Odds in favour of an event A are 2 to 1 and odds in favour of A ∪ B are 3 to 1. Consistent with this information the smallest and largest values for the probability of event B are given by

$\frac{1}{12}≤ P(B)≤ \frac{3}{4}$

We have,

$P(A) =\frac{2}{3}, P(A ∪ B) =\frac{3}{4}$

$∴ P(A ∪ B) = P(A) +P(B) -P( A ∩ B)$

$⇒ \frac{3}{4}=\frac{2}{3} + P(B) -P(A ∩ B)$

$⇒ P(A ∩ B)= P(B) -\frac{1}{12}$

Now, 

$0 ≤ P (A ∩ B) ≤ P(A)$

$⇒ 0 ≤P(B)-\frac{1}{12} ≤ P(A)⇒ \frac{1}{12} ≤ P(B) ≤ \frac{3}{4}$