# If $P(A)=0.65\: P(B)=0.80$ then $P(A \cap B)$ lies in the interval

$\begin {array} {1 1} (A)\;[.30,80] & \quad (B)\;[0.35, 0.75] \\ (C)\;[0.4, 70] & \quad (D)\;[ 0.45, 0.65] \end {array}$

$P(A \cap B) \leq min. {P(A),P(B)}\: =\: min.{0.65, 0.80}$
$= 0.65$
$\therefore P(A \cap B ) \leq 0.65$
Also $P(A \cap B)= P(A)+P(B)-P( A \cup B) \geq 0.65 + 0.80-1$
$( \because P(A \cup B) \leq 1 )$
$= 0.45 \: \: \: \: \therefore -P ( A \cup B ) \geq -1$
$\therefore 0.45 \leq P( A \cap B ) \leq 0.65$
Ans : (D)