An unbiased cubical die is thrown $2n$ times, and let $P$ denote the probability of getting even numbers at least $n$ times, then

Let $p=$probability of getting an even number on a single throw of the die

and let $X=$number of times an even number is obtained,

then $X~B(2n\cdot p)$

Now, $P=(X\ge n)$

           $\begin{array}{l}=\sum _{k-n}^{2n} P(X=k)=\sum _{k-n}^{2n} \left({ }^{2n}{C}_k\frac{1}{2^{2n}}\right)\\ =\frac{A}{2^{2n}}\end{array}$

where $A{=}^{2n}{C}_n{+}^{2n}{C}_{n+1}+\cdots {+}^{2n}{C}_{2n} \left(1\right)$

         $A{=}^{2n}{C}_n{+}^{2n}{C}_{n-1}+\cdots {+}^{2n}{C}_0 \left(2\right)$

Adding (1) and (2) we get

     $$\begin{gathered} 2A=\sum _{k=0}^{2n} \left({ }^{2n}{C}_k\right){+}^{2n}{C}_n=2^{2n}{+}^{2n}{C}_n \\ A=2^{2n-1}+\frac{1}{2}\left({ }^{2n}{C}_n\right) \end{gathered}$$

Thus, $P=\frac{1}{2}+\frac{1}{2^{2n+1}}\left({ }^{2n}{C}_n\right)\Rightarrow P>\frac{1}{2}$

    Also, ${ }^{2n}{C}_n=\frac{\left(2n\right)!}{n!n!}$

          $\begin{array}{r}=\frac{\left(1\right)\left(2\right)\left(3\right)\left(4\right)\dots (2n-1)\left(2n\right)}{n!n!}\\ =\frac{2^n(\mathrm{n}!)\left[\right(1\left)\right(3\left)\right(5)\dots (2\mathrm{n}-1\left)\right]}{n!n!}\end{array}$

         $\begin{array}{l}=\frac{2^{2n-1}}{n}\left(\frac{3}{2}\right)\left(\frac{5}{4}\right)\left(\frac{7}{6}\right)\cdots \left(\frac{2n-1}{2n-2}\right)\\ \ge \frac{2^{2n-1}}{n}\end{array}$

From (3), we get

      $P\ge \frac{1}{2}+\frac{1}{4n}$

For  $\begin{array}{r}n=4,\mathrm{P}=\frac{1}{2}+\frac{1}{2^9}\left({ }^8{C}_4\right)\\ =\frac{1}{2}+\frac{70}{2^9}=\frac{163}{256}\end{array}$

and for $n=5,P=\frac{1}{2}+\frac{1}{2^{11}}\left({ }^{10}{C}_5\right)=\frac{319}{512}$