Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If}\mathrm{P}\left(B\right)=\frac{3}{5},\mathrm{P}\left(A|B\right)=\frac{1}{2}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{4}{5},\mathrm{then}\mathrm{P}\left(\overline{A\cup B}\right)+\mathrm{P}\left(\overline{A}\cup B\right)= \\ \left(\mathrm{a}\right)\frac{1}{5}\left(\mathrm{b}\right)\frac{4}{5}\left(\mathrm{c}\right)\frac{1}{2}\left(\mathrm{d}\right)1 \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(B\right)=\frac{3}{5},\mathrm{P}\left(A|B\right)=\frac{1}{2}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{4}{5} \\ \mathrm{As},\mathrm{P}\left(A|B\right)=\frac{1}{2} \\ \Rightarrow \frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(B\right)}=\frac{1}{2} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{2}\times \mathrm{P}\left(B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{2}\times \frac{3}{5} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{3}{10} \\ \mathrm{As},\mathrm{P}\left(A\cup B\right)=\frac{4}{5} \\ \Rightarrow \mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)=\frac{4}{5} \\ \Rightarrow \mathrm{P}\left(A\right)+\frac{3}{5}-\frac{3}{10}=\frac{4}{5} \\ \Rightarrow \mathrm{P}\left(A\right)+\frac{3}{10}=\frac{4}{5} \\ \Rightarrow \mathrm{P}\left(A\right)=\frac{4}{5}-\frac{3}{10} \\ \Rightarrow \mathrm{P}\left(A\right)=\frac{5}{10} \\ \Rightarrow \mathrm{P}\left(A\right)=\frac{1}{2} \\ \mathrm{Now}, \\ \mathrm{P}\left(\overline{A\cup B}\right)+\mathrm{P}\left(\overline{A}\cup B\right)=\left[1-\mathrm{P}\left(A\cup B\right)\right]+\left[1-\mathrm{P}\left(\mathrm{Only}A\right)\right] \\ =\left[1-\frac{4}{5}\right]+1-\left[\mathrm{P}\left(A\right)-\mathrm{P}\left(A\cap B\right)\right] \\ =\frac{1}{5}+1-\left[\frac{1}{2}-\frac{3}{10}\right] \\ =\frac{6}{5}-\frac{2}{10} \\ =\frac{10}{10} \\ =1 \end{gathered}$$

Hence, the correct alternative is option (d).