The experiment is drawing three cards successively without replacement. 

Since, total numbers of kings are 4 in the pack of 52 cards. 

Therefore, the probability of getting king in first drawn is P(E₁ ) = \(\frac{4}{52}\) = \(\frac{1}{13}\) . 

Since, the card king is not replaced after first draw, therefore, total numbers of 51 cards are left in the pack of cards in which only 3 king card present. 

Therefore, the probability of getting king in second draw is P(E₂ ) = \(\frac{3}{51}\) = \(\frac{1}{17}\) .

 Now, since, both king cards which are drawing in first two draw is not replaced, therefore, total numbers of 50 cards are left in the pack of cards in which 4 ace cards are present. 

Therefore, the probability of getting ace in third draw is P(E₃ ) = \(\frac{4}{50}\) = \(\frac{2}{25}\) . 

Since, all three events E₁, E₂ and E₃ are independent. 

Therefore, the probability of getting first two cards as king and the third card as ace is

P(E₁ )P(E₂ )P(E₃ ) = \(\frac{1}{13} \times \frac{1}{17} \times \frac{2}{25}\) = \(\frac{2}{5525}\).

Hence, the probability that first two cards are kings and the third card drawn is an ace is \(\frac{2}{5525}\).