Question

Choose the correct alternative in the following question:

Associated to a random experiment two events A and B are such that $\mathrm{P}\left(B\right)=\frac{3}{5},\mathrm{P}\left(A|B\right)=\frac{1}{2}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{4}{5}$. The value of P(A) is

$\left(\mathrm{a}\right)\frac{3}{10}\left(\mathrm{b}\right)\frac{1}{2}\left(\mathrm{c}\right)\frac{1}{10}\left(\mathrm{d}\right)\frac{3}{5}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(B\right)=\frac{3}{5},\mathrm{P}\left(A|B\right)=\frac{1}{2}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{4}{5} \\ \mathrm{As},\mathrm{P}\left(A|B\right)=\frac{1}{2} \\ \Rightarrow \frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{1}{2} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{2}\times \mathrm{P}\left(B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{2}\times \frac{3}{5} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{3}{10} \\ \mathrm{Now}, \\ \mathrm{P}\left(A\cup B\right)=\frac{4}{5} \\ \Rightarrow \mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)=\frac{4}{5} \\ \Rightarrow \mathrm{P}\left(A\right)+\frac{3}{5}-\frac{3}{10}=\frac{4}{5} \\ \Rightarrow \mathrm{P}\left(A\right)+\frac{6-3}{10}=\frac{4}{5} \\ \Rightarrow \mathrm{P}\left(A\right)+\frac{3}{10}=\frac{4}{5} \\ \Rightarrow \mathrm{P}\left(A\right)=\frac{4}{5}-\frac{3}{10} \\ \Rightarrow \mathrm{P}\left(A\right)=\frac{8-3}{10} \\ \Rightarrow \mathrm{P}\left(A\right)=\frac{5}{10} \\ \therefore \mathrm{P}\left(A\right)=\frac{1}{2} \end{gathered}$$

Hence, the correct alternative is option (b).