Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{Let}A\mathrm{and}B\mathrm{be}\mathrm{two}\mathrm{events}.\mathrm{If}\mathrm{P}\left(A\right)=0.2,\mathrm{P}\left(B\right)=0.4,\mathrm{P}\left(A\cup B\right)=0.6,\mathrm{then}\mathrm{P}\left(A|B\right)\mathrm{is}\mathrm{equal}\mathrm{to} \\ \left(\mathrm{a}\right)0.8\left(\mathrm{b}\right)0.5\left(\mathrm{c}\right)0.3\left(\mathrm{d}\right)0 \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=0.2,\mathrm{P}\left(B\right)=0.4\mathrm{and}\mathrm{P}\left(A\cup B\right)=0.6 \\ \mathrm{As},\mathrm{P}\left(A\cup B\right)=0.6 \\ \Rightarrow \mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)=0.6 \\ \Rightarrow 0.2+0.4-\mathrm{P}\left(A\cap B\right)=0.6 \\ \Rightarrow 0.6-\mathrm{P}\left(A\cap B\right)=0.6 \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=0.6-0.6 \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=0 \\ \mathrm{Now}, \\ \mathrm{P}\left(A|B\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{0}{0.4}=0 \end{gathered}$$

Hence, the correct alternative is option (d).