One of the two events, A and B must occur. If P (A) = (2/3) P (B), the odds in favour of B are

Concept:

Let an event A, Probability of occurring of A be P (A) and not occurring of A be P (A^c)

• \({\rm{Odds\;in\;favour\;of\;A\;}} = \frac{{{\rm{P\;}}\left( {\rm{A}} \right)}}{{{\rm{\;P\;}}\left( {\bar A} \right)}}\)
• \({\rm{Odds\;in\;Against\;of\;A\;}} = \frac{{{\rm{P\;}}\left( {\bar A} \right)}}{{{\rm{\;P\;}}\left( {\rm{A}} \right)}}\)

Calculation:

Given: One of the two events A and B must occur,

So: P (A) + P (B) = 1

⇒ (2/3) P (B) + P (B) = 1

∴ P (B) = 3/5

Now, \({\rm{P\;}}\left( {{\rm{\bar B}}} \right) = 1 - {\rm{P\;}}\left( {\rm{B}} \right) = 1{\rm{\;}} - \frac{3}{5} = \frac{2}{5}\)