Question

If A and B are independent events, then write expression for P(exactly one of A, B occurs).

Answer 1

$$\begin{gathered} \mathrm{As},A\mathrm{and}B\mathrm{are}\mathrm{independent}\mathrm{events}. \\ \mathrm{So},\mathrm{P}\left(A\cap B\right)=\mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right).....\left(\mathrm{i}\right) \\ \mathrm{Now}, \\ \mathrm{P}\left(\mathrm{exactly}\mathrm{one}\mathrm{of}A,B\mathrm{occurs}\right)=\mathrm{P}\left(\mathrm{only}A\right)+\mathrm{P}\left(\mathrm{only}B\right) \\ =\left[\mathrm{P}\left(A\right)-\mathrm{P}\left(A\cap B\right)\right]+\left[\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)\right] \\ =\left[\mathrm{P}\left(A\right)-\mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right)\right]+\left[\mathrm{P}\left(B\right)-\mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right)\right]\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ =\mathrm{P}\left(A\right)\times \left[1-\mathrm{P}\left(B\right)\right]+\mathrm{P}\left(B\right)\times \left[1-\mathrm{P}\left(A\right)\right] \\ =\mathrm{P}\left(A\right)\times \mathrm{P}\left(\overline{B}\right)+\mathrm{P}\left(B\right)\times \mathrm{P}\left(\overline{\mathit{A}}\right) \end{gathered}$$