Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If}A\mathrm{and}B\mathrm{are}\mathrm{such}\mathrm{that}\mathrm{P}\left(A\cup B\right)=\frac{5}{9}\mathrm{and}\mathrm{P}\left(\overline{A}\cup \overline{B}\right)=\frac{2}{3},\mathrm{then}\mathrm{P}\left(\overline{A}\right)+\mathrm{P}\left(\overline{B}\right)= \\ \left(\mathrm{a}\right)\frac{9}{10}\left(\mathrm{b}\right)\frac{10}{9}\left(\mathrm{c}\right)\frac{8}{9}\left(\mathrm{d}\right)\frac{9}{8} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\cup B\right)=\frac{5}{9}\mathrm{and}\mathrm{P}\left(\overline{A}\cup \overline{B}\right)=\frac{2}{3} \\ \mathrm{As},\mathrm{P}\left(\overline{A}\cup \overline{B}\right)=\frac{2}{3} \\ \Rightarrow \mathrm{P}\left(\overline{A\cap B}\right)=\frac{2}{3} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=1-\frac{2}{3} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{3} \\ \mathrm{Also}, \\ \mathrm{P}\left(A\cup B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ \Rightarrow \mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)=P\left(A\cup B\right)+P\left(A\cap B\right) \\ =\frac{5}{9}+\frac{1}{3} \\ \Rightarrow \mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)=\frac{8}{9}.....\left(\mathrm{i}\right) \\ \mathrm{Now}, \\ \mathrm{P}\left(\overline{A}\right)+\mathrm{P}\left(\overline{B}\right)=1-\mathrm{P}\left(A\right)+1-\mathrm{P}\left(B\right) \\ =2-\left[\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)\right] \\ =2-\frac{8}{9}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ =\frac{10}{9} \end{gathered}$$

Hence, the correct alternative is option (b).