Question

A coin is tossed '$2n$' times. The chance that the number of times one gets head is not equal to the number of times one gets tail is

• A. $\frac{(2n!)}{(n!{)}^2}.{\left(\frac{1}{2}\right)}^{2n}$
• B. $1-\frac{(2n!)}{(n!{)}^2}$
• C. $1-\frac{(2n!)}{(n!{)}^2}.\frac{1}{4^n}$
• D. $\frac{(2n!)}{(n!{)}^2}\frac{1}{4^n}$

Answer 1

The correct option is C $1-\frac{(2n!)}{(n!{)}^2}.\frac{1}{4^n}$

Probability that the number of heads is equal to the number of tails (i.e. each occurs $n$ times) = $\frac{{(}_n^{2n}C)}{\left(2^n\right)\times \left(2^n\right)}=\frac{\left(2n\right)!}{{(n!)}^2}\times \frac{1}{4^n}$
Therefore, the required probability = $1-\frac{\left(2n\right)!}{{(n!)}^2}.\frac{1}{4^n}$