According to the question, Five cards are drawn from a pack of 52 cards.

By using the formula of probability, we get,

P (E) = favourable outcomes / total possible outcomes

Five cards are drawn at random, so the total possible outcomes will be ${}^{52}C_5$

n (S) = 2598960

(i) Let E be the event that exactly only one ace is present

$n (E) = {}^4C_1{}^{48}C_4 = 778320$

P (E) = n (E) / n (S)

= 778320 / 2598960

= 3243/10829

(ii) Let E be the event that at least one ace is present

E = {1 or 2 or 3 or 4 ace(s)}

$n (E) = {}^4C_1{}^{48}C4+{}^4C_2{}^{48}C_3+{}^4C_3{}^{48}C_2+{}^4C_4{}^{48}C_1 = 886656$

P (E) = n (E) / n (S)

= 886656 / 2598960

= 18472/54145