Answer
\[P(A)=\frac{2}{3}\]
\[P(B)=\frac{1}{3}\]
Work Step by Step
Given,
\[P(A|B)=0.6\]
\[\text{P(At least one of the events occurs)}=P(A\cup B)=0.8\]\[\text{P(Exactly one of the events occurs)}=P(A)+P(B)=0.6\]
\[\begin{align}
& P(A\cup B)=P(A)+P(B)-P(A\cap B) \\
& P(A\cap B)=P(A)+P(B)-P(A\cup B) \\
& P(A\cap B)=0.8-0.6 \\
& P(A\cap B)=0.2 \\
\end{align}\]
\[\begin{align}
& P(A|B)=\frac{P(A\cap B)}{P(B)} \\
& 0.6=\frac{0.2}{P(B)} \\
& P(B)=\frac{1}{3} \\
\end{align}\]
\[\begin{align}
& P(A\cup B)=P(A)+P(B)-P(A\cap B) \\
& 0.8=P(A)+\frac{1}{3}-0.2 \\
& P(A)=\frac{2}{3} \\
\end{align}\]
