Official solution
Given E1 , E2 , E3 are pairwise indepedent events
so P(E1 \[ \cap \] E2 ) = P(E1 ).P(E2 )
and P(E2 \[ \cap \] E3 ) = P(E2 ).P(E3 )
and P(E3 \[ \cap \] E1 ) = P(E3 ).P(E1 )
and P(E1 \[ \cap \] E2 \[ \cap \] E3) = 0
Now \[P\left( {{{E_2^C \cap E_3^C} \over {{E_1}}}} \right)\]
\[ = {{P\left[ {{E_1} \cap \left( {E_2^C \cap E_3^C} \right)} \right]} \over {P\left( {{E_1}} \right)}}\]
\[ = {{P\left( {{E_1}} \right) - \left[ {P\left( {{E_1} \cap {E_2}} \right) + P\left( {{E_1} \cap {E_3}} \right) - P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)} \right]} \over {P\left( {{E_1}} \right)}}\]
\[ = {{P\left( {{E_1}} \right) - P\left( {{E_1}} \right).P\left( {{E_2}} \right) - P\left( {{E_1}} \right).P\left( {{E_3}} \right) - 0} \over {P\left( {{E_1}} \right)}}\]
\[ = 1 - P\left( {{E_2}} \right) - P\left( {{E_3}} \right)\]
\[ = 1 - P\left( {{E_3}} \right) - P\left( {{E_2}} \right)\]
\[ = P\left( {E_3^C} \right) - P\left( {{E_2}} \right)\]