## Wednesday, April 22, 2015

### Probability: Coin Tossing

A while back I helped a student with a probability problem, and I took the problem, generalized it a bit, and wish to post it here. Here is the problem:

A coin has the probability $p$ of turning up heads when tossed. Suppose we toss the coin $2n$ times, where $n$ is a natural number. Compute the probability that the total number of heads is even.

Let $P(X)$ be the probability that the total number of heads is even and $P(Y)$ be the probability that the total number of heads is odd.

Because we know it is certain that the total number of heads is even OR odd, we may state:

$P(X)+P(Y)=1$

Using the binomial probability formula, and the binomial theorem, we get:

$P(X)-P(Y)=\sum_{k=0}^{2n}\left[{2n \choose k}p^{2n-k}(p-1)^k \right]=(2p-1)^{2n}$

$2P(X)=1+(2p-1)^{2n}$
$P(X)=\frac{1}{2}\left(1+(2p-1)^{2n} \right)$