$A\text{ and }B\text{ are }2\text{ events such that }P\left(A\right)=\frac{3}{4}\text{ and }P\left(B\right)=\frac{5}{8}\text{. If }a\text{ and }b\text{ are the possible minimum and maximum values }$

$\text{ of }P(A\cap B)\text{, then the value of }a+b\text{ is }$

$\begin{array}{l}A\cap B\subseteq B\\ P(A\cap B)\le P\left(B\right)\\ P(A\cap B)\le \frac{5}{8}----\left(1\right)\\ \mathrm{Now}P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\\ P(A\cup B)\le 1\\ \Rightarrow P\left(A\right)+P\left(B\right)-P(A\cap B)\le 1\\ P(A\cap B)\ge \frac{3}{4}+\frac{5}{8}-1\\ \text{ i.e., }P(A\cap B)\ge \frac{3}{8}-----\left(2\right)\end{array}$

From (i) and (ii) 

$\begin{array}{l}\frac{3}{8}\le P(A\cap B)\le \frac{5}{8}\\ \therefore x+y=\frac{3}{8}+\frac{5}{8}=1\end{array}$