Question

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

• A. $3:5$
• B. $2:5$
• C. $3:2$
• D. None of these

Answer 1

Answer: (C)

$3:2$

Let the given events be $E_1$ and $E_2$
Then $P\left(E_1\right)=\frac{2}{3}P\left(E_2\right)$

The events $E_1$ and $E_2$ are exhaustive
$\therefore P(E_1\cup E_2)=1$

Since $E_1$ and $E_2$ are mutually exclusive
$P\left(E_1\right)+P\left(E_2\right)=1$

$\Rightarrow \frac{2}{3}P\left(E_2\right)+P\left(E_1\right)=1\Rightarrow P\left(E_2\right)=1\Rightarrow P\left(E_1\right)=\frac{3}{5}$

$\therefore P\left(E_2\right)=\frac{3}{5}$ and $P\left(E_1\right)=\frac{2}{3}\times \frac{3}{5}=\frac{2}{5}$

Since $P\left(E_2\right)=\frac{3}{5},$ odds in favour of $E_2$ are $3:5-3\Rightarrow 3:2.$