Question

If A and B are independent events such that P(A) = p, P(B) = 2p and P(Exactly one of A and B occurs) = $\frac{5}{9}$, then find the value of p.

Answer 1

$$\begin{gathered} \mathrm{As},A\mathrm{and}B\mathrm{are}\mathrm{independent}\mathrm{events}. \\ \mathrm{So},\mathrm{P}\left(A\cap B\right)=\mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right).....\left(\mathrm{i}\right) \\ \mathrm{Now}, \\ \mathrm{P}\left(\mathrm{Exactly}\mathrm{one}\mathrm{of}A\mathrm{and}B\mathrm{occurs}\right)=\frac{5}{9} \\ \Rightarrow \mathrm{P}\left(\mathrm{Only}A\right)+\mathrm{P}\left(\mathrm{Only}B\right)=\frac{5}{9} \\ \Rightarrow \left[\mathrm{P}\left(A\right)-\mathrm{P}\left(A\cap B\right)\right]+\left[\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)\right]=\frac{5}{9} \\ \Rightarrow \left[\mathrm{P}\left(A\right)-\mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right)\right]+\left[\mathrm{P}\left(B\right)-\mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right)\right]=\frac{5}{9}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow \mathrm{P}\left(A\right)\times \left[1-\mathrm{P}\left(B\right)\right]+\mathrm{P}\left(B\right)\times \left[1-\mathrm{P}\left(A\right)\right]=\frac{5}{9} \\ \Rightarrow p\left[1-2p\right]+2p\left[1-p\right]=\frac{5}{9} \\ \Rightarrow p-2p^2+2p-2p^2=\frac{5}{9} \\ \Rightarrow 3p-4p^2=\frac{5}{9} \\ \Rightarrow 27p-36p^2=5 \\ \Rightarrow 36p^2-27p+5=0 \\ \mathrm{So},\mathrm{using}\mathrm{quadratic}\mathrm{formula},\mathrm{we}\mathrm{get} \\ p=\frac{-\left(-27\right)\pm \sqrt{{\left(-27\right)}^2-4\times 36\times 5}}{2\times 36} \\ =\frac{27\pm \sqrt{729-720}}{72} \\ =\frac{27\pm \sqrt{9}}{72} \\ =\frac{27\pm 3}{72} \\ \Rightarrow p=\frac{27+3}{72}\mathrm{or}p=\frac{27-3}{72} \\ \Rightarrow p=\frac{30}{72}\mathrm{or}p=\frac{24}{72} \\ \therefore p=\frac{5}{12}orp=\frac{1}{3} \end{gathered}$$