$\text{ Suppose }A\text{ and }B\text{ are two events with }P\left(A\right)=0.5\text{ and }$$P(A\cup B)=0.8\text{. Let }P\left(B\right)=p\text{ if }A\text{ and }B\text{ are mutually }$

$\text{ exclusive and }P\left(B\right)=q\text{ if }A\text{ and }B\text{ are independent events then the value of }q/p\text{ is }$

$\text{ When }A\text{ and }B\text{ are mutually exclusive, }P(A\cap B)=0$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}P(A\cup B)=P\left(A\right)+P\left(B\right)----\left(1\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0.8=0.5+p\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}p=0.3\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}P(A\cup B)=P\left(A\right)+P\left(B\right)----\left(2\right)\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=P\left(A\right)+P\left(B\right)-P(A\cap B)\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0.8=0.5+q-\left(0.5\right)q\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0.3=\frac{q}{2}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}q=0.6-----\left(3\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}q/p=2\end{array}$