Step 1:

Let $X$ denote the no of spade cards in three draws.

$\therefore x$ can take values 0,1,2,3

$P(X=0)$=Probability of not getting a spade card.

Probability of getting a spade card is $P(S)=\large\frac{13}{52}$

$\Rightarrow \large\frac{1}{4}$

Probability of not getting a spade card is $P(F)=1-\large\frac{1}{4}$

$\Rightarrow \large\frac{3}{4}$

Step 2:

$P(X=0)=P(FFF)$

$\qquad\quad\;\;=\large\frac{3}{4}\times\large\frac{3}{4}\times\large\frac{3}{4}$

$\qquad\quad\;\;=\large\frac{27}{64}$

$P(X=1)=P(SFF)+P(FSF)+P(FFS)$

$\qquad\quad\;\;=\large\frac{1}{4}\times\large\frac{3}{4}\times\large\frac{3}{4}+\large\frac{3}{4}\times\large\frac{1}{4}\times\large\frac{3}{4}+\large\frac{3}{4}\times\large\frac{3}{4}\times\large\frac{1}{4}$

$\qquad\quad\;\;=3\times\large\frac{3^2}{4^3}$

$\qquad\quad\;\;=\large\frac{27}{64}$

$P(X=2)=P(SSF)+P(FSS)+P(SFS)$

$\qquad\quad\;\;=\large\frac{1}{4}\times\large\frac{1}{4}\times\large\frac{3}{4}+\large\frac{3}{4}\times\large\frac{1}{4}\times\large\frac{1}{4}+\large\frac{1}{4}\times\large\frac{3}{4}\times\large\frac{1}{4}$

$\qquad\quad\;\;=3\times\large\frac{3^3}{4^3}$

$\qquad\quad\;\;=\large\frac{9}{64}$

$P(X=3)=P(SSS)$

$\qquad\quad\;\;=\large\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}$

$\qquad\quad\;\;=\large\frac{1}{64}$

Step 3:

Hence the required probability is