1.
In total there are C(52, 13) ways that we can pick a hand, that is

2.
P(a hand contains at least 3 kings)
=P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)
3.
first let's find P(a hand contains exactly 3 kings):
P(a hand contains exactly 3 kings)
=n(a hand contains exactly 3 kings)/C(52, 13)
n(a hand contains exactly 3 kings)=C(4, 3)*C(48, 10)
where C(4,3) is the total number of ways we can pick 3 out of 4 kings,
C(48, 10) is the number of picking 10 letters to complete a hand, out of the 52-4=48 non-king cards.
so P(a hand contains exactly 3 kings)=[C(4, 3)*C(48, 10)]/C(52, 13)
4. with the same reasoning as in step 3:
P(a hand contains 4 kings)=n(a hand contains 4 kings)/C(52, 13)
= [C(4, 4)*C(48, 9]/C(52, 13)
5.
P(a hand contains at least 3 kings)
=P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)
=[C(4, 3)*C(48, 10)]/C(52, 13)+ [C(4, 4)*C(48, 9)]/C(52, 13)
=

=

simplify by 38! in the denominators and 48! in the numerators :

now simplify by 9! in all denominators:
