Solution :
A= last card is spade
B= not spade
$P(A) = \large\frac{13}{52} =\frac {1}{4}$
$P(B) = \large\frac{39}{52} =\frac {3}{4}$
$\in $= 2 cards drawn are spade
$P \bigg( \large\frac{\in}{A} \bigg) =\bigg( \large\frac{12}{51} \bigg) \times \bigg( \large\frac{11}{50} \bigg) =\large\frac{132}{2550}$
$P \bigg( \large\frac{\in}{B} \bigg) =\bigg( \large\frac{13}{51} \bigg) \times \bigg( \large\frac{12}{50} \bigg) =\large\frac{156}{2550}$
$P\bigg(\large\frac{A}{\in}\bigg)= \large\frac{P(\Large\frac{\in}{A}) P(A)}{P(A)P(\Large\frac{\in}{A})+ P(\Large\frac{\in}{B} ) \normalsize P(B)}$
$\qquad= \large\frac{ \Large\frac{132}{2550} \times \large\frac{1}{4}}{\Large\frac{132}{2550 \times 4 } + \Large\frac{156 \times 3 }{2550 \times 4}}$
$\qquad= \large\frac{132}{132+468}$
$\qquad= \large\frac{132}{600}$
$\qquad= \large\frac{11}{50}$