If you flip a fair coin four times, what is the probability that you get heads at least twice?

Consider a general task of flipping N coins and the probability of exactly K times the heads are up. Let's use a symbol `P(N,K)` for this probability.
Knowing this, we can use the result to evaluate
`P(4,2)+P(4,3)+P(4,4)`
which will answer the question of what is the probability of getting heads at lease 2 times out of flipping a coin 4 times.

Since there are only `2` outcomes from a single flip, head or tail, for N flips we can get `2^N` different outcomes.
The outcomes we are interested in are those that contain exactly `K` heads and `N-K` tails in any order. That is where combinatorics will come handy.

Any outcome of the random experiment of flipping a coin N times can be represented as a string of N characters, each one being a letter H (to designate that the corresponding flip resulted in a head) or T (if it was a tail).

The number of outcomes with exactly `K` heads out of `N` flips is the number of strings of the length N consisting of characters H and T, where H occurs `K` times and T occurs `N-K` times in any order.
This number is, obviously, a number of combinations of K items out of N, which symbolically is represented as `C_N^K` (there are other notations as well) and is equal to
`C_N^K = (N!)/(K!*(N-K)!)`
For all the theory behind this and other formulas of combinatorics we can refer you to a corresponding part of the advanced course of mathematics for high school at Unizor.

The probability of having `K` heads out of `N` flips is equal to the ratio of the number of "successful" outcomes (those with exactly `K` heads) to a total number of outcomes mentioned above:
`P(N,K) = C_N^K/2^N = (N!)/(K!*(N-K)!*2^N)`

Now we can calculate the probability of at least two heads out of four flips (don't forget that `0! =1` by definition):
`P(4,2)+P(4,3)+P(4,4) =`

`= 1/2^4*[(4*3*2*1)/((1*2)*(1*2))+(4*3*2*1)/((1*2*3)*(1))+(4*3*2*1)/((1*2*3*4)*1)] =`

`=(6+4+1)/16 = 11/16`