Let A and B be two events.

As, out of 2 events, 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.

So, by definition of mutually exclusive events we know that:

P (A ∪ B) = P (A) + P (B)

According to question one event must happen.

A or B is a sure event.

So, P (A ∪ B) = P (A) + P (B) = 1 … Equation (1)

Given: P (A) = (2/3) P (B)

We have to find the odds in favour of B.

$\begin{array}{l}
\mathrm{P}(\mathrm{B})+\frac{2}{3} \mathrm{P}(\mathrm{B})=1 \\
\frac{5}{3} \mathrm{P}(\mathrm{B})=1 \\
\mathrm{P}(\mathrm{B})=3 / 5 \\
\text { So, } \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-3 / 5 \\
=2 / 5
\end{array}$

$\therefore$ Odd in favour of B:

$\frac{\mathrm{P}(\mathrm{B})}{\mathrm{P}(\overline{\mathrm{B}})}=\frac{3 / 5}{2 / 5}=\frac{3}{2}$