Question

Let A and B be two events such that $P\left(A\right)\ne 0,P\left(B\right)\ne 1\mathrm{and}P\left(\overline{A}/\overline{B}\right)=\frac{1-k}{P\left(\overline{B}\right)},$ then k = ______________.

Answer 1

For two events, given $P\left(A\right)\ne 0,P\left(B\right)\ne 1\mathrm{and}P\left(\overline{A}/\overline{B}\right)=\frac{1-k}{P\left(\overline{B}\right)},$
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{P\left(\overline{A}\cap \overline{B}\right)}{P\left(\overline{B}\right)}=\frac{1-k}{P\left(\overline{B}\right)}\left[\mathrm{By}\mathrm{defination}\mathrm{of}\mathrm{conditional}\mathrm{probability}\right] \\ \mathrm{i}.\mathrm{e}P\left(\overline{A}\cap \overline{B}\right)=1-k \\ \mathrm{i}.\mathrm{e}P\left(\overline{A\cup B}\right)=1-k \\ \mathrm{i}.\mathrm{e}1-P\left(A\cup B\right)=1-k \\ \mathrm{i}.\mathrm{e}k=P\left(A\cup B\right) \end{gathered}$$