Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If}A\mathrm{and}B\mathrm{are}\mathrm{two}\mathrm{events}\mathrm{such}\mathrm{that}\mathrm{P}\left(A|B\right)=p,\mathrm{P}\left(A\right)=p,\mathrm{P}\left(B\right)=\frac{1}{3}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{5}{9},\mathrm{then}p= \\ \left(\mathrm{a}\right)\frac{2}{3}\left(\mathrm{b}\right)\frac{3}{5}\left(\mathrm{c}\right)\frac{1}{3}\left(\mathrm{d}\right)\frac{3}{4} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A|B\right)=p,\mathrm{P}\left(A\right)=p,\mathrm{P}\left(B\right)=\frac{1}{3}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{5}{9} \\ \mathrm{As},\mathrm{P}\left(A|B\right)=p \\ \Rightarrow \frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=p \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=p\times \mathrm{P}\left(B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=p\times \frac{1}{3} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{p}{3} \\ \mathrm{Now}, \\ \mathrm{P}\left(A\cup B\right)=\frac{5}{9} \\ \Rightarrow \mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)=\frac{5}{9} \\ \Rightarrow p+\frac{1}{3}-\frac{p}{3}=\frac{5}{9} \\ \Rightarrow p-\frac{p}{3}=\frac{5}{9}-\frac{1}{3} \\ \Rightarrow \frac{2p}{3}=\frac{2}{9} \\ \Rightarrow p=\frac{2}{9}\times \frac{3}{2} \\ \therefore p=\frac{1}{3} \end{gathered}$$

Hence, the correct alternative is option (c).