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.
$$\frac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}.{\left( {\frac{1}{2}} \right)^{2n}}$$
B.
$$1 - \frac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}$$
C.
$$1 - \frac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}.\frac{1}{{{4^n}}}$$
D.
none of these
Answer :
$$1 - \frac{{\left( {2n} \right)!}}{{{{\left( {n!} \right)}^2}}}.\frac{1}{{{4^n}}}$$
Solution :
The required probability
$$ = 1 - $$ probability of equal number of heads and tails
$$\eqalign{
& = 1 - {}^{2n}{C_n}.{\left( {\frac{1}{2}} \right)^n}.{\left( {\frac{1}{2}} \right)^{2n - n}} \cr
& = 1 - \frac{{\left( {2n} \right)!}}{{\left( {n!} \right)\left( {n!} \right)}}.{\left( {\frac{1}{4}} \right)^n} \cr} $$