What is the probability that both cards are aces if you do not replace that first card before drawing the second?

Suppose two cards are drawn from a standard 52 card deck without replacement. Assuming all cards are equally likely to be selected, what is the probability that both cards are not aces?

My Solution

A = Event that first card is an ace

B = Event that second card is an ace given that first is an ace

C = Event that both cards are aces

D = Event that both cards are not aces $$P(A) = \frac{4}{52}$$ $$P(B) = \frac{3}{51}$$ $$P(C) = \frac{4}{52}*\frac{3}{51} = \frac{1}{221}$$ $$P(D) = 1 - P(C) = 1 - \frac{1}{221} = \frac{220}{221}$$

Actual Solution

A = Event that first card is not an ace

B = Event that second card is not an ace given that first is not an ace

C = Event that both cards are not aces $$P(A) = \frac{48}{52}$$ $$P(B) = \frac{47}{51}$$ $$P(C) = \frac{48}{52}*\frac{47}{51} = \frac{188}{221}$$

Why my solution that P(D) = 1 - P(C) is wrong?