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

Given,

P(E) = 2/3 P(F)

Also, P(E ∪ F) = 1

P(E) + P(F) = 1

⇒ P(F){2/3 + 1} = 1

∴ P(F) = 3/5

And P(F’) = 1 – 3/5 = 2/5

We have to find

∴ Odds in favour of F = 3/2

As our answer matches only with option (d)

∴ Option (d) is the only correct choice.