Question

If A and B are two events such that P $\left(\overline{A}\cup \overline{B}\right)=\frac{2}{3}\mathrm{and}P\left(A\cup B\right)=\frac{5}{9},\mathrm{then}P\left(\overline{A}\right)+P\left(\overline{B}\right)$ = ______________.

Answer 1

Given, for two events A and B $P\left(\overline{A}\cup B\right)=\frac{2}{3}\mathrm{and}P\left(A\cup B\right)=\frac{5}{9}$
$$\begin{gathered} \mathrm{Since}P\left(\overline{A}\cup \overline{B}\right)=P\left(\overline{A}\right)+P\left(\overline{B}\right)-P\left(\overline{A}\cap \overline{B}\right) \\ =P\left(\overline{A}\right)+P\left(\overline{B}\right)-P\left(\left(\overline{A\cup B}\right)\right) \\ \mathrm{i}.\mathrm{e}P\left(\overline{A}\cup \overline{B}\right)=P\left(\overline{A}\right)+P\left(\overline{B}\right)-\left[1-P\left(A\cup B\right)\right] \\ \mathrm{i}.\mathrm{e}P\left(\overline{A}\right)+P\left(\overline{B}\right)=P\left(\overline{A}\cup \overline{B}\right)+1-P\left(A\cup B\right) \\ =\frac{2}{3}+1-\frac{5}{9} \\ =\frac{6+9-5}{9} \\ \mathrm{i}.\mathrm{e}P\left(\overline{A}\right)+P\left(\overline{B}\right)=\frac{10}{9} \end{gathered}$$