Answer
\[{{p}_{2}}=\frac{5}{8}\]
Work Step by Step
${{A}_{1}}$ and ${{A}_{2}}$ exist such that ${{A}_{1}}\cup {{A}_{2}}=S$ and ${{A}_{1}}\cap {{A}_{2}}=\varnothing $
\[\begin{align}
& P({{A}_{1}})={{p}_{1}} \\
& P({{A}_{2}})={{p}_{2}} \\
& 3{{p}_{1}}-{{p}_{2}}=\frac{1}{2} \\
\end{align}\]
We know that,
\[\begin{align}
& P({{A}_{1}})+P({{A}_{2}})=P(S) \\
& {{p}_{1}}+{{p}_{2}}=1 \\
\end{align}\]
So, put\[{{p}_{1}}=1-{{p}_{2}}\] in the equation \[3{{p}_{1}}-{{p}_{2}}=\frac{1}{2}\] we get
\[\begin{align}
& \frac{1}{2}=3(1-{{p}_{2}})-{{p}_{2}} \\
& \frac{1}{2}=3-3{{p}_{2}}-{{p}_{2}} \\
& \frac{1}{2}=3-4{{p}_{2}} \\
& {{p}_{2}}=\frac{5}{8} \\
\end{align}\]
