Official solution
Consider the events,
E1 = missing card is spade
E2 = missing card is not a spade
A = Two spade cards are drawn
\[P\left( {{E_1}} \right) = {1 \over 4}\]
\[P\left( {{E_2}} \right) = {3 \over 4}\]
\[P\left( {{A \over {{E_1}}}} \right) = {{{}^{12}{C_2}} \over {{}^{51}{C_2}}}\]
\[P\left( {{A \over {{E_2}}}} \right) = {{{}^{13}{C_2}} \over {{}^{51}{C_2}}}\]
\[P\left( {{{{E_2}} \over A}} \right) = {{P\left( {{A \over {{E_2}}}} \right).P\left( {{E_2}} \right)} \over {P\left( {{A \over {{E_1}}}} \right).P\left( {{E_1}} \right) + P\left( {{A \over {{E_2}}}} \right).P\left( {{E_2}} \right)}}\]
\[ = {{{{{}^{13}{C_2}} \over {{}^{51}{C_2}}}.{3 \over 4}} \over {{{{}^{12}{C_2}} \over {{}^{51}{C_2}}}.{1 \over 4} + {{{}^{13}{C_2}} \over {{}^{51}{C_2}}}.{3 \over 4}}}\]
= \[{{39} \over {50}}\]