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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