Answer
\[P(E|(A\cup B))=0.75\]
Work Step by Step
Let A and B be two events such that
\[\begin{align}
& P{{(A\cup B)}^{C}}=0.6 \\
& P(A\cup B)=1-P{{(A\cup B)}^{C}} \\
& P(A\cup B)=1-0.6 \\
& P(A\cup B)=0.4 \\
\end{align}\]
\[P(A\cap B)=0.1\]
Event E is the event that exactly one occurs.
The portion of A, included in E is \[A\cap {{B}^{C}}\]. Similarly, the portion of B included in E is \[B\cap {{A}^{C}}\] It follows that E can be written as a union:
\[E=(A\cap {{B}^{C}})\cup (B\cap {{A}^{C}})\]
\[\begin{align}
& P(E)=P(A\cap {{B}^{C}})+P(B\cap {{A}^{C}}) \\
& P(E)=P(A\cup B)-P(A\cap B) \\
& P(E)=0.4-0.1 \\
& P(E)=0.3 \\
\end{align}\]
\[\begin{align}
& P(E|(A\cup B))=\frac{P(E\cap (A\cup B))}{P(A\cup B)} \\
& P(E|(A\cup B))=\frac{P((A\cap {{B}^{C}})\cup (B\cap {{A}^{C}})\cap (A\cup B))}{P(A\cup B)} \\
& P(E|(A\cup B))=\frac{P((A\cap {{B}^{C}})\cup (B\cap {{A}^{C}}))}{P(A\cup B)} \\
& P(E|(A\cup B))=\frac{P(E)}{P(A\cup B)} \\
& P(E|(A\cup B))=\frac{0.3}{0.4} \\
& P(E|(A\cup B))=0.75 \\
\end{align}\]
Therefore \[P(E|(A\cup B))=0.75\]
