Statement-1: The probability of getting a tail most of  the times in 10 tosses of a unbiased coin is $\frac{1}{2}\begin{Bmatrix}1-\frac{10!}{2^{10}5!5!}\end{Bmatrix}$

Statement-2: ${^{2n}C}_0+{^{2n}C}_1 + {^{2n}C}_2 +...+ {^{2n}C}_n = 2^{2n-1}, n \,\, ∈ \,\, N $

Statement 1 is True, Statement 2 is False.

Let p be the probability of getting a tail in a single trial. Then, $p= \frac{1}{2}=q.$ Let X denote the number of tails in 10 trials. Then, 

$P(X=r)= {^{10}C}_r\left(\frac{1}{2}\right)^{10}$

∴ Probability of getting a tail most of the times $= P(X ≥ 6)$

$= \sum\limits^{10}_{r=6}P(X=r) = \left(\frac{1}{2}\right)^{10}\begin{Bmatrix} {^{10}C}_6 + {^{10}C}_7+{^{10}C}_8 +{^{10}C}_9 +{^{10}C}_{10}\end{Bmatrix}$

$= \left(\frac{1}{2}\right)^{11}\begin{Bmatrix} {^{10}C}_0 + {^{10}C}_1+... +{^{10}C}_{10} +{^{10}C}_{5}\end{Bmatrix}$

$= \left(\frac{1}{2}\right)^{11}\begin{Bmatrix}2^{10}-\frac{10!}{5!5!}\end{Bmatrix}=\frac{1}{2}\begin{Bmatrix}2^{10}-\frac{10!}{2^{10}5!5!}\end{Bmatrix}$

So, statement-1 is true.

We have,

${^{2n}C}_0 + {^{2n}C}_1 + {^{2n}C}_2 +... + {^{2n}C}_n + {^{2n}C}_{n+1} + ...+ {^{2n}C}_{2n} = 2^{2n}$

$⇒2 \left({^{2n}C}_0 + {^{2n}C}_1 + {^{2n}C}_2 +... + {^{2n}C}_{n-1}\right) + {^{2n}C}_{2n} = 2^{2n}$

$⇒ {^{2n}C}_0 + {^{2n}C}_1 + {^{2n}C}_2 +... + {^{2n}C}_{n-1}= 2^{2n-1}-\frac{1}{2}{^{2n}C}_n $

So, staement-2 is true.