Total number of ways of drawing one card from $52$ cards

$={}^{52}{C}_1 \left\{{\because }^n{C}_r=\frac{n!}{r!(n-r)!}\right\}$

$=\frac{52!}{1! 51!}=52$

Total number of ways drawing one card of ace from $4$ ace cards$={}^{4}C_1=\frac{4!}{1! 3!}=4$

For second card,

Total number of ways of drawing one card from $51$ cards

$= {}^{51}{C}_1=51$ and

Total number of ways drawing one card of ace from $3$ ace cards$={}^3{C}_1=3$

Probability of both events $=\frac{4}{52}\times \frac{3}{51}$

$=\frac{1}{13}\times \frac{1}{17}=\frac{1}{221}$