head exactly or 0 head exactly. But how does one calculate the latter, without some fancy formula? 
I know that the probability of one "string" is $(\frac{1}{2})^n$ and so presumably one would have to multiply this probability with the number of times that string occurs, yet? But how do I generalize how many times I get $0$ heads in $n$ throws and $1$ head in $n$ throws? 
I've looked at the net, and it seems something like the binomial coefficient it used... I just don't get how? How does one intuitively explain that the formula for the binomial coefficient gives me how many times I'll get x heads in n throws?
 A: To answer your question you basically have to answer how to get probability for event $m$ heads in $n$ throws. To do this think of each throw as a variable $X_{i}$, if head on $i$th throw $X_{i}=1$ and $0$ if tails. Now let $Y=\sum_{i=1}^{n}X_{i}$. Here $Y$ counts how many heads there are. Now we want to ask "What is probability of seeing $m$ heads in $n$ throws" we can translate that to finding $P(Y=m)$.
Now in order to answer this question notice that since each throw is independent, that for any particular outcome with $m$ heads (for example let $n=5$ $m=2$, particular outcome could be $HTTHT$ also stated as $X_1=1,X_2=0,X_3=0,X_4=1,X_5=0$) the probability is $\left(\frac{1}{2}\right)^{m}\left(\frac{1}{2}\right)^{n-m}$. Now all we essentially have to do is count how many different combinations of $n$ coin tosses have $m$ heads each with probability $\left(\frac{1}{2}\right)^{m}\left(\frac{1}{2}\right)^{n-m}$ of occuring which is essentially asking out of $n$ tosses how many ways can we choose $m$ of the tosses to be heads. Well this can be found by $n \choose m$. Thus finally we have 
$$P(Y=m)={n\choose m}\left(\frac{1}{2}\right)^{m}\left(\frac{1}{2}\right)^{n-m}$$ 
Now if you want at least 2 heads you would just sum probability of getting 2 heads, 3 heads, 4 heads, ..., n heads. Or you could do 1 minus sum of probability of getting 0 heads and 1 head.  
A: The probability of exactly $k$ heads in $n$ throws is $\frac{\binom{n}{k}}{2^n}$.
Therefore:


*

*The probability of $0$ heads in $n$ throws is $\frac{\binom{n}{0}}{2^n}=\frac{1}{2^n}$.

*The probability of $1$ head  in $n$ throws is $\frac{\binom{n}{1}}{2^n}=\frac{n}{2^n}$.


So the probability of at least $2$ heads in $n$ throws is $1-\frac{1}{2^n}-\frac{n}{2^n}$.
A: The binomial coefficient ${n\choose m}$ is defined to be the number of ways you can choose $m$ things from a set of $n$ things.
The reason it appears in the formula for the probability of getting exactly $m$ heads in $n$ flips is because you need to $choose$ which of the $n$ flips results in a head. In other words, there are $n\choose m$ ways to get exactly $m$ heads in $n$ flips.
A: You have the right idea.
There are $2^n$ outcomes of $n$ coin tosses.
There is one outcome with no heads.
There are $n$ outcomes with exactly one head: the first toss, or the second toss, ..., or the $n$th toss.
The rest have two or more heads.
So the probability of getting zero or one head is:
$$P(0,1) = \frac{1}{2^n} + \frac{n}{2^n}$$
and the probability of getting two or more heads, therefore, is:
$$P(\geq 2) = 1 - P(0,1) = \frac{2^n - n - 1}{2^n}.$$
$P(0)$ and $P(1)$ above are special cases; the others can be gotten with binomial coefficients:
$$P(0 \leq k \leq n) = \frac{{n \choose k}}{2^n}.$$
This can be read as "$n$ choose $k$".  Intuitively, you're choosing $k$ coins of the $n$ you have to get the heads.  If $k=2$, then you can count them explicitly as follows:


*

*First coin and any one of the following $(n-1)$ coins plus

*Second coin and any one of the following $(n-2)$ coins plus

*Third coin and any one of the following $(n-3)$ coins plus

*...

*... plus

*The $(n-1)$th coin and the $n$th coin.


This corresponds to
$$1 + 2 + 3 + ... + (n-1) = \frac{n(n-1)}{2} = {n \choose 2}$$
possibilities.
What about $k=3$?  Just build upon the expression for $k=2$:


*

*Coin 1 and any two of the remaining $n-1$ coins plus

*Coin 2 and any two of the remaining $n-2$ coins plus

*...

*... plus

*Coin $n-2$, coin $n-1$, and coin $n$


This total is
$$\frac{1}{2}\left[(n-1)(n-2) + (n-2)(n-3) + (n-3)(n-4) + ... + (2)(1)\right] \\
= \frac{1}{2}\left[(n^2 - 3n + 2) + (n^2 - 5n + 6) + (n^2 - 7n + 12) + ... + \left((n^2 - (n-2)(n-1)n + (n-2)(n-1)\right)\right] \\ 
= \frac{1}{2}\left[(n-2)n^2 - ((n-1)^2-1)n + \frac{(n-2)(n-1)n}{3}\right] \\
= \frac{n(n-1)(n-2)}{6} = {n \choose 3}.$$
We could do more, and the result would be:
$${n \choose k} = \frac{n!}{k!(n-k)!}.$$
