Let A and B are two events.

As, out of 2 events A and B 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) = (\( \dfrac{2}{3}\)) P (B)

We have to find the odds in favour of B.

P (B) = \( \dfrac{3}{5}\)

So, P (B′) = 1 – \( \dfrac{3}{5}\)

=\( \dfrac{2}{5}\)

Odd in favour of B: \( \dfrac{3}{2}\)

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