A die is thrown (2n+1) times. The probability of getting 1 or 3 or 4 atmost n times is

Given a dice is thrown (2n+1) times.

The Probability of getting 1, 3 or 4 could be $\mathrm{p}=\frac{3}{6}=\frac{1}{2}$

The Probability of not getting 1, 3 or 4 would be $\mathrm{q}=\frac{3}{6}=\frac{1}{2}$

The Probability of getting 1, 3 or 4 at most n times is

$$\begin{gathered} ={}^{2\mathrm{n}+1}{\mathrm{c}}_0 {\left(\mathrm{p}\right)}^0 {\left(\mathrm{q}\right)}^{2\mathrm{n}+1-0}+{}^{2\mathrm{n}+1}{\mathrm{c}}_1 {\left(\mathrm{p}\right)}^1 {\left(\mathrm{q}\right)}^{2\mathrm{n}} \\ +.....+{}^{2\mathrm{n}+1}{\mathrm{c}}_{\mathrm{n}} {\mathrm{p}}^{\mathrm{n}} {\left(\mathrm{q}\right)}^{2\mathrm{n}+1-\mathrm{n}}+{}^{2\mathrm{n}+1}{\mathrm{c}}_{2\mathrm{n}+1} {\mathrm{p}}^{2\mathrm{n}+1} {\mathrm{q}}^{2\mathrm{n}+1-(2\mathrm{n}+1)} \\ ={}^{2\mathrm{n}+1}{\mathrm{c}}_0 {\left(\frac{1}{2}\right)}^0 {\left(\frac{1}{2}\right)}^{2\mathrm{n}+1}+{}^{2\mathrm{n}+1}{\mathrm{c}}_1 {\left(\frac{1}{2}\right)}^1 {\left(\frac{1}{2}\right)}^{2\mathrm{n}} \\ +.......+{}^{2\mathrm{n}+1}{\mathrm{c}}_{2\mathrm{n}+1} {\left(\frac{1}{2}\right)}^{2\mathrm{n}+1} {\left(\frac{1}{2}\right)}^0 \\ ={\left(\frac{1}{2}\right)}^{2\mathrm{n}+1} \left({}^{2\mathrm{n}+1}{\mathrm{c}}_0+{}^{2\mathrm{n}+1}{\mathrm{c}}_1+......+{}^{2\mathrm{n}+1}{\mathrm{c}}_{\mathrm{n}}\right) \\ =\frac{1}{2^{2\mathrm{n}+1}}\times {\left(2\right)}^{2\mathrm{n}} \\ =\frac{1}{2} \end{gathered}$$