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

Oct 16, 2015

$\frac{11}{16}$

#### Explanation:

Consider a general task of flipping N coins and the probability of exactly K times the heads are up. Let's use a symbol $P \left(N , K\right)$ for this probability.
Knowing this, we can use the result to evaluate
$P \left(4 , 2\right) + P \left(4 , 3\right) + P \left(4 , 4\right)$
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):
$P \left(4 , 2\right) + P \left(4 , 3\right) + P \left(4 , 4\right) =$

$= \frac{1}{2} ^ 4 \cdot \left[\frac{4 \cdot 3 \cdot 2 \cdot 1}{\left(1 \cdot 2\right) \cdot \left(1 \cdot 2\right)} + \frac{4 \cdot 3 \cdot 2 \cdot 1}{\left(1 \cdot 2 \cdot 3\right) \cdot \left(1\right)} + \frac{4 \cdot 3 \cdot 2 \cdot 1}{\left(1 \cdot 2 \cdot 3 \cdot 4\right) \cdot 1}\right] =$

$= \frac{6 + 4 + 1}{16} = \frac{11}{16}$