If $M$ and $N$ are any two events, the probability that the exactly one of them occurs is

$P$(exactly one of $M,N$ occurs)

$=P\left\{ \left( M\cap { N }^{ C } \right) \cup \left( { M }^{ C }\cap N \right) \right\}=P\left( M\cap { N }^{ C } \right) +\left( { M }^{ C }\cap N \right)$

$=P\left( M \right) -P\left( M\cap N \right) +P\left( N \right) -P\left( M\cap N \right)=P\left( M \right) +P\left( N \right) -2P\left( M\cap N \right)$

also,$P$(exactly one of them occurs)

$=\left\{ 1-P\left( { M }^{ C }\cap { N }^{ C } \right) \right\} \left\{ 1-P\left( { M }^{ C }\cup { N }^{ C } \right) \right\}$

$=P\left( { M }^{ C }\cup { N }^{ C } \right) -P\left( { M }^{ C }\cap { N }^{ C } \right)=P\left( { M }^{ C } \right) +P\left( { N }^{ C } \right) -2P\left( { M }^{ C }\cap { N }^{ C } \right)$