Question

If A and B are two events such that P(A/B) = p, P(A) = p, $P\left(B\right)=\frac{1}{3}\mathrm{and}P\left(A\cup B\right)=\frac{5}{9},$ then p = _____________.

Answer 1

for events A and B, Given $P\left(\frac{A}{B}\right)=P,P\left(A\right)=P,P\left(B\right)=\frac{1}{3}\mathrm{and}P\left(A\cup B\right)=\frac{5}{9}$
$$\begin{gathered} \mathrm{i}.\mathrm{e}P\left(\frac{A}{B}\right)=P \\ \mathrm{i}.\mathrm{e}\frac{P\left(A\cap B\right)}{P\left(B\right)}=P\left[\mathrm{By}\mathrm{defination}\mathrm{of}\mathrm{conditional}\mathrm{probability}\right] \\ \mathrm{i}.\mathrm{e}P\left(A\cap B\right)=P\left(P\left(B\right)\right) \\ \therefore P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right) \\ \mathrm{i}.\mathrm{e}\frac{5}{9}=P+\frac{1}{3}-PP\left(B\right) \\ \mathrm{i}.\mathrm{e}\frac{5}{9}=P+\frac{1}{3}-P\times \frac{1}{3} \\ \mathrm{i}.\mathrm{e}\frac{5}{9}-\frac{1}{3}=P-\frac{P}{3} \\ \frac{5-3}{9}=\frac{2P}{3} \\ \mathrm{i}.\mathrm{e}\frac{2}{9}=\frac{2P}{3} \\ \mathrm{i}.\mathrm{e}P=\frac{1}{3} \end{gathered}$$