Question

One of the two exclusive events must occur. If the chance of one is $\frac{2}{3}$ of the other, then odds in favour of the other are:

• A. 1 : 3
• B. 3 : 1
• C. 2 : 3
• D. 3 : 2

Answer 1

The correct option is D 3 : 2

Let P (B) = x, P (A)=$\frac{2x}{3}$
$P\left(A\right)+P\left(B\right)=x+\frac{2x}{3}=1\Rightarrow \frac{5x}{3}=1\Rightarrow x=\frac{3}{5}.P\left(A\right)=\frac{2}{5},P\left(B\right)=\frac{3}{5}$.
Odds in favour of $B=\frac{\frac{3}{5}}{1-\frac{3}{5}}=\frac{3}{2}$ i.e., 3 : 2