Q. A coin is tossed 2$n$ 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) None of these
Explanation
(c)
The required probability $= 1 -$ probability of equal number of heads and tails.
Out of $2n$ tossed $n$ times heads and $n$ times tails.
$= 1 - {\ ^{2n}C}_{n}\left( \frac{1}{2} \right)^{n}\left( \frac{1}{2} \right)^{2n - n\ }$
$= 1 - \frac{(2n)!}{n!n!}\left( \frac{1}{2} \right)^{2n} = 1 - \frac{(2n)!}{(n!)^{2}} \bullet \frac{1}{4^{n}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.