A fair coin is tossed $2n$ times. The probability of getting as many heads in the first $n$ tosses as in the last $n$ is
- $\displaystyle \frac{2n_{C_{n}}}{2^{2n}}$
- $\displaystyle \frac{2n_{C_{n-1}}}{2^{n}}$
- $\displaystyle \frac{n}{2^{n}}$
- $\displaystyle \frac{n^{2}}{2^{n}}$
The total number of outcomes is 2^2n. The number of ways to have n heads in the first n tosses and n heads in the last n tosses is (nCn) * (nCn) = 1. However, the question implies having an equal number of heads in both halves, which is equivalent to choosing n positions out of 2n total positions for the heads, represented by 2nCn.
Let X be the number of heads in the first n tosses and Y be the number of heads in the last n tosses, both of which follow a binomial distribution. The required probability is the sum of the probabilities that X equals Y for all values k from 0 to n, calculated as the sum of (nCk / 2^n) * (nCk / 2^n). Using the identity that the sum of the squares of binomial coefficients (nCk)^2 equals 2nCn, this sum simplifies to 2nCn / 2^(2n).