The chance of one event happeing is the square of the chance of a second event, but the odds against the first are the cube of the odds against the second. The chance of the first event is :

Solution :
Let the two events be $${E_1}$$ and $${E_2}$$
Let their chances be $$p$$ and $$q$$ respectively.
Then $$p = {q^2}......\left( {\text{i}} \right)$$
The chances of not happening of the events are $$1 - p$$  and $$1 - q$$  respectively.
Odds against the first event $$ = \frac{{1 - p}}{p}$$
Odds against the second event $$ = \frac{{1 - q}}{q}$$
$$\eqalign{ & {\text{Given,}} \cr & \frac{{1 - p}}{p} = {\left( {\frac{{1 - q}}{q}} \right)^3} \cr & \Rightarrow \frac{{1 - {q^2}}}{{{q^2}}} = \frac{{{{\left( {1 - q} \right)}^3}}}{{{q^3}}} \cr & \left[ {{\text{From eqution }}\left( {\text{i}} \right)} \right] \cr & \Rightarrow \left( {\frac{{1 - q}}{{{q^2}}}} \right)\left[ {\left( {1 + q} \right) - \frac{{{{\left( {1 - q} \right)}^2}}}{q}} \right] = 0 \cr & \because \,q \ne 1{\text{ and }}q \ne 0 \cr & \therefore \,q\left( {1 + q} \right) = 1 - 2q + {q^2} \Rightarrow q = \frac{1}{3} \cr & \therefore {\text{ from equation }}\left( {\text{i}} \right)\,p = {q^2} = \frac{1}{9} \cr & \therefore \,p\left( {{E_1}} \right) = p = \frac{1}{9} \cr} $$