The p.m.f of a random variable \(\mathrm{X}\) is
\(P(X=x)=\frac{1}{2^5}\left(\begin{array}{l}5 \ x\end{array}\right), \quad x=0,12345\) then \(=0 \quad\) otherwise,

\(\mathrm{P}(\mathrm{X}=\mathrm{x})=\frac{1}{32}{ }^5 \mathrm{C}_{\mathrm{x}} \text {, where } \mathrm{x}=0,1,2,3,4,5 \)
\( =0, \text { otherwise } \)
\( \therefore \mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \)
\( =\frac{1}{32}\left[{ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_1+{ }^5 \mathrm{C}_2\right]=\frac{1}{32}(1+5+10)=\) \(\frac{16}{32}=\frac{1}{2} \)
\( \mathrm{P}(\mathrm{X} \geq 3)=\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \)
\( =\frac{1}{32}\left[{ }^5 \mathrm{C}_3+{ }^5 \mathrm{C}_4+{ }^5 \mathrm{C}_5\right]=\frac{1}{32}(10+5+1)=\) \(\frac{16}{32}=\frac{1}{2}\)