Question

The chance of an events happening is the square of the chance of the a second events but the odds against the fist are the cube of the odds against the second . Then , the of the first events happening is

• A. $\frac{1}{9}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{8}$

Answer 1

The correct option is A $\frac{1}{9}$

Let the chance of second event be $p$
$\therefore$ The chance of first event be $p^2$.
Now odds against the first event $=\frac{q_1}{p_1}=\frac{1-p^2}{p^2}$
odds against the second event $=\frac{q_2}{p_2}=\frac{1-p}{p}$
Since the odds against the first are the cubes of the odds against the second therefore
$\frac{1-p^2}{p^2}={\left(\frac{1-p}{p}\right)}^3\Rightarrow \frac{(1-p)(1+p)}{p^2}=\frac{{(1-p)}^3}{p^3}\Rightarrow p(1+p)={(1-p)}^2\Rightarrow p^2+p=p^2-2p+1$
Therefore the chance of the first event is $p^2=\frac{1}{9}$