A and B are two events such that odds against A are 2:1 odds in favour of A ∪  B are 3 : 1. If x ≤ P(B) ≤ y, then the ordered pair (x,y) is

$(5/12, 3/4)$

We have,

$P(A)=\frac{1}{3}$ and $P(A ∪ B)=\frac{3}{4}$

$∴ P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$

$⇒ \frac{3}{4}=\frac{1}{3}+P(B) -P(A ∩ B)$

$⇒ \frac{5}{12}=P(B)-P(A ∩ B)$

$⇒ P(B) =\frac{5}{12}+P(A ∩ B) ⇒ P(B) ≥ 5/12$ .....(i)

Again,

$P(B)=\frac{5}{12} + P(A ∩ B)$

$⇒ P(B) ≤ 5/12 + P(A)$           $[∵P(A∩B) ≤ P(A)]$

$⇒ P(B) ≤ \frac{5}{12}+\frac{1}{3}=\frac{3}{4}$ .......(ii)

From (i) and (ii), we obtain   $5/12 ≤ P(B) ≤ 3/4$

Hence,  $x = 5/12$ and $y = 3/4$