Let E and F be the two events such that one must occur 

Given, 

P(E) = \(\frac{2}{3}\) P(F) 

Also, P(E∪F) = 1 P(E) + P(F) = 1 

⇒ P(F){ \(\frac{2}{3}\) + 1} = 1 

∴ P(F) =   \(\frac{3}{5}\) 

And P(F’) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)

We have to find \(\frac{P(F)}{P(\bar F)}\) = \(\frac{3}{\frac{5}{\frac{2}{5}}}\) = \(\frac{3}{2}\)

∴ Odds in favour of F = \(\frac{3}{2}\)

As our answer matches only with option (d) 

∴ Option (d) is the only correct choice.