The p.m.f. of a random variable \(X\) is given by
\(\begin{aligned}
\mathrm{P}[\mathrm{X}=x] & =\frac{\binom{5}{x}}{2^5}, \text { if } x=0,1,2,3,4,5 =0, \text { otherwise }
\end{aligned}\)
Then which of the following is not correct?

\(\begin{aligned} \mathrm{P}(\mathrm{X} \leq 1) & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1) \\ & =\frac{{ }^5 \mathrm{C}_0}{2^5}+\frac{{ }^5 \mathrm{C}_1}{2^5}=\frac{6}{2^5} \\ \mathrm{P}(\mathrm{X} \leq 2) & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\ & =\frac{{ }^5 \mathrm{C}_0}{2^5}+\frac{{ }^5 \mathrm{C}_1}{2^5}+\frac{{ }^5 \mathrm{C}_2}{2^5}=\frac{16}{2^5}\end{aligned}\)
\(\begin{aligned}
\mathrm{P}(\mathrm{X} \geq 3) & =\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\
& =\frac{{ }^5 \mathrm{C}_3}{2^5}+\frac{{ }^5 \mathrm{C}_4}{2^5}+\frac{{ }^5 \mathrm{C}_5}{2^5}=\frac{16}{2^5} \\
\mathrm{P}(\mathrm{X} \leq 3) & =\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\
& +\mathrm{P}(\mathrm{X}=3) \\
& =\frac{{ }^5 \mathrm{C}_0}{2^5}+\frac{{ }^5 \mathrm{C}_1}{2^5}+\frac{{ }^5 \mathrm{C}_2}{2^5}+\frac{{ }^5 \mathrm{C}_3}{2^5}=\frac{26}{2^5} \\
\mathrm{P}(\mathrm{X} \geq 4) & =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\
& =\frac{{ }^5 \mathrm{C}_4}{2^5}+\frac{{ }^5 \mathrm{C}_5}{2^5}=\frac{6}{2^5}
\end{aligned}\)
\(\therefore \quad \mathrm{P}(\mathrm{X} \leq 2)\gt\mathrm{P}(\mathrm{X} \geq 3)\) is not true.