Question

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

• A. $\frac{2n_{C_n}}{2^{2n}}$
• B. $\frac{2n_{C_{n-1}}}{2^n}$
• C. $\frac{n}{2^n}$
• D. $\frac{n^2}{2^n}$

Answer 1

The correct option is A $\frac{2n_{C_n}}{2^{2n}}$

Let there be '$k$' heads in the first $n$ tosses, where $k=1,2,...n$
Thus, required probability

$=\frac{{{(}^n{C}_0)}^2}{2^n\times 2^n}+\frac{{{(}^n{C}_1)}^2}{2^n\times 2^n}+\frac{{{(}^n{C}_2)}^2}{2^n\times 2^n}+\frac{{{(}^n{C}_3)}^2}{2^n\times 2^n}+...+\frac{{{(}^n{C}_n)}^2}{2^n\times 2^n}$

$=\frac{{{(}^n{C}_0)}^2+{{(}^n{C}_1)}^2+...+{{(}^n{C}_n)}^2}{2^n\times 2^n}=\frac{{}^{2n}{C}_n}{2^{2n}}$

Hence, option 'A' is correct.