Answer
0.7
Work Step by Step
Let A and B be any two events defined on S. Then,
\[\begin{align}
& P(A)=0.4 \\
& P(B)=0.5 \\
& P(A\cap B)=0.1 \\
\end{align}\]
We have to find the probability that A or B but not both occur; that means we need to find\[P((A\cup B)\cap {{(A\cap B)}^{C}})\]as:
\[\begin{align}
& P((A\cup B)\cap {{(A\cap B)}^{C}})=P((A\cup {{B}^{C}})\cup (B\cap {{A}^{C}})\cup (A\cap {{A}^{C}})\cup (B\cap {{B}^{C}})) \\
& P((A\cup B)\cap {{(A\cap B)}^{C}})=P(A\cup {{B}^{C}})+(B\cap {{A}^{C}})+(A\cap {{A}^{C}})+(B\cap {{B}^{C}}) \\
& P((A\cup B)\cap {{(A\cap B)}^{C}})=[P(A)-P(A\cap B)]+[P(B)-P(A\cap B)]+0+0 \\
& P((A\cup B)\cap {{(A\cap B)}^{C}})=P(A)+P(B)-2P(A\cap B) \\
& P((A\cup B)\cap {{(A\cap B)}^{C}})=0.4+0.5-2(0.1) \\
& P((A\cup B)\cap {{(A\cap B)}^{C}})=0.7 \\
\end{align}\]
Hence, the probability is 0.7.
