Question

One of the two events must happen. Given that the chance of one is two-third of the other, find the odds in favour of the other.

Answer 1

Let the given events be A and B
Now,

\[P\left( A \right) = \frac{2}{3}P\left( B \right)\]

Let P(B) = x

\[\therefore P\left( A \right) = \frac{2}{3}x\]

The events A and B are exhaustive.
∴ P(A) or P(B) = 1
⇒ P(A) + P(B) = 1                (∵ A and B are mutually exclusive)

 \[\Rightarrow \frac{2}{3}x + x = 1\]

\[\Rightarrow \frac{5x}{3} = 1\]

\[\Rightarrow x = \frac{3}{5}\]

\[\therefore P\left( B \right) = \frac{3}{5}\]

This implies that the odds in favour of B are 3 : (5 – 2), i.e. 3 : 2.