Let A and B are two events. 

As, out of 2 events A and B only one can happen at a time which means no event have anything common. 

∴ We can say that A and B are mutually exclusive events. 

By definition of mutually exclusive events we know that: 

P(A ∪ B) = P(A) + P(B) 

According to question one event must happen. 

This implies A or B is a sure event. 

∴ P(A ∪ B) = P(A) + P(B) = 1 …Equation 1 

Given, P(A) = (2/3)P(B) 

To find: odds in favour of B

∴ P(B’) + \(\frac{2}{3}\) P(B) = 1 

⇒ \(\frac{5}{3}\) P(B) = 1 ⇒ P(B) = \(\frac{3}{5}\)

∴ P(B’) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)

∴ Odd in favour of B = \(\frac{P(B)}{P(\bar B)}\) = \(\frac{3}{\frac{5}{\frac{2}{5}}}\) = \(\frac{3}{2}\)