Answer
The probability that either A or B but not both will occur is 0.3.
Work Step by Step
Let A and B be any two events defined on S such that,
\[\begin{align}
& P{{(A\cup B)}^{C}})=0.5 \\
& P(A\cap B)=0.2 \\
\end{align}\]
By DeMorgan’s law,
\[\begin{align}
& P{{(A\cup B)}^{C}}=1-P(A\cup B) \\
& P(A\cup B)=1-P{{(A\cup B)}^{C}} \\
& P(A\cup B)=1-0.5 \\
& P(A\cup B)=0.5 \\
\end{align}\]
We have to find the probability that either A or B but not both will occur; that means to find \[P((A\cup B)\cap {{(A\cap B)}^{C}})\]as:
\[\begin{align}
& P((A\cup B)\cap {{(A\cap B)}^{C}})=P(A\cup {{B}^{C}})+P(B\cap {{A}^{C}}) \\
& P((A\cup B)\cap {{(A\cap B)}^{C}})=P(A)+P(B)-2P(A\cap B) \\
\end{align}\]
Since,
\[\begin{align}
& P(A\cup B)=P(A)+P(B)-P(A\cap B) \\
& 0.5=P(A)+P(B)-0.2 \\
& P(A)+P(B)=0.7 \\
\end{align}\]
Therefore,
\[\begin{align}
& P((A\cup B)\cap {{(A\cap B)}^{C}})=P(A)+P(B)-2P(A\cap B) \\
& P((A\cup B)\cap {{(A\cap B)}^{C}})=0.7-2(0.2) \\
& P((A\cup B)\cap {{(A\cap B)}^{C}})=0.3 \\
\end{align}\]
The probability that either A or B but not both will occur is 0.3.
