Answer
The probability of winning exactly once is 0.30.
Work Step by Step
Let A be the event that State’s football team wins Saturday’s game and let B be the event that State’s football team wins two weeks from now.
We are given that,
\[\begin{align}
& P(A)=0.10 \\
& P(B)=0.30 \\
\end{align}\]
And if \[({{A}^{C}}\cap {{B}^{C}})\]means losing both games, then
\[P({{A}^{C}}\cap {{B}^{C}})=0.65\]
We want to find the probability of winning exactly once; that means we need to find
\[P((A\cap {{B}^{C}})\cup (B\cap {{A}^{C}}))\].
\[P((A\cap {{B}^{C}})\cup (B\cap {{A}^{C}}))=P(A\cup B)-P(A\cap B)\]
By DeMorgan’s law,
\[P({{A}^{C}}\cap {{B}^{C}})=P{{(A\cup B)}^{C}}=0.65\]
\[\begin{align}
& P{{(A\cup B)}^{C}}=1-P(A\cup B) \\
& P(A\cup B)=1-0.65 \\
& P(A\cup B)=0.35 \\
\end{align}\]
\[\begin{align}
& P(A\cap B)=P(A)+P(B)-P(A\cup B) \\
& P(A\cap B)=0.10+0.30-0.35 \\
& P(A\cap B)=0.05 \\
\end{align}\]
Therefore,
\[\begin{align}
& P((A\cap {{B}^{C}})\cup (B\cap {{A}^{C}}))=P(A\cup B)-P(A\cap B) \\
& P((A\cap {{B}^{C}})\cup (B\cap {{A}^{C}}))=0.35-0.05 \\
& P((A\cap {{B}^{C}})\cup (B\cap {{A}^{C}}))=0.30 \\
\end{align}\]
The required probability is 0.30.
