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

$\begin{array}{l}\mathrm{P}\left[\right(\mathrm{M}\cap \overline{\mathrm{N}})\cup (\overline{\mathrm{M}}\cap \mathrm{N}\left)\right]=\mathrm{P}(\mathrm{M}\cap \overline{\mathrm{N}})+\mathrm{P}(\overline{\mathrm{M}}\cap \mathrm{N})\to \left(1\right)\\ [\because \mathrm{M}\cap \overline{\mathrm{N}},\overline{\mathrm{M}}\cap \mathrm{N}\text{ are disjoint events }]\\ =\left[\mathrm{P}\right(\mathrm{M})-\mathrm{P}(\mathrm{M}\cap \mathrm{N}\left)\right]+\left[\mathrm{P}\right(\mathrm{N})-\mathrm{P}(\mathrm{M}\cap \mathrm{N}\left)\right]\\ =\mathrm{P}\left(\mathrm{M}\right)+\mathrm{P}\left(\mathrm{N}\right)-2\mathrm{P}(\mathrm{M}\cap \mathrm{N})\to \left(2\right)\end{array}$
$\begin{array}{l}\text{ Also }\mathrm{P}\left(\overline{\mathrm{M}}\right)+\mathrm{P}\left(\overline{\mathrm{N}}\right)-2\mathrm{P}(\overline{\mathrm{M}}\cap \overline{\mathrm{N}})\\ =(1-\mathrm{P}(\mathrm{M}\left)\right)+(1-\mathrm{P}(\mathrm{N}\left)\right)-2.[1-\mathrm{P}(\mathrm{M}\cup \mathrm{N}\left)\right]\\ [\because \overline{\mathrm{M}}\cap \overline{\mathrm{N}}=(\overline{\mathrm{M}\cup \mathrm{N}}\left)\right]\\ =2\left[\mathrm{P}\right(\mathrm{M}\cup \mathrm{N}\left)\right]-\mathrm{P}\left(\mathrm{M}\right)-\mathrm{P}\left(\mathrm{N}\right)\\ =\mathrm{P}\left(\mathrm{M}\right)+\mathrm{P}\left(\mathrm{N}\right)-2\mathrm{P}(\mathrm{M}\cap \mathrm{N})\text{ by addition theorem }\\ \text{ of probability }\end{array}$